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MACHINE COMPUTATION
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ELEMENTARY STATISTICS
Katharine Pease:
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MACHINE COMPUTATION
ELEMENTARY STATISTICS
With Special Reference to the Friden, Marchant, and
Monroe Calculating Machines
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KATHARINE PEASE
Department of Psychology
Barnard College, Columbia University
24
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1949
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20-
CHARTWELL HOUSE, INC.
NEW YORK
N. Y.
275

R. P. YouTZ, EDITOR FOR PSYCHOLOGY
QA
75
.P36
Copyright
1949
CHARTWELL HOUSE, INC.
280 Madison Avenue
New York 16
N. Y.
PRINTED IN THE UNITED STATES OF AMERICA

Sociology
Nahr
10-9-52
80325
2 сар.
Without a doubt, technical facility is a first requisite for valuable
mental activity: we shall fail to appreciate the rhythm of Milton, or the
passion of Shelley, so long as we find it necessary to spell the words
and are not quite certain of the forms of the individual letters. In this
sense there is no royal road to learning. But it is equally an error to
confine attention to technical processes, excluding consideration of
general ideas.
A. N. WHITEHEAD
•
:
PREFACE
This manual is for students learning to use computing machines
in connection with courses in elementary statistical methods. It is
set up to be self-teaching, so that the student, by following the
procedures in sequence, may learn to use the machines with a
minimum amount of help from the instructor. This sequence has
been established on the basis of trials for two semesters in the
psychometrics laboratory at Barnard College, and for three
semesters in the statistics laboratory of Teachers College, Columbia
University.
Only those basic techniques which are needed in the computation
of common elementary statistics are presented here. Procedures are
outlined for the four fundamental operations of addition, subtrac-
tion, multiplication, and division. Practice is given in other opera-
tions used in computing measures of central tendency, variability,
trends, indices, and correlation. These include complements, ac-
cumulative and subtractive multiplication, multiplication by a
constant, reciprocals, and square root.
A questionnaire concerning the need for an elementary statistics
computing machine manual was sent out in March 1948 to more
than two hundred teachers of statistics in more than one hundred
colleges and universities throughout the country. Results indicate
clearly that there is great need for such a book written for the
student, which brings together instructions for different types and
makes of calculating machines.
The several makes of calculating machines vary in structure and
design. Consequently, no one manual can set forth all details of
operation for all types of machines without becoming bulky and
confusing. Therefore, the three makes reported by statistics in-
structors as being used most often, i.e., Friden, Marchant, and
Monroe, have been chosen, and specific directions have been written
for certain models of these makes.
MOND
Detailed instructions are included for both automatic and hand-
บ
vi
Preface
operated calculators: the Friden ST-10, Marchant ACT-10M, and
Monroe CAA (fully automatic models); Monroe M (semiauto-
matic); and the Monroe Educator and K (hand-operated) models.
In certain places notation is made concerning procedures on other
models, i.e., Marchant D and Monroe A-1 and AA-1 machines.
The student need only know whether the machine he is about to
use is automatic or hand-operated, and its make; operating princi-
ples are the same regardless of differences in operational details.
It must be stated at the outset that no attempt is made here to
exhaust the possibilities of operation of the calculating machine.*
An effort is made to describe only those basic techniques which are
specifically needed in the computation of elementary statistics by
students in economics, psychology, education, sociology, and biol-
ogy. These descriptions suggest methods which have been found to
be effective; each instructor and computer may prefer other meth-
ods and other procedures. The author welcomes criticisms and
comments by both instructors and students who use this book.
A project such as this can be accomplished only with the generous
cooperation of many people. To these the author owes thanks:
first to the Barnard College students in psychometric methods for
whom the original draft was written, and especially to Miss Joanne
Webber who prepared and sent out the questionnaire appraising the
need for such a manual in college statistical laboratories.
The author is deeply indebted to her teachers, Professor Helen
M. Walker and Professor Irving Lorge of Teachers College, Colum-
bia University, under whose active interest and inspiration the
book took shape and grew to its present form; and to Professor
R. P. Youtz, Executive Officer, Department of Psychology,
Barnard College, whose personal encouragement has been a con-
stant source of stimulation in its writing. To those busy col-
leagues in colleges and universities throughout the country whose
answers to the questionnaire proved of inestimable help, the
author is most grateful.
Dr. Joseph Lev, formerly of Rutgers University, offered encour-
agement to the project in its early stages; and Professor Dudley J.
* See References, p. 188, for references to manuals and methods published by
various manufacturers of calculators, which were found to be of particular help
in outlining the procedures of this book. Dunlap's excellent manual including a
comprehensive treatment of computational techniques for descriptive statis-
tics, with special reference to the MATHEMATON, is unfortunately out of print.
Preface
vii
Cowden, of the University of North Carolina, gave kind permission
to hear his lectures during the 1948 Summer Session of Columbia
University.
Special thanks also are due Mr. Sidney Cohen of the Statistics
Laboratory of Teachers College, Columbia University, who used a
preliminary draft in his teaching and gave many helpful comments;
and to Mrs. Irene D. Jaworski and Mr. Robert J. Shockley who
worked through another draft in the summer of 1948 and made
further useful suggestions which are incorporated here.
Particular gratitude is expressed to Mr. L. E. Krackenberger of
the New York office of the Friden Calculating Machine Company,
and to Miss Ann Curran and Mr. Herman Hovemann of the New
York office of the Marchant Calculating Machine Company, whose
excellent knowledge of their respective machines served as a con-
stant reservoir of constructive information freely given. Miss Olive
Brown and Mr. C. G. Schmucki of the New York office of the
Monroe Calculating Machine Company offered valuable sugges-
tions concerning Monroe models. The figures were prepared by
Professor Frank H. Lee, Department of Drafting, Columbia Uni-
versity, with the assistance of Mr. George Weidmann.
Barnard College
Columbia University
K. P.
•
PREFACE.
TO THE STUDENT
•
CONTENTS
A. Addition
CHAPTER 2
DESCRIPTION OF CALCULATING MACHINES
A. Friden Model ST-10
B. Marchant Model ACT-10M
•
C. Marchant Model ACR-8D .
D. Monroe CAA Models
E. Monroe Model AA-1
F. Monroe M Models
G. Monroe K and Educator Models
CHAPTER 1
CHAPTER 3
ADDITION AND SUBTRACTION
B. Subtraction
C. Complements
PART ONE: BASIC OPERATIONS
(a) Addition of whole
numbers
(b) Addition of numbers
with decimals.
Friden Marchant
27
29
40
44
30
32
41
46
•
33
Monroe
Semi- Hand-
Auto- auto- oper-
matic matic ated
35
42
48
33
35
42
48
36
page
V
39
43
50
1
LO
5
5
8
10
12
16
18
21
27
ix
X
Contents
MULTIPLICATION
A. Multiplication of Whole
Numbers .
(a) Entering multipliers
from left . .
(b) Entering multipliers
from right
(c) Multiplication from
right..
(d) Multiplication from
left
(e) Multiplication by
multipliers with dig-
its 6, 7, 8, 9
CHAPTER 4
B. Multiplication of Num-
bers with Decimals
(a) Multiplication
(b) Preset decimals
of
numbers with deci-
mals ..
Friden Marchant
•
C. Accumulative and Neg-
ative Multiplication .
(a) Accumulative multi-
plication
(b) Negative multiplica-
tion
•
(c) Accumulative and
negative multiplica-
tion
(d) Sum and sum of
squares
52
60
60
61
70
71
71
73
D. Multiplication by a
Constant.
E. Use of Tables of Products
53
89
53
54
60
71 75
62
64
70
75
76
77
90
Monroe
Semi- Hand-
Auto- auto- oper-
matic matic ated
56 56
55
66
67
58
56
79
56
58
68
69
58
60 60 60
56
80 84
81 86
56
91 92
58
59
70 70 70
83
83
68
69
80 84 84
page
84
86
92,
52
93
Contents
xi
DIVISION .
A. Division of Whole Num-
bers
(a) Quotient with no re-
mainder
(b) Quotient with re-
mainder
B. Division of Numbers
with Decimals
•
(a) Reciprocal of a num-
CHAPTER 5
Friden Marchant
(a) Numbers in extreme
left of keyboard . . 99
SQUARE ROOT.
A. Extracting the Square
Root of a Number by
Subtraction of Succes-
sive Odd Numbers
(a) Roots of whole num-
bers
94
(b) Constant or preset
decimal setup .
100
C. Division by a Constant 111
D. Reciprocals
95
B. Extracting the Square
Root of a Number by
Approximation by Divi-
99
(b) Reciprocal of a num-
ber less than 1
113
E. Use of Tables of Reciprocals .
F. Use of Tables of Quotients.
95
123
(b) Roots of decimals. 126
96
ber greater than 1. 112 114
99
102
103
111
115
CHAPTER 6
128
131
141
sion.
143
C. Use of Tables of Squares and Square Roots
Monroe
Semi- Hand-
Auto- auto- oper-
matic matic ated
96 96 97
97 97 99
99 99 99
105
106
111
105 108
118
106 109
111
111
117 117
118
119
145 145
120
•
132 132 137
135 135
139
147
page
94
122
122
123
€ 149
xii
Contents
PART TWO: HOW TO OBTAIN
CERTAIN COMMON STATISTICS
CHAPTER 7
THE MEAN AND STANDARD DEVIATION FROM A
GROUPED FREQUENCY DISTRIBUTION USING
AN ARBITRARY ORIGIN
. 153
CHAPTER 8
THE PRODUCT-MOMENT CORRELATION COEFFI-
CIENT, MEAN AND STANDARD DEVIATION FROM
RAW SCORES
•
1
CHAPTER 9
OTHER COMMON STATISTICS . .
A. Percentiles .
B. Standard Scores
REFERENCES
COMPUTING SHEETS
GLOSSARY OF SYMBOLS
ANSWER KEY
APPENDIX A
•
APPENDIX B
APPENDIX C
page
-
161
181
181
183
188
. 191
. 195
. 203
CHAPTER 1
TO THE STUDENT
Introduction
This book is designed to be used by many of you in many differ-
ent elementary statistics laboratories where there may be one or
several types of calculators. It is written in such a way that you
will be able to work on any of several types of calculating machines
at any time and in any sequence, with a minimum of confusion.
It is necessary that you first become familiar with the structure
and parts of the machines you are going to use. Various models
and makes of calculating machines are described in detail in
Chapter 2. If the model you are using is not mentioned specifically,
find the one nearest to it, and study that description. For example,
your machine may have only eight banks of keys in the keyboard
instead of the ten described; and eight upper and seventeen lower
dials in the carriage instead of twenty upper and twenty lower
dials. This means that your machine cannot handle as large num-
bers as the one referred to in the book; you will find that the proce-
dures for operation are the same, and that you will read the results in
the same sets of dials.
Computational procedures are divided into two parts: PART ONE
outlines in detail the basic operations of addition, subtraction,
multiplication, division, and square root. This is done separately
for automatic and hand-operated machines, and for three makes,
Friden, Marchant, and Monroe. PART TWO gives specific procedures
and computing sheets for obtaining percentiles, the arithmetic
mean, standard deviation, regression coefficient, product-moment
correlation coefficient, and standard scores.
PART ONE: Basic Operations
1. The general principles of the operation are set forth; e.g.,
multiplication is successive addition. In all cases these are
simple, brief, but essential to your understanding of the
process.
1
2
Machine Computation of Elementary Statistics
2. Specific directions for operation of each type of machine are
given separately, using an EXAMPLE to demonstrate various
steps in the operation. It is important that you follow the
process step by step and verify the values recorded in the
example.
3. PROBLEMS are given for you to work out, with space provided
for answers and checks.
4. Since the problems are the same for each machine, the answer
key (Appendix C) is useful for any set of calculations on any
model.
In order to secure maximum benefit from the book, you should
master each set of operations before going on to the next one. In many
instances you may wish to solve further problems in order to be
sure of your mastery of the techniques involved. And you will un-
doubtedly have class assignments for which these procedures will
provide methods of quick computation with checks.
All the way through the book, checks are stressed. Sometimes
these seem unnecessarily laborious, as in the case of the computa-
tion of the product-moment correlation coefficient. The use of
checks serves not only to assure the accuracy of your final results,
but also in the last analysis, to save labor.
PART TWO: How to Obtain Certain Common Statistics
Procedures are outlined for obtaining the following statistics:
the mean and standard deviation from a grouped frequency dis-
tribution using an arbitrary origin; the Pearson product-moment
correlation coefficient; mean, standard deviation, and regression
coefficients from raw scores; percentiles; and standard scores.
Symbols used in the formulas are explained, and are also listed
in the Glossary (Appendix A).
The two computing sheets for standard deviation and correlation
coefficient included in Appendix B are only given as samples. For
practical use these sheets should be 8 by 11 inches.
General Orientation
Position of the machine. It is suggested that the machine is
placed to your right at a slight angle and your work sheets directly
in front of you as in Figure 1.

WORK SHEETS
COMPUTING MACHINE

OPERATOR
FIGURE 1. Position of the machine
3
4
Machine Computation of Elementary Statistics
Operation of the machine. Use only the fore and middle fingers
of the right hand. You will find it convenient to grasp a pencil
between thumb and palm of the operating hand. It is then unneces-
sary to lay down and pick up the pencil when recording results on
work sheets. All operations, as far as possible, should be carried out
with one hand. In hand-operated machines, the left hand is used to
shift the carriage; in all machines, to set numbers on the keyboard
in finding square root; and, on the Marchant, to hold down the
reverse key in negative multiplication.*
Guides. You will find it helpful to use an index card, varying in
size with the span of work to be covered, to guide you in reading
from your work sheets. When several columns of figures are on one
sheet, it is wise to fold strips of paper and lay them over the
columns which are not being used in a particular operation.
Negative numbers. Minus signs are a common source of error in
statistical calculation. It is suggested that you use a red pencil for
recording all negative numbers, particularly in more extensive
problems.
Care of the machine. Always clear the machine before starting
to operate and when finishing a set of operations. Always be sure
it is cleared and covered when leaving it. Never operate any lever
or key or depress keyboard keys while the machine is in operation,
unless the directions so state. Always allow the machine to complete
one operation before you begin the next one. When something goes
wrong in its operation, report the trouble to the instructor at once.
* Some operators prefer to place the machine on the left, use the left hand to
operate it, leaving the right hand free to write.
CHAPTER 2
DESCRIPTION OF CALCULATING MACHINES
Before starting to operate a calculating machine, it is desirable
for the student to become familiar with its arrangement of keys and
dials. Descriptions are given here for the following makes and
models: FRIDEN ST-10, MARCHANT ACT-10M and ACR-
8D,* and MONROE CAA, AA-1, MA-7, K, and Educator. Figures
2 to 8 show the layout of keys and dials for these machines.
A. FRIDEN MODEL ST-10 (1, 2, 3)+
1. The keyboard. There are ten columns of ten keys each. In
each column, the keys are numbered from 1 to 9, with a blank key
at the foot of each column to represent 0. The first column on the
extreme right may be considered the units position, the second
column the tens position, the third column the hundreds position,
and so forth. Keys are operated by depressing them until they catch
and stay down. If a second key in a column is depressed, the first
will automatically be released. If two keys in the same column
are pushed down simultaneously, only the larger number will
register in the machine. Decimal markers appear between each two
columns of keys. The keyboard may be cleared by a touch of the
CLEAR OF RETURN CLEAR keys. The arrow above the first column
(carriage position indicator) indicates the position of the carriage
with respect to the first keyboard column.
2. The carriage. There are two sets of dials in the movable
carriage, designated by their position as "upper" and "lower."
Results of addition, subtraction, and multiplication appear in the
upper dials; those of division appear in the lower dials. The re-
mainder in division registers in the upper dials; the value of the
* See special note on p. 23.
† Numbers in parentheses refer to numbered items in the References,
page 188.
5
CARRIAGE
MULTIPLIER
PROOF REGISTER
MULTIPLIER
CORRECTION KEY
NEGATIVE
MULTIPLY KEY
MULTIPLIER KEYBOARD-
ACCUMULATIVE
MULTIPLY KEY
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ㅁㄷ
​SELECTIVE TABULATOR 0000000000
LOWER DIALS -
KEYBOARD
MULTIPLIER UNIT
OOO
OOO
DOO
80
MULTIPLY KEY
KEYS
YS
M
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
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¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
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FIGURE 2. FRIDEN Model ST-10
0
0
0
QO
KEYBOARD DECIMAL MARKER
UPPER DIAL
ACCUMULATOR
LOCK
DIAL TWIRLERS
UPPER DIALS
LOWER DIAL
ACCUMULATOR
TOR
LOCK
COUNTER CONTROL KEY
DIVISION STOP KEY
DIVIDEND
TABULATING KEY
MINUS BAR
CARRIAGE
POSITION KEYS
PLUS BAR
ADD KEY
DIVIDE KEYS
POSITIVE
KEYBOARD LOCK
RETURN CLEAR KEY
CLEAR KEY

Description of Calculating Machines
multiplier and the number of addition and subtraction operations
appear in the lower dials. Dial twirlers above the upper dials allow
numbers to be set in by hand; selective tabulator keys below the
lower dials permit automatic tabulating of the carriage in any se-
lected position in division.
Decimal markers appear below the upper dials and above the
lower dials. Touching carriage position keys moves the carriage
either to right or left as indicated by the arrow on each key. The
carriage automatically returns to the extreme left (position one)
with a touch of the RETURN CLEAR key.
3. Plus and minus bars. The plus bar is used for addition, and
the minus bar for subtraction. When either the plus or minus bar
is touched, the indicated operation is performed (see special sec-
tions on addition and subtraction for details); when the ADD key is
down, the entire keyboard clears with each touch of the plus or
minus bar; when the ADD key is up, the keys on the keyboard re-
main depressed or "repeat."
4. Multiplier. This is a special unit for multiplication only.
Numbers punched in the unit appear in the register directly above
the numeral keys. There are four operating keys, three for multipli-
cation processes (see special sections on multiplication for details)
and one for correction of numbers entered in the unit.
7
5. Division keys. The dividend tabulating key (DIVD TAB)
automatically tabulates the carriage to the selected position, enters
the dividend in the upper dials, and clears the keyboard; when the
divide keys (÷) are touched, division is performed.
6. Other keys. The division STOP key is used to terminate a
division in any desired carriage position; the counter control key
(marked "G" on some machines) is used to control the direction of
counting of lower dials in addition and subtraction; the positive
keyboard lock prevents the accidental clearance of any portion of
the keyboard values; the dial accumulator locks can be used to
prevent the upper and lower dials from clearing.
For the operations described in this manual the normal positions
of these keys are as follows: division STOP key in the middle of its
slot; counter control key up; positive keyboard lock down; dial
accumulator locks with white line facing left.
8
Machine Computation of Elementary Statistics
B. MARCHANT MODEL ACT-10M (4
1. The keyboard. There are ten columns of ten keys each. In
each column, the keys are numbered from 1 to 9. The bottom row
of red keys, numbered from right to left from 1 to 10, automatically
controls the position of the carriage in multiplication and division.
The first keyboard column on the extreme right may be considered
the units position, the second column the tens position, the third
column the hundreds position, and so forth. Keys are operated by
depressing them until they catch and stay down. If a second key in
a column is depressed, the first will automatically be released.
Decimal markers appear between each two columns of keys. The
keyboard and keyboard dials may be cleared by a touch of the
KEYBOARD DIAL clear key. The keyboard dials along the upper part
of the keyboard show each item set in the keyboard.
2. The carriage. There are two sets of dials in the movable
carriage, designated by their position as "upper" and "middle."
Results of addition, subtraction, and multiplication and the re-
mainder in division appear in the middle dials; the multiplier in
multiplication, the quotient in division, and the count of items in
addition and subtraction appear in the upper dials. The carriage
position indicator (a red arrow) indicates the position of the car-
riage with respect to the first keyboard column, i.e., when the
arrow points to upper dial one, that dial is directly above keyboard
column one, and the carriage is said to be in position one, or the
first position.
Decimal markers appear below each set of dials. The carriage is
moved either to the right or to the left by touching the carriage
shift keys on which the arrow points to the right or to the left.
When no selective tabulator keys are depressed, the carriage auto-
matically returns to position one (extreme left) with a touch of the
CLEAR TAB key.
3. Plus and minus bars. When the plus bar or the minus bar is
touched, the indicated operation is performed (see special sections
on addition and subtraction for details) and the keyboard auto-
matically clears.
4. Multiplier. The row of black keys to the right of the keyboard,
numbered from 0 to 9, is the automatic multiplier keyboard. The
green carriage shift control keys control direction of automatic
KEYBOARD
SELECTIVE
TABULATOR KEYS
CARRIAGE.
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​ㅁ
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0000000000
KEYBOARD DECIMAL
MARKER
ㅁㅁㅁㅁ
​KEYBOARD
DIALS
B
☐
FIGURE 3. MARCHANT Model ACT-10M
+
UPPER DIALS
-MIDDLE DIALS
AUTOMATIC DIVISION KEY
DIVISION STOP KEY
COUNTER CONTROL LEVER
MINUS BAR
CARRIAGE SHIFT
CONTROL KEYS
CARRIAGE SHIFT KEY
PLUS BAR
DIVISION CLEAR-RETURN KEY
CARRIAGE SHIFT KEY
REVERSE KEY
AUTOMATIC
MULTIPLIER
KEYBOARD
-CLEARANCE AND
TABULATION KEYS
KEYBOARD LOCK

6
10
Machine Computation of Elementary Statistics
movement of the carriage in multiplication; the red NONSHIFT key
is for the purpose of holding the carriage stationary in multiplica-
tion. The REVERSE key is for use in negative multiplication. (See
section on multiplication for operational details.)
5. Division keys. When the automatic division key (AUTO) is
touched, division is performed. Touching the division STOP key
automatically stops division when the carriage is in any desired
position, and serves also to release the REVERSE key in the event of
its being stuck.
The division clear-return key, when in normal (vertical) posi-
tion, causes the middle and keyboard dials to clear and the carriage
to return to the original position at the end of a division operation;
if the lever is inclined away from the operator, this clear-return
operation does not take place, and the quotient appears in the
middle dials.
6. Dial clearance and direction keys. The middle and upper
dials are cleared by depression of the MIDDLE DIAL and UPPER DIAL
clearance keys. When the manual counter control key is down, the
upper dials count backward in addition and forward in subtraction;
when up (normal), the upper dials count forward in addition and
backward in subtraction. The keyboard lock locks the keyboard
dial clearance key so it cannot be depressed.
C. MARCHANT MODEL ACR-8D (4)
1. The keyboard. There are eight columns of ten keys each.
(In 10D models, there are ten columns of ten keys each.) In each
column, the keys are numbered from 1 to 9, with a blank key at the
foot of each column to represent zero. The first column on the ex-
treme right may be considered the units position, the second
column the tens position, the third the hundreds position, and so
forth. Keys are operated by depressing them until they catch and
stay down. If a second key in a column is depressed, the first will
automatically be released. Decimal markers appear between each
two columns of keys. The keyboard and keyboard dials may be
cleared by a touch of the KEYBOARD DIAL clearance key. The key-
board dials along the upper part of the keyboard show each item
set in the keyboard.
2. The carriage. There are two sets of dials in the movable car-
riage, designated by their position as "upper" and "middle."
Results of addition, subtraction, and multiplication and the re-
000000 ☐☐☐ 000000
SELECTIVE TABULATOR KEYS
ㅁㅁ
​CARRIAGE
KEYBOARD
☐
0000ㅁㅁㅁㅁ
​☐☐☐☐☐☐☐
¯¯¯¯¯¯OO
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00000000
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KEYBOARD DECIMAL MARKERS
0
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KEYBOARD
DIALS
F
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FIGURE 4. MARCHANT Model ACR-8D
UPPER DIALS
-MIDDLE DIALS
-AUTOMATIC DIVISION KEY
CARRIAGE TABULATION CONTROL
DIVISION STOP KEY
-COUNTER CONTROL LEVER
·MINUS BAR
SHORT CUT BAR
CARRIAGE SHIFT KEYS
MULTIPLICATION BAR
DIVISION CLEAR- RETURN KEY
PLUS BAR
CLEARANCE AND
TABULATION KEYS
KEYBOARD LOCK

11
12
Machine Computation of Elementary Statistics
mainder in division appear in the middle dials; the multiplier in
multiplication, the quotient in division, and the count of items in
addition and subtraction appear in the upper dials. The carriage
position indicator (an arrow) indicates the position of the carriage
with respect to the first keyboard column, i.e., when the arrow
points to upper dial one, that dial is directly above keyboard
column one, and the carriage is said to be in position one, or in the
first position.
Decimal markers appear below each set of dials. The carriage is
moved either to the right or left by touching the carriage shift keys
on which the arrow points to the right or to the left. When no
selective tabulator keys are depressed, the carriage automatically
returns to position one (extreme left) with a touch of the CLEAR
RETURN key.
3. Plus and minus bars. When the plus bar or the minus bar is
touched, the indicated operation is performed (see special sections
on addition and subtraction for details) and the keyboard auto-
matically clears.
4. Multiplication bars. The multiplication (X) bar is for use in
multiplication and the SHORT CUT bar for negative multiplication.
(See special sections on multiplication for operational details.)
5. Division keys. When the automatic division (AUTO ÷) key is
touched, division is performed. Touching the division STOP key
automatically stops division when the carriage is in any desired
position. The division clear-return key, when in normal (vertical)
position, causes the middle and keyboard dials to clear and the
carriage to return to the original position at the end of a division
operation; if the lever is inclined away from the operator, this clear-
return operation does not take place, and the quotient appears in
the middle dials.
6. Dial clearance keys. The middle and upper dials are cleared
by depression of the MIDDLE DIAL and UPPER DIAL clearance keys.
The keyboard lock locks the keyboard dial clearance key so that it
cannot be depressed.
D. MONROE CAA MODELS (11)
1. The keyboard. There are ten columns of ten keys each. In each
column, the keys are numbered from 0 to 9. The first column on the
extreme right may be considered the units position, the second
UPPER
DIALS LOCK
LOWER
DIALS LOCK
LOWER DIALS
400
TRANSFER SLIDE
TAB STOP KEYS
UPPER DIALS
CARRIAGE
KEYBOARD
MULTIPLIER SETUP DIALS—
KEYBOARD DECIMAL
MARKERS
DIVIDE STOP KEY
DIVIDE KEY
NEGATIVE MULTIPLY KEY.
MULTIPLY KEY
O
O
000000000
O
0
FIGURE 5. MONROE Model CAA-10
나
​D-
– CONSTANT
MULTIPLIER LEVER
TRANSFER LEVER
MINUS BAR
CARRIAGE SHIFT KEYS
PLUS BAR
SETUP KEY
CHANGE LEVER
NON-REPEAT KEY
REPEAT KEY
DIVIDEND TAB KEY
DIAL CLEAR KEYS
CHANGE LEVER LOCK
CLEAR KEY
SETUP TAB KEY

13
14
Machine Computation of Elementary Statistics
column the tens position, the third column the hundreds position,
and so forth. Keys are operated by depressing them until they catch
and stay down. If a second key in a column is depressed, the first
will automatically be released. In the event two are depressed
simultaneously, only the larger number will register in the machine.
The zero key also serves to lock a key in a column. Decimal markers
are provided to be set between any two columns of keys. The key-
board may be cleared by a touch of the CLEAR key. When the non-
repeat key is depressed, the keys depressed in the keyboard auto-
matically clear after each operation; when the repeat (R) key is
depressed, the keys depressed on the keyboard remain so until
cleared either by the operator or by the fully automatic action of
the machine.
2. The carriage. There are three sets of dials in the movable
carriage designated as upper, lower, and multiplier setup dials.
Results of addition, subtraction, and multiplication and the re-
mainder in division appear in the lower dials; the multiplier in
multiplication, the quotient in division, and the count of items in
addition and subtraction appear in the upper dials. The multiplier
setup dials store multipliers from either the keyboard or the lower
dials. (See sections on multiplication for operational details.) The
upper dials lock and lower dials lock control the clearance of the
upper and lower dials respectively. The upper dials lock has two
positions: when pushed in, the upper dials can be cleared; when
pulled out, the figures in the upper dials remain there in spite of
any clearance operations. The lower dials lock has three positions:
when pushed all the way in, the lower dials can be cleared; when
pulled all the way out, the figures in the lower dials remain there in
spite of any clearance operations; when half-way in, all the lower
dials from the ninth through the first are locked against clearance,
while the dials from the tenth all the way to the left can be cleared
(split dial clearance operation).
Decimal markers appear below both upper and lower dials. The
carriage is moved either to the right or to the left by touching the
carriage shift keys on which the arrow points to the right or to the
left. When any of the nine tab stop keys is depressed, the carriage
position in multiplication and division is predetermined; tab stop
keys that are depressed are cleared by depressing the zero tab stop
key.
Description of Calculating Machines
15
3. Plus and minus bars. When the plus bar or the minus bar is
touched, the indicated operation is performed. (See special sections
on addition and subtraction for details.)
4. Multiplication. On the fully automatic model, there are two
setup keys, the SETUP key and the SETUP TAB key. The SETUP key,
when depressed with the carriage in position one, transfers the key-
board amount into the multiplier setup dials; when the carriage is
in any other position, this operation transfers the keyboard amount
into the lower dials. In either case, the keyboard clears and the
upper dials are not affected. The SETUP key is also a squaring lock.
If it is depressed and held down until the setup action is completed,
the keyboard amount will both transfer into the multiplier setup
dials and remain on the keyboard (only when the carriage is in the
first position). Depressing the SETUP TAB key clears the upper and
lower dials, shifts the carriage to the extreme left or to a tab stop
position, transfers the keyboard amount into the multiplier setup
dials, and clears the keyboard; when the carriage shifts to a tab
stop position the keyboard amount also enters the lower dials.
The multiply (X) key when depressed (the carriage must be in
the first position) causes the upper and lower dials to clear and the
amount set in the keyboard to be multiplied by the amount in the
multiplier setup dials. The negative multiply key (X) when de-
pressed (the carriage must be in the first position) causes negative
multiplication. (See special sections on multiplication for opera-
tional details.)
5. Division keys. When the divide key is depressed, division is
performed. Touching the divide stop key stops the division at any
desired position of the carriage and clears the keyboard. A simple
method of determining whether or not electric current is on is to
depress the divide stop key. Depression of the divide tab key auto-
matically tabulates the carriage to a selected position, clears the
upper and lower dials, enters the keyboard amount into the lower
dials, clears the 1 from the upper dials, and clears the keyboard.
6. Other keys. Upper and lower dials are cleared by depression
of the UPPER and LOWER keys. The change lever automatically con-
trols the direction of rotation of the upper dials; normally it is in
the up (X) position for multiplication and in the lower (÷) posi-
tion for division. The transfer slide is for use in automatic tabula-
tion of the carriage in certain multiplication operations; the trans-
16
Machine Computation of Elementary Statistics
fer lever is used where it is desired to transfer the lower dials
amount into the multiplier setup dials. The constant multiplier
lever locks an amount in the multiplier setup dials so that it may be
retained as a constant multiplier. The change lever lock is used
only in special operations for locking the change lever in either the
X or position.
E. MONROE MODEL AA-1 (12)
1. The keyboard. There are ten columns of ten keys each. In
each column, the keys are numbered from 0 to 9. The first column
on the extreme right may be considered the units position, the
second column the tens position, the third column the hundreds
position, and so forth. Keys are operated by depressing them until
they catch and stay down. If a second key in a column is depressed,
the first will automatically be released. In the event two are de-
pressed simultaneously, only the larger number will register in the
machine. Decimal markers appear between each two columns of
keys. The keyboard may be cleared by a touch of the KEYBOARD
CLEAR key (marked 0 on some machines). When the nonrepeat key
is depressed, the keys depressed in the keyboard automatically
clear after each operation; when the repeat (R) key is down, the
keys depressed in the keyboard remain so until cleared by a touch
of the KEYBOARD CLEAR key.
2. The carriage. There are two sets of dials in the movable car-
riage designated by their position as "upper" and "lower." Results
of addition, subtraction, and multiplication and the remainder in
division appear in the lower dials; the multiplier in multiplication,
the quotient in division, and the count of items in addition and sub-
traction appear in the upper dials. The upper dials are split in the
middle into a right and a left bank. The upper dials control knob,
or lock lever, when pushed in, locks the right bank of upper dials so
they do not clear when the dials clearance key is touched. The
upper dials are cleared by depressing the upper dials clearance
key () and the lower dials by depressing the lower dials clearance
key ().
MA
Decimal markers appear above both the upper and the lower
dials. The carriage is moved either to the right or to the left by
touching the carriage shift keys on which the arrow points to the
right or to the left.
☐ ☐ ☐ ☐ ☐|☐ ☐ ☐☐☐☐☐☐☐☐
םם
םםם
CARRIAGE
][
KEYBOARD
DIVIDE KEY
CONSTANT
CONS:
LEVERS
MULTIPLY KEY
MULTIPLIER
SETUP BAR
ㅁ
​ㅁㅁ
​8
ㅁㅁ ​ㅁ
​100
D
ㅁㅁ
​OOOOOOOOOO
OOOOOOOOOO
□0000000000
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯0000
OOOOOOOOOO
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯ ¯
OOOOOOOOOO
J
FIGURE 6. MONROE Model AA-1
UPPER DIALS
UPPER DIALS
LOCK LEVER
LOWER DIALS
KEY
MARKE MAL
KEYBOARD DECIMAL
COUNTER CONTROL LEVER
MULTIPLICATION LEVER
MINUS BAR
CAR
CARRIAGE
SHIFT KEYS
PLUS BAR
REPEAT KEY
DIVIDE KEY
SQUARING LOCK
NON-REPEAT KEY
KEYBOARD CLEAR KEY
UPPER DIALS CLEARANCE KEY
LOWER DIALS CLEARANCE KEY

17
18
Machine Computation of Elementary Statistics
3. Plus and minus bars. When the plus bar or minus bar is
touched, the indicated operation is performed.
4. Multiplication keys. The multiplier setup bar (X) is de-
pressed to set the multiplier in the machine from the keyboard,
and to clear the keyboard; the multiply key (=) when depressed
causes multiplication to take place automatically, and the carriage
to return to position one. The two constant levers (CONS) are used
to lock a constant multiplier in the machine. (See special section
for operational details.)
5. Division keys. There are two divide keys (÷) for use in auto-
matic division. Either one, depending on the convenience of the
operator, may be pushed up to start automatic division. Operating
either of the divide keys automatically depresses the R key. If it
is desirable to stop the division before the carriage has reached the
final position, the divide key should be pushed back to its original
position.
6. Other keys. A counter control lever, sometimes called the
upper dials reversing lever, governs the direction of the right upper
dials in addition and subtraction. When the lever is at X, the dials
count forward in addition and backward in subtraction; when at
÷, backward in addition and forward in subtraction.
The carriage position indicator is an arrow directly above the
first keyboard column, pointing to the dials directly above that
column. When the carriage is at the extreme left, the arrow points.
to the first dials and the carriage is said to be in position one, or
first position. If the arrow points to the second dials, the carriage.
is in position two, and so forth.
F. MONROE MODEL MA-7 (6)
1. The keyboard. There are ten columns of ten keys each. In
each column, the keys are numbered from 0 to 9. The first column
on the extreme right may be considered the units position, the
second column the tens position, the third column the hundreds.
position, and so forth. Keys are operated by depressing them until
they catch and stay down. If a second key in a column is depressed,
the first will automatically be released. If two keys in the same col-
umn are pushed down simultaneously, only the larger number will
register in the machine. Decimal markers appear between each two
"
ㅁ
​CARRIAGE
☐☐☐☐☐☐☐
KEYBOARD
☐ ☐
DIVIDE KEY
☐☐☐☐☐☐☐☐ ☐ ☐|☐ ☐
ㅁㅁ ​☐☐
DO ---
ㅁㅁㄷ
​--- ---
O O O O O O O O O O
¯¯¯¯¯¯¯¯¯¯
□0000000000
OOOOOOOOOO
OOOOOOOOOO
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
[
FIGURE 7. MONROE Model MA-7
UPPER DIALS
UPPER DIALS
LOCK LEVER
LOWER DIALS
HANOSHIFT KNOB
MA
KEYBOARD DECIMAL
MARKERS
COUNTER CONTROL LEVER
MINUS BAR
CARRIAGE
SHIFT KEYS
Ers
PLUS BAR
REPEAT KEY
NON-REPEAT KEY
KEYBOARD CLEAR KEY
UPPER DIALS CLEARANCE KEY
LOWER DIALS CLEARANCE KEY

19
20
Machine Computation of Elementary Statistics
columns of keys. The keyboard may be cleared by a touch of the
KEYBOARD CLEAR key, marked 0 on some machines. The arrow
above the first column (carriage position indicator) indicates the
position of the carriage with respect to the first keyboard column.
When the NONREPEAT key is depressed, the keys depressed in the
keyboard automatically clear after each operation; when the repeat
(R) key is down, the keys depressed in the keyboard remain so
until cleared by a touch of the KEYBOARD CLEAR key.
2. The carriage. There are two sets of dials in the movable car-
riage, designated by their position as "upper" and "lower." The
symbol | appears beside the upper dials and | beside the lower
dials. Results of addition, subtraction, and multiplication and the
remainder in division appear in the lower dials; the multiplier in
multiplication, the quotient in division, and the count of items in
addition and subtraction appear in the upper dials. On certain
models, the upper dials are split in the middle into a right and a
left bank. On such machines, there is a dial lock lever which, when
engaged, locks the right bank of upper dials so they do not clear
when the dials clearance key is touched. The upper dials are cleared
by depressing the upper dials clearance key (||), and the lower dials
by depressing the lower dials clearance key (1).
Movable decimal markers appear below both upper and lower
dials. The carriage is moved either to the right or to the left by the
carriage shift keys on which the arrow points to the right or to the
left.
3. Plus and minus bars. When the plus bar or the minus bar is
touched, the indicated operation is performed. On semiautomatic
machines, the plus and minus bars are used in multiplication oper-
ations. (See special sections.)
4. Division key. When the division key is moved up in its slot,
division automatically takes place.
5. Other keys. On certain machines, a counter control lever gov-
erns the direction of the right upper dials in addition and sub-
traction. When the lever is at X, the dials count forward in addi-
tion and backward in subtraction; when at ÷ backward in addition
and forward in subtraction.
The carriage position indicator is an arrow directly above the
first keyboard column, pointing to the dials directly above that
column. When the carriage is at the extreme left, the arrow points
Description of Calculating Machines
21
to the first dials, and the carriage is said to be in position one, or
first position. If the arrow points to the second dials, the carriage is
in position two, and so forth.
G. MONROE K AND EDUCATOR MODELS (7, 14)
1. The keyboard. There are ten (sometimes eight or seven) col-
umns of ten keys each. In each column, the keys are numbered from
0 to 9. The first column to the right may be considered the units
position, the second column the tens position, the third column the
hundreds position, and so forth. Keys are operated by depressing
them until they catch and stay down. If a second key in a column
is depressed, the first will automatically be released. In the event
two are depressed simultaneously, only the larger number will
register in the machine. Decimal markers appear between each
two columns of keys. The keyboard may be cleared by a touch of
the clear key. When the nonrepeat key is depressed, the keys de-
pressed in the keyboard automatically clear after each operation;
when the repeat (R) key is depressed, the keys depressed in the
keyboard remain so until cleared by a touch of the clear key.
2. The carriage. There are two sets of dials in the movable car-
riage, designated as "upper" and "lower." Results of addition, sub-
traction, and multiplication and the remainder in division appear
in the lower dials; the multiplier in multiplication, the quotient in
division, and the count of items in addition and subtraction appear
in the upper dials. On certain models with twenty upper dials, the
dials are split in the middle into a right and left bank of ten dials
each. On such machines, there is a dial lock lever which, when
pushed to the left, locks the right bank of upper dials so they do not
clear when the dial clearance crank is turned. One revolution for-
ward of the dial clearance crank clears the upper dials; one revolu-
tion backward clears the lower dials.
Movable decimal markers appear below both the upper and the
lower dials. The carriage is moved either to the right or to the left
by the carriage shift lever; one half turn moves the carriage one
position. If it is necessary to shift the carriage several places, it
may be raised by the hand shifting knob on the right of the carriage
and moved to the desired position.
3. Plus and minus bars. On electric models, when the plus bar
(+) or the minus bar (-) is touched, the indicated operation is
CARRIAGE
ㅁ
​BELL LEVER
KEYBOARD
I
|
J ☐ ☐ ☐☐ ☐
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
OOOOOOOOOO
OOOOOOOOOO
OOOOOOOOOO
¯¯¯¯¯¯¯¯¯¯
оооооооооо
¯¯¯¯¯¯¯¯¯¯
¯¯¯¯¯¯¯¯¯¯
ooooooo 000
口 ​☐ ☐
►
ㅁㅁㅁ
​с
CARRIAGE_SHIFT LEVER
FIGURE 8. MONROE Model KA-203
MINUS BAR.
PLUS BAR
UPPER DIALS
UPPER DIALS
LOCK LEVER
COUNTER CONTROL LEVER
RANCE.
DIAL CLEARANCE.
HAND SHIFT KNOB
LOWER DIALS
NON-REPEAT KEY
REPEAT KEY
CRANK
MARKERS
KEYBOARD CLEAR KEY
KEYBOARD DECIMA
MAL

22
Description of Calculating Machines
23
performed. (See special sections on addition and subtraction for
details.)
4. Operating crank. On hand-operated models, when the oper-
ating crank is turned away from the operator (forward) addition
takes place; when turned toward the operator (backward) subtrac-
tion takes place. Complete forward or backward turns of the crank
must be made from and to the neutral position (stopping place).
5. Other keys. On certain models, a counter control lever gov-
erns the direction of the right upper dials in addition and subtrac-
tion. When the lever is at X, the dials count forward in addition
and backward in subtraction; when at÷, backward in addition
and forward in subtraction.
The carriage position indicator is an arrow directly above the
first keyboard column, pointing to the dials directly above that
column. When the carriage is at the extreme left, the arrow points
to the first dials, and the carriage is said to be in position one, or
first position. If the arrow points to the second dials, the carriage
is in position two, and so forth.
On some models, it is possible to control the ringing of the bell
by a bell lever at the left side of the keyboard. (See special sections
for operational details.)
Special Note: After this manuscript went to press, the
Marchant Calculating Machine Company announced a new
model, the Figuremaster (8 FA and 10 FA). Directions written
here for the ACT-10M model may be followed in operating the
10 FA machine; with one exception: on the new machine, multi-
plication is done only from the left. On the Figuremaster, it is also
possible to depress and use more than one selective tabulator key
at one time.
PART ONE
BASIC OPERATIONS
CHAPTER 3
ADDITION AND SUBTRACTION
A. ADDITION
Addition can be accomplished with the carriage in any position.
It is best always to have the carriage shifted to the extreme left
(position one). On most machines, the carriage is in position one
when the carriage position indicator (arrow) points to the first dial.
Amounts should be set up in the extreme right of the keyboard.
In working with whole numbers, units digits are set in the first
column on the right, tens digits in the second, hundreds in the
third, and so on.
For any extensive addition and subtraction operations, an add-
ing machine* should be used. A machine of the listing type, which
prints the numbers being added and subtracted and the totals,
provides the best method of checking these operations.
Friden
(a) Addition of whole numbers
EXAMPLE: 124 + 360
484
1. Clear the machine by depressing the CLEAR (to clear the
keyboard) and RETURN CLEAR (to clear the carriage dials) keys
simultaneously with the fore and middle fingers.
2. Position the carriage by touching the upper carriage shift
key on which the arrow points to the left until the carriage position
indicator on the upper right of the keyboard panel points to the
first lower dial (position one).
=
3. Depress the add key so that the keyboard will clear after
each depression of the plus bar.
* Burroughs, Clary, Monroe, Remington's Sundstrand, etc.
27
28
Machine Computation of Elementary Statistics
4. Set 124 in the keyboard by depressing 1 in the third column
from the right, 2 in the second, and 4 in the first. Be sure to depress
only one key in each column. Put the numbers in the keyboard in
the same order you would write them on paper.
5. In case of depressing the wrong key in a column, correct by
depressing the zero key. Depressing the zero key releases any key
in the column.
6. Tap the plus bar, being sure to release it immediately.
124 appears in the upper dials. The keyboard clears. The number
1 appears in the lower dials to indicate that one item has been
added; if the number 2 or a higher number appears, then the plus
bar has been held down too long, and an incorrect count of items
added will result. It usually takes some practice to acquire the
right touch. Try this several times, clearing the dials after each
operation.
7. Set 360 in the keyboard by depressing 3 in the third column,
6 in the second, and doing nothing in the first. Depressing the zero
key serves only to clear the column in the event that there is a key
already depressed in it.
8. Touch the plus bar. The total 484 appears in the upper dials
directly above the keyboard columns in which the numbers were
depressed. The keyboard clears as before. The number 2 in the
lower dials indicates that two items have been added.
9. To clear the dials, touch the RETURN CLEAR key; or shift the
two DIAL CLEARANCE keys to the right.
PROBLEMS: Addition of whole numbers. Check by readding; place
check answer in second answer space. When two digits of a number
are the same, or close as 3 and 4, they may be set in the keyboard
simultaneously by using the forefinger to depress the first and the
middle finger to depress the second.
(1) 489
426
(3) 915
243
(2) 106
976
(4) 821
765
531
Addition and Subtraction
29
(5) 202
111
999
(7) 1,441
3,943
2,843
5,760
(9) 45,838
147
6,343
71,106
35
9
(6) 711
258
(8) 6,735
2,607
901
13,839
(10)
362
647
-
6
6,290
69,000
722
19
33,869
988,460
(b) Addition of numbers with decimals
EXAMPLE: 3.45 + 15.3 + 1.006 + 0.04
19.796
1. Inspect the numbers to be added to find the largest number
of decimal places needed. In this case the number 1.006 has the
largest, three.
2. Turn over the keyboard decimal between columns three and
four; move the upper dial decimal marker until it is between upper
dials three and four, directly above the keyboard decimal.
3. Set 345 in keyboard columns four, three, and two, so that
the decimal appears between the 3 and the 4.
4. Add. The number 3450 appears in the upper dials, with the
decimal between the 3 and the 4. The 1 in the lower dials indicates
that one item has been added.
5. Set 153 in keyboard columns five, four, and three so that the
decimal appears between the 5 and the 3. Add. The number 18750
appears in the upper dials, with the decimal between the 8 and the
7. The 2 in the lower dials indicates that two items have been
added.
30
Machine Computation of Elementary Statistics
6. Set 1006 in keyboard columns four to one, so that the deci-
mal appears between the 1 and the first 0. Add. The number
19.756 appears in the upper dials, correctly pointed off.
7. Set 4 in keyboard column two. Add. The total 19796 appears
in the upper dials with the decimal between the first 9 and the 7.
The final answer is then read as 19.796. The 4 in the lower dials
indicates that four items have been added.
PROBLEMS: Addition of numbers with decimals. Be sure to reset.
both keyboard and dial decimals for each problem. Check by re-
adding. Practice setting in two digits at a time where feasible.
(12)
(11) 58.7500
6.2100
2.3800
14.8000
5.0099
166.5000
432.0000
0.4870
(13) 197.0498
122.0000
3.0608
6,308.3600
99.8900
726.0040
4.9902
0.0800
S
=
5.26000
2.35000
0.00900
Marchant
369.09000
56.93300
4,000.80000
77.00000
9.00037
(14) 8,239.008
668.008
1,360.920
10.007
362.887
3,872.004
1.590
87.000
(a) Addition of whole numbers
EXAMPLE: 124 + 360
484
1. Clear the machine by depressing the KEYBOARD DIAL, MIDDLE
DIAL, and UPPER DIAL keys simultaneously with the second, third,
and fourth fingers.
2. Position the carriage by touching the upper black carriage
shift key on which the arrow points to the left until the carriage
Addition and Subtraction
31
position indicator above the carriage points to the first upper dial
(position one). Be sure no carriage tab key is depressed.
3. Set 124 in the keyboard by depressing 1 in the third column
from the right, 2 in the second, and 4 in the first. Put the numbers
in the keyboard in the same order you would write them on paper.
The number 124 will appear in keyboard dials three, two, and one
as a check on the numbers put in the keyboard.
4. In case of depressing the wrong key in a column, correct by
depressing the correct key. If a key is depressed by mistake when
a zero is desired, correct by tapping lightly another key in the
column without actually depressing it far enough to keep it down.
In certain models there are zero keys instead of selective tab keys
at the keyboard base. Depressing a zero key releases any other key
in the column and registers as 0 in the keyboard dials.
5. Touch the add bar. The number 124 appears in the middle
dials. The keyboard clears. The number 1 appears in the upper
dials to indicate that one item has been added.
6. Set 360 in the keyboard by depressing 3 in the third column,
6 in the second, and doing nothing in the first. Note that 360 ap-
pears in keyboard dials three to one.
7. Touch the add bar. The total 484 appears in the middle
dials directly above the keyboard columns in which the numbers
were depressed. The keyboard clears as before. The number 2 in
the upper dials indicates that two items have been added.
8. To clear the dials touch the MIDDLE DIAL and UPPER DIAL
keys simultaneously with the fore- and middle fingers.
PROBLEMS: Addition of whole numbers. Check by readding; place
check answer in second answer space. When two digits of a number
are the same, or close as 3 and 4, they may be set in the keyboard
simultaneously by using the forefinger to depress the first and the
middle finger to depress the second.
(1) 489
426
(3) 915
243
(2) 105
976
(4) 821
765
531
32
Machine Computation of Elementary Statistics
(6) 711
258
362
647
(5) 202
111
999
(7) 1,441
3,943
2,843
5,760
(9) 45,838
147
6,343
71,106
35
9
(8) 6,735
2,607
901
13,839
(10)
6
6,290
69,000
=
722
19
33,869
988,460
(b) Addition of numbers with decimals
EXAMPLE: 3.45 + 15.3 + 1.006 + 0.04
19.796
1. Inspect the numbers to be added to find the largest number
of decimal places needed. In this case the number 1.006 has the
largest, three.
2. Turn over the keyboard decimal between columns three and
four; move the middle dial decimal marker until it is between
middle dials three and four, directly above the keyboard decimal.
3. Set 345 around the keyboard decimal, in keyboard columns
four, three, and two, so that the decimal appears between 3 and 4
on both the keyboard and keyboard dials. Add. The number 3450
appears in the middle dials with the decimal between the 3 and 4.
The 1 in the upper dials indicates that one item has been
added.
4. Set 153 in keyboard columns five, four, and three so that the
decimal appears between the 5 and the 3. Add. The number 18750
appears in the middle dials with the decimal between the 8 and
the 7. The 2 in the upper dials indicates that two items have been
added.
•
Addition and Subtraction
33
5. Set 1006 in keyboard columns four to one so that the decimal
appears between the 1 and the first 0. Add. The number 19.756
appears in the middle dials correctly pointed off.
6. Set 4 in keyboard column two. The keyboard dial reads
0.040. Add. The total 19796 appears in the middle dials with the
decimal between the first 9 and the 7. The final answer is then read
as 19.796. The 4 in the upper dials indicates that four items have
been added.
PROBLEMS: Addition of numbers with decimals. Check by readding.
Practice setting in two digits at a time where feasible.
(12)
(11) 58.7500
6.2100
2.3800
14.8000
5.0099
(13)
166.5000
432.0000
0.4870
197.0498
122.0000
3.0608
6,308.3600
99.8900
726.0040
4.9902
0.0800
5.26000
2.35000
0.00900
369.09000
56.93300
4,000.80000
77.00000
9.00037
(14) 8,239.008
668.008
1,360.920
10.007
362.887
3,872.004
1.590
87.000
Monroe: Automatic and Semiautomatic
(a) Addition of whole numbers
EXAMPLE: 124 + 360 484
1. Clear the machine by depressing UPPER (to clear the upper
dials), LOWER (to clear the lower dials), and CLEAR (to clear the
keyboard) keys simultaneously with the fore- and middle fingers.
2. Position the carriage by touching the upper carriage shift
key on which the arrow points to the left until the carriage is moved
34
Machine Computation of Elementary Statistics
to the extreme left (position one). In this position, the first lower
and upper dials are directly above the first keyboard column.
3. Depress the nonrepeat key so that the keyboard will clear
after each depression of the plus bar.
4. Set 124 in the keyboard by depressing 1 in the third column
from the right, 2 in the second, and 4 in the first. Be sure to depress
only one key in each column. Put the numbers in the keyboard in
the same order you would write them on paper.
5. In case of depressing the wrong key in a column, correct by
depressing the correct key. Depressing the zero key releases any
key in the column.
6. Touch the plus bar. The number 124 appears in the lower
dials. The keyboard clears. The number 1 appears in the upper
dials to indicate that one item has been added.
7. Set 360 in the keyboard by depressing 3 in the third column,
6 in the second, and doing nothing in the first. Depressing the zero
key serves only to clear the column in the event there is a key al-
ready depressed there.
8. Touch the plus bar. The total 484 appears in the lower dials
directly above the keyboard columns in which the numbers were
depressed. The keyboard clears as before. The number 2 in the
upper dials indicates that two items have been added.
9. To clear the dials, touch the UPPER and LOWER keys simul-
taneously.
PROBLEMS: Addition of whole numbers. Check by readding; place
check answer in second answer space. When two digits of a number
are the same, or close as 3 and 4, they may be set in the keyboard
simultaneously by using the forefinger to depress the first and the
middle finger to depress the second.
(1) 489
426
(3) 915
243
(2) 106
976
(4) 821
765
531
Addition and Subtraction
35
(5) 202
111
999
(7) 1,441
3,943
2,843
5,760
(9) 45,838
147
6,343
71,106
35
9
(6) 711
258
362
647
(8) 6,735
2,607
901
13,839
(10)
6
6,290
69,000
722
19
33,869
988,460
(b) Addition of numbers with decimals
EXAMPLE: 3.45 + 15.3 + 1.006 + 0.04
19.796
1. Inspect the numbers to be added to find the largest number
of decimal places needed. In this case the number 1.006 has the
largest, three.
2. Place the keyboard decimal between columns three and four;
move the lower dial decimal until it is between lower dials three
and four, directly above the keyboard decimal.
3. Set 345 around the keyboard decimal, in keyboard columns
four, three, and two, so that the decimal appears between the 3
and the 4. Add. The number 3450 appears in the lower dials, with
the decimal between the 3 and the 4. The number 1 in the upper
dials indicates that one item has been added.
4. Set 153 in keyboard columns five, four, and three so that
the decimal appears between the 5 and the 3. Add. The number
18750 appears in the lower dials, with the decimal between the 8
and the 7. The 2 in the upper dials indicates that two items have
been added.
36
Machine Computation of Elementary Statistics
5. Set 1006 in keyboard columns four to one, so that the deci-
mal appears between the 1 and the first 0. Add. The number 19.756
appears in the lower dials correctly pointed off.
6. Set 4 in keyboard column two. Add. The total 19796 appears
in the lower dials, with the decimal between the first 9 and the 7.
The final answer is then read as 19.796. The 4 in the upper dials
indicates that four items have been added.
PROBLEMS: Addition of numbers with decimals. Check by readding.
Practice setting in two digits at the same time where feasible.
(12)
(11) 58.7500
6.2100
2.3800
14.8000
5.0099
166.5000
432.0000
0.4870
(13) 197.0498
122.0000
3.0608
6,308.3600
99.8900
726.0040
4.9902
0.0800
5.26000
2.35000
0.00900
369.09000
56.93300
4,000.80000
77.00000
9.00037
(14) 8,239.008
668.008
1,360.920
10.007
362.887
3,872.004
=
1.590
87.000
•
Monroe: Hand-operated
On hand-operated Monroe models, the operating principles are
the same whether the machine has operating cranks or a motor.
Directions are written for models with hand cranks; remarks in
parentheses refer to machines with motors. (See point 4 below.)
(a) Addition of whole numbers
EXAMPLE: 124 + 360 484
1. Clear by turning the dial clearance crank one turn in each
direction, i.e., away from the operator, then toward the operator.
Each revolution should be a complete one; test the crank to see
Addition and Subtraction
37
that it hangs freely in neutral position by working it gently to and
fro in this position. The carriage will be raised while each set of
dials is being cleared, then settle into position when the crank is in
neutral.
2. Clear the keyboard by depressing the keyboard clearance key.
3. Put the carriage in position one-extreme left-by turning
the carriage position lever to the left. When the carriage is in posi-
tion one, the first dials are in line with the first keyboard column.
4. Depress the nonrepeat key so the keyboard will clear after
each revolution of the operating crank (after each depression of
the plus bar). Be sure the operating crank is in neutral-up-posi-
tion.
5. Set 124 in the keyboard by depressing 1 in the third column
from the right, 2 in the second, and 4 in the first. Be sure to depress
only one key in each column. Put the numbers in the keyboard in
the same order you would write them on paper.
6. In case of depressing the wrong key in a column, correct by
depressing the correct key or the zero key. Depressing the zero key
releases any key in the column.
7. Turn the operating crank away from the front of the machine
one complete revolution, and stop it at the top of the turn. (Tap
the plus bar, being sure to release it immediately.) The number 124
appears in the lower dials. The keyboard clears. The number 1
appears in the upper dials to indicate that one item has been added.
(In models with motors, if the number 2—or a higher number-
appears, the plus bar has been held down too long and the incorrect
count of items will result. It usually takes some practice to acquire
the right touch.)
8. Set 360 in the keyboard by depressing 3 in the third column,
6 in the second, and doing nothing in the first. Depressing the zero
key serves only to clear the column in the event there is a key al-
ready depressed there.
9. Turn the operating crank away from the front of the machine
one complete revolution and stop at the top of the turn. (Tap the
plus bar once, being sure to release it immediately.) The total 484
appears in the lower dials directly above the keyboard columns in
which the numbers were depressed. The keyboard clears as be-
fore. The number 2 in the upper dials indicates that two items have
been added.
38
Machine Computation of Elementary Statistics
10. On Educator machines, the upper dials count only to 9, then
reverse. On other machines, there is a carry-over feature; that is,
the count in the upper dials carries over from the 9 in the first dial
(units) to 1 in the second dial (tens), so that the upper dials read
10 when 10 items have been added, and not a red 9, as in the
Educator.
·
PROBLEMS: Addition of whole numbers. Check by readding; place
check answer in second answer space. When two digits of a number
are the same, or close as 3 and 4, they may be set in the keyboard
simultaneously by using the forefinger to depress the first and the
middle finger to depress the second.
(1) 489
426
(3) 915
243
(5) 202
111
999
(7) 1,441
3,943
2,843
5,760
(9) 45,838
147
6.343
71,106
35
9
(2) 106
976
(4) 821
765
531
(6) 711
258
362
647
(8) 6,735
2,607
901
13,839
(10)
6
6,290
69,000
722
19
33,869
988,460
Addition and Subtraction
39
(b) Addition of numbers with decimals
EXAMPLE: 3.45 + 15.3 + 1.006 + 0.04
19.796
1. Inspect the numbers to be added to find the largest number
of decimal places needed. In this case, 1.006 has the largest, three.
2. Turn over the keyboard decimal marker between columns
three and four; move the lower dial decimal marker until it is be-
tween lower dials three and four, directly above the keyboard
decimal.
=
3. Set 345 around the decimal in keyboard columns four, three,
and two, so that the decimal appears between the 3 and the 4.
4. Add. The number 3450 appears in the lower dials, with the
decimal between the 3 and the 4. The number 1 in the upper dials
indicates that one item has been added.
5. Set 153 in keyboard columns five, four, and three so that the
decimal appears between the 5 and the 3. Add. The number 18750
appears in the lower dials, with the decimal between the 8 and the 7.
The number 2 in the upper dials indicates that two items have
been added.
(11) 58.7500
6.2100
2.3800
14.8000
5.0099
6. Set 1006 in keyboard columns four through one, so that the
decimal appears between the 1 and the first 0. Add. The number
19.756 appears in the lower dials correctly pointed off.
7. Set 4 in keyboard column two. Add. The total 19796 appears
in the lower dials, with the decimal between the first 9 and the 7.
The final answer is then read as 19.796. The 4 in the upper dials
indicates that four items have been added.
166.5000
432.0000
0.4870
PROBLEMS: Addition of numbers with decimals. Check by readding.
Practice setting in two digits at a time where feasible.
(12)
5.26000
2.35000
0.00900
369.09000
56.93300
4,000.80000
77.00000
9.00037
40
Machine Computation of Elementary Statistics
(13) 197.0498
122.0000
3.0608
(14) 8,239.008
668.008
1,360.920
6,308.3600
99.8900
726.0040
4.9902
0.0800
B. SUBTRACTION
Addition and subtraction are inverse operations; the effect of one
operation cancels the effect of the other.
567
-489
Friden
EXAMPLE: 484 124
= 360
1. Clear the machine and move the carriage to position one.
2. Depress the add key.
3. Set the minuend 484 in the extreme right of the keyboard.
4. Touch the plus bar. The number 484 appears in the upper
dials, the keyboard clears, and a number 1 appears in the lower dials
to indicate that one item has been added.
5. Set the subtrahend 124 in the extreme right of the keyboard
directly under 484 in the upper dials.
6. Touch the minus bar, being sure to release it immediately.
The remainder 360 appears in the upper dials, the keyboard clears,
and a 0 appears in the first dial to indicate that the second item
was subtracted from the first.
(4) 1,275
185
10.007
362.887
3,872.004
PROBLEMS: Subtraction. Check by adding subtrahend to remainder.
(1)
(2)
1,090
585
. (3) 867
28
·
1.590
87.000
(5) 7,568
-1,298
་
Addition and Subtraction
41
PROBLEMS: Addition and subtraction of more than two numbers.
Check by redoing.
(6)
(7)
4,678
- 1,987
780
(8) 3,655.09
3.90
837.54
8,843.21
5,542.00
-8,077.19
W
(4) 1,275
185
Marchant
(5)
7,650
452
-7,568
(9) 89.99*
19.01
438.78
92.63
- 236.01
- 186.40
EXAMPLE: 484
124 =
360
1. Clear the machine and move the carriage to position one.
2. Set the minuend 484 in the extreme right of the keyboard.
3. Touch the plus bar. The number 484 appears in the middle
dials, the keyboard clears, and a number 1 appears in the upper
dials to indicate that one item has been added.
7,568
- 1,298
J
G
4. Set the subtrahend 124 in the extreme right of the keyboard
directly under the 484 in the middle dials.
5. Touch the minus bar. The remainder 360 appears in the
middle dials, the keyboard clears, and a 0 appears in the first upper
dial to indicate that the second item was subtracted from the first.
PROBLEMS: Subtraction. Check by adding subtrahend to remainder.
(1)
(2)
567
-489
1,090
585
(3)
867
- 28
* When 89.99 is subtracted first, a bell rings and 1001 preceded by a series of
9's appears in the upper dials; a series of 9's appears in the lower dials. This
indicates that a negative number is in the machine, which is satisfactory.
Continue with the operations as indicated, and record the final answer.
42
Machine Computation of Elementary Statistics
PROBLEMS: Addition and subtraction of more than two numbers.
Check by redoing.
(6)
4,678
-1,987
780
(8) 3,655.09
3.90
837.54
8,843.21
5,542.00
-8,077.19
G
(7)
(1) 567
-489
(9)
7,650
452
-7,568
1,090
585
89.99*
19.01
438.78
92.63
Monroe: Automatic and Semiautomatic
EXAMPLE: 484 124 =
360
1. Clear the machine and move the carriage to position one.
2. Depress the nonrepeat key.
3. Set the minuend 484 in the extreme right of the keyboard.
4. Touch the plus bar. The number 484 appears in the lower
dials, the keyboard clears, and a number 1 appears in the upper dials
to indicate that one item has been added.
-
-236.01
- 186.40
5. Set the subtrahend 124 in the extreme right of the keyboard
directly under the 484 in the lower dials.
6. Touch the minus bar. The remainder 360 appears in the
lower dials, the keyboard clears, and a 0 appears in the first upper
dial to indicate that the second item was subtracted from the first.
PROBLEMS: Subtraction. Check by adding subtrahend to remainder.
(2)
(3) 867
28
* When 89.99 is subtracted first, 1001 preceded by a series of 9's appears in
the middle dials; a series of 9's appears in the upper dials. This indicates that a
negative number is in the machine, which is satisfactory. Continue with the
operations as indicated and record the final answer.
Addition and Subtraction
43
(4) 1,275
185
4,678
- 1,987
780
(5) 7,568
-1,298
PROBLEMS: Addition and subtraction of more than two numbers.
Check by redoing.
(6)
(8) 3,655.09
3.90
837.54
8,843.21
5,542.00
-8,077.19
(7) 7,650
452
-7,568
(9)
- 89.99*
19.01
438.78
92.63
- 236.01
- 186.40
-
Monroe: Hand-operated
EXAMPLE: 484 124 = 360
1. Clear the machine and move the carriage to position one.
2. Depress the nonrepeat key.
3. Set the minuend 484 in the extreme right of the keyboard.
4. Add. The number 484 appears in the lower dials, the key-
board clears, and a number 1 appears in the upper dials to indicate
that one item has been added.
5. Set the subtrahend 124 in the extreme right of the keyboard
directly under the 484 in the lower dials.
6. Turn the operating crank toward the front of the machine
one complete revolution, and stop it at the top of the turn. (Tap
the minus bar, being sure to release it immediately.) The re-
mainder 360 appears in the lower dials, the keyboard clears, and a 0
appears in the first upper dial to indicate that the second item was
subtracted from the first.
* When 89.99 is subtracted first, 1001 preceded by a series of 9's appears in
the lower dials; a series of 9's appears in the upper dials. This indicates that a
negative number is in the machine, which is satisfactory. Continue with the
operations as indicated and record the final answer.
44
Machine Computation of Elementary Statistics
PROBLEMS: Subtraction. Check by adding subtrahend to remainder.
(1) 567
-489
(2)
(3)
(4) 1,275
185
4,678
- 1,987
780
(5)
(8) 3,655.09
3.90
837.54
8,843.21
5,542.00
-8,077.19
1,090
585
7,568
- 1,298
PROBLEMS: Addition and subtraction of more than two numbers.
Check by redoing.
(6)
(7)
.
(9)
Friden
G
C. COMPLEMENTS
Problems often arise in statistics where the result of a sum of
numbers is negative. The most effective method of handling this
situation on the calculating machine is to make use of complements.
7,650
452
-7,568
867
28
89.99*
19.01
438.78
92.63
- 236.01
- 186.40
C
EXAMPLE: 372 + 149 – 697
A series of numbers such as this should be treated as the sum of
signed numbers,† and all operations carried through to the end as
in the example.
- 176
* When 89.99 is subtracted first, a bell rings and 1001 preceded by a series of
9's appears in the lower dials; a series of 9's appears in the upper dials. This
indicates that a negative number is in the machine, which is satisfactory.
Continue with the operations as indicated and record the final answer.
† Numbers preceded by a plus or a minus sign.
Addition and Subtraction
45
1. Clear the machine and move the carriage to position one.
2. Depress ADD key.
3. Add 372 into the machine.
4. Add 149. The sum 521 appears in the upper dials.
5. Subtract 697. The upper dials show 824 on the right, pre-
ceded by several 9's. The 9's indicate that the result of the sub-
traction is a negative number. The answer is negative, and is the
complement of 824.
6. There are two methods of finding the complement.
(a) (1) Pull ADD key up.
(2) Depress in column one that figure which when added
to 4 (the figure in the first upper dial) will equal 10.
In all other columns after the first, depress that key
whose value together with the number in the correspond-
ing upper dial will equal 9. Thus, 6 should be set in
column one (4 + 6 10); 7 in column two (2 + 7 = 9);
and 1 in column three (8+ 1 = 9). Zero goes in all other
columns (0 + 9 = 9).
=
(3) Add. If the proper figures have been set in the key-
board, the upper dials will read O's. The answer (the
complement) is in the keyboard; it is negative, 176.
Add the amount into the machine to check the keyboard
reading. Remember it is a negative number, and when
written is preceded by a minus sign.
(b) (1) Pull ADD key up.
(2) On the keyboard enter the numbers in the first three
upper dials, preceded by two 9's, 99824.
(3) Subtract twice. The number 176 preceded by two
0's, an 8, and several 9's appears in the upper dials.
(4) The answer (complement) is -176. The numbers
to the left of the O's in the upper dials may be disre-
garded. Remember the answer is a negative number, and
when written is preceded by a minus sign.
PROBLEMS: Complements. Check by the following method, adding
positive and negative numbers separately, changing signs, taking
the difference of the sums, and giving the difference a sign opposite
to that of the larger value. For instance, for the EXAMPLE, (a)
372 +149521; (b) give this a negative sign -521; (c) change
46
Machine Computation of Elementary Statistics
-697 to +697; (d) take the difference 697-521 = 176; (e) give
the difference the sign opposite to that of the larger value 176.
(2)
(1)
687
-543
-267
(4) -389.00
438.78
111.22
0.05
34.66
100.00
1.96
-956.98
877.11
9.00
597.06
-986.45
279
-281
219
- 538
—
(3)
(5) 9,876.490
820.950
9.006
11.620
1.033
-9,740.002
86.987
0.080
972.290
2.398
Marchant
-4,290.009
20.090
1,478
263
-1,581
C
EXAMPLE: 372 + 149 697
-176
A series of numbers such as this should be treated as the sum of
signed numbers, and all operations carried through to the end as
in the example.
1. Clear the machine and move the carriage to position one.
2. Add 372 into the machine.
3. Add 149. The sum 521 appears in the middle dials.
4. Subtract 697. The middle dials read 9's in all but the first
three on the right, which read 824. The 9's indicate that the result
of the subtraction is a negative number. The answer is negative,
and is the complement of 824.
5. There are two methods of finding the complement.
(a) (1) Depress in column one that figure which when added
to 4 (the figure in the first middle dial) will equal 10.
In all other columns after the first, depress that key
whose value together with the number in the correspond-
Addition and Subtraction
47
(1)
ing middle dial will equal 9. Thus, 6 should be set in
column one (4 + 6 = 10); 7 in column two (2 + 7 = 9);
and 1 in column three (8 + 1 = 9). Zero goes in all
other columns (0 +99).
(2) Depress the 1 key in the multiplier keyboard. (In D
models, depress the multiplier bar.) If the proper figures
have been set in the keyboard, the middle dials will read
all O's. The answer (the complement) is in the keyboard;
it is negative, -176. Depress the 1 key in the multiplier
keyboard a second time (or the multiplier bar in D
models); 176 appears in the middle dials as a check on
the keyboard reading. Remember it is a negative num-
ber, and when written is preceded by a minus sign.
(b) (1) On the keyboard enter the numbers in the first three
middle dials, preceded by two 9's in columns four
and five.
687
-543
-267
(2) Depress the REVERSE bar, then the 1 key in the
multiplier keyboard. (In D models, depress the SHORT-
CUT bar.) Five O's preceded by several 9's appear in the
middle dials.
PROBLEMS: Complements. Check by the following method, adding
positive and negative numbers separately, changing signs, taking
the difference of the sums, and giving the difference a sign opposite
to that of the larger value. For instance, for the EXAMPLE, (a)
372 + 149 = 521; (b) give this a negative sign -521; (c) change
-697 to 697; (d) take the difference 697 521176; (e) give
the difference the sign opposite to that of the larger value — 176.
(2)
(3) Repeat step two. The number 176 preceded by two
O's, an 8, and several 9's appears in the middle dials.
(4) The answer (complement) is -176. The numbers to
the left of the O's in the middle dials may be disregarded.
Remember the answer is a negative number, and when
written is preceded by a minus sign.
G
279
-281
219
-538
(3)
1,478
263
- 1,581
48
Machine Computation of Elementary Statistics
(5)
9,876.490
820.950
(4) -389.00
438.78
111.22
0.05
34.66
100.00
1.96
-956.98
877.11
9.00
597.06
-986.45
9.006
11.620
1.033
EXAMPLE: 372 + 149 — 697
-9,740.002
86.987
0.080
972.290
2.398
-4,290.009
20.090
Monroe: Automatic and Semiautomatic.
697 = - 176
A series of numbers such as this should be treated as the sum of
signed numbers, and all operations carried through to the end as
in the example.
1. Clear the machine and move the carriage to position one.
2. Depress nonrepeat key.
3. Add 372 into the machine.
4. Add 149. The sum 521 appears in the lower dials.
5. Subtract 697. The lower dials read 824 on the right, preceded
by several 9's. The 9's indicate that the result of the subtraction
is a negative number. The answer is negative, and is the comple-
ment of 824.
6. There are two methods of finding the complement.
(a) (1) Depress repeat key.
(2) Depress in column one that figure which when added.
to 4 (the figure in the first lower dial) will equal 10.
In all other columns after the first, depress that key
whose value together with the number in the corre-
sponding lower dial will equal 9. Thus, 6 should be set
in column one (4 + 6 = 10); 7 in column two (2 + 7 =
9); and 1 in column three (8+ 1 = 9). Zero goes in all
other columns (0+ 9 = 9).
(3) Add. If the proper figures have been set in the key-
?
Addition and Subtraction
49
(1)
board, the lower dials will read all O's. The answer (the
complement) is in the keyboard; it is negative, –176.
Add the amount into the machine to check the keyboard
reading. Remember it is a negative number and when
written is preceded by a minus sign.
(b) (1) Depress repeat key.
(2) On the keyboard enter the numbers in the first three
upper dials, preceded by two 9's, 99824.
(3) Subtract twice. The number 176 preceded by two
O's, an 8, and several 9's appears in the lower dials.
(4) The answer (complement) is 176. The numbers
to the left of the O's in the lower dials may be disre-
garded. Remember the answer is a negative number, and
when written is preceded by a minus sign.
PROBLEMS: Complements. Check by the following method, adding
positive and negative numbers separately, changing signs, taking
the difference of the sums, and giving the difference a sign opposite
to that of the larger value. For instance, for the EXAMPLE, (a)
372 + 149 521; (b) give this a negative sign -521; (c) change
697 to +697; (d) take the difference 697 521 176; (e) give
the difference the sign opposite to that of the larger value - 176.
=
=
687
-543
-267
(4) -389.00
438.78
111.22
0.05
34.66
100.00
1.96
-956.98
877.11
9.00
597.06
-986.45
(2)
279
- 281
219
-538
(5)
-
M
(3) 1,478
263
-1,581
9,876.490
820.950
9.006
11.620
1.033
-9,740.002
86.987
0.080
972.290
2.398
-4,290.009
20.090
50
Machine Computation of Elementary Statistics
Monroe: Hand-operated
-176
EXAMPLE: 372 + 149 - 697
A series of numbers such as this should be treated as the sum of
signed numbers, and all operations carried through to the end as
in the example.
=
1. Clear the machine and move the carriage to position one.
2. Depress nonrepeat key.
3. Add 372 into the machine.
4. Add 149. The sum 521 appears in the lower dials.
5. Subtract 697. A bell will ring to indicate that a negative
number is in the machine. The lower dials read 824 on the right
preceded by several 9's. The 9's indicate that the result of the sub-
traction is a negative number. The answer is negative, and is the
complement of 824.
+
6. There are two methods of finding the complement.
(a) (1) Depress repeat key.
=
(2) Depress in column one that figure which when added
to 4 (the figure in the first lower dial) will equal 10. In
all other columns after the first, depress that key whose
value together with the number in the corresponding
lower dial will equal 9. Thus, 6 should be set in column
one (4+ 6 10); 7 in column two (2 +7 9); and
1 in column three (8+ 1 = 9). Zero goes in all other
columns (0 + 9 = 9).
(3) Add. If the proper figures have been set in the key-
board, a bell will ring to indicate that the negative num-
ber has been taken out of the machine. The lower dials
will read all O's. The answer (the complement) is in the
keyboard; it is negative, -176. Add the amount into
the machine and read from the lower dials to check the
keyboard reading. Remember, it is a negative number,
and when written is preceded by a minus sign.
(b) (1) Depress repeat key.
(2) On the keyboard, enter the numbers in the first
three upper dials, preceded by two 9's, 99824.
(3) Subtract twice. The number 176, preceded by two
O's, an 8, and several 9's appears in the lower dials.
Addition and Subtraction
51
(1) 687
-543
-267
PROBLEMS: Complements. Check by the following method, adding
positive and negative numbers separately, changing signs, taking
the difference of the sums, and giving the difference a sign opposite
to that of the larger value. For instance, for the EXAMPLE, (a)
372 + 149 = 521; (b) give this a negative sign -521; (c) change
-697 to +697; (d) take the difference 697 — 521 176; (e) give
the difference the sign opposite to that of the larger value -176.
=
(2)
(4) The answer (complement) is -176. The numbers
to the left of the O's in the lower dials may be dis-
regarded. Remember that the answer is a negative
number, and when written is preceded by a minus sign.
(4) -389.00
438.78
111.22
Kj
S
0.05
34.66
100.00
1.96
-956.98
877.11
9.00
597.06
-986.45
279
-281
219
-538
(3) 1,478
263
-1,581
(5) 9,876.490
820.950
9.006
11.620
1.033
-9,740.002
86.987
0.080
972.290
2.398
-4,290.009
20.090
CHAPTER 4
MULTIPLICATION
The process of multiplication is based on the principle of suc-
cessive addition. In hand-operated and semiautomatic models, the
shifting of the carriage from the units position to the tens, from the
tens to the hundreds, and so forth, must be done by hand. In
automatic machines, this is a fully automatic process.
A. MULTIPLICATION OF WHOLE NUMBERS
Friden
EXAMPLES: (a) 246 X 38 = 9,348
×
(b) 6,789 X 5,309 36,042,801
×
=
1. Depress ADD key.
2. Set 246 in the right of the keyboard.
3. Enter 38 in the multiplier unit by touching the keys of the
unit in the order in which they are read, i.e., from left to right,
3 and 8. Verify the entry by reading the proof register above the
multiplier keys.
4. Touch the multiply (MULT) key. The carriage automatically
clears and returns to position one; the multiplication takes place,
the keyboard and multiplier clear, and the product 9,348 appears
in the upper dials. The multiplier 38 appears in the lower dials.
5. Without clearing the dials or returning the carriage to posi-
tion, set the second multiplicand 6,789 on the right of the keyboard
and enter 5,309 in the multiplier unit. Note that the 0 must be
entered.
6. Touch the MULT key. The carriage automatically clears and
returns to position one; the multiplication takes place, the key-
board and multiplier clear, and the product 36,042,801 appears in
the upper dials.
52
Multiplication
53
7. In the event of an error in entering a number in the multi-
plier, depress the multiplier correction (MULT CORR) key. This
clears the multiplier without affecting the dial entries. If the add
key is down, the keyboard also will clear; if up, the keyboard is
unaffected by the MULT CORR operation.
PROBLEMS: Multiplication of whole numbers. Check by multiplying
again, reversing the multipliers and multiplicands.
(1) 234 X 506 =
(2) 1,290 × 1,060
(3) 607 X 432 =
(4) 207 X 891 =
(5) 2,199 X 1,999
(6) 6,219 × 4,866
(7) 5,682 × 5,682
(8) 25 × 43,602 =
(9) 290 × 893,992
(10) 7,777 × 63,900
=
}
=
=
Marchant
EXAMPLES: (a) 246 X 38 = 9,348
(b) 6,789 × 5,309 = 36,042,801
Multiplication may be done by entering multipliers from either
left or right.
(a) Entering multipliers from the left
1. Depress upper green shift control key.
2. Set 246 in keyboard columns three to one.
3. Since there are two digits in the multiplier 38, press selective
tab key two so that the carriage position indicator points to upper
dial two. (In D models use lower carriage shift bar to move carriage
to this position.)
4. On the multiplier keyboard, touch keys 3 and 8 in that order.
The multiplication takes place; the product 9,348 appears in the
middle dials; the multiplier 38 appears in the upper dials; and the
multiplicand 246 remains in the keyboard and may be checked by
reading the keyboard dials.
[In D models touch multiplier (X) bar to build up 3 in upper
dial two; touch upper carriage shift key to move carriage one
place to the left; build up 8 in upper dial one.]
54
Machine Computation of Elementary Statistics
5. To prepare for the next multiplication operation, depress
the clearance keys simultaneously. Note that the carriage returns
to position two, since selective tab key two is still depressed.
6. Set 6,789 in keyboard columns four to one.
7. Since there are four digits in the multiplier, press selective
tab key four so that the carriage position indicator is above upper
dial four. (In D models use lower carriage shift bar to move
carriage to this position.)
8. On the multiplier keyboard, touch keys 5, 3, 0, and 9 in that
order. The product 36,042,801 appears in the middle dials; the
multiplier 5,309 in the upper dials; and the multiplicand 6,789
remains in the keyboard and may be checked by reading the key-
board dials.
(In D models touch X bar to build up 5 in upper dial four;
tap upper carriage shift key to move the carriage one place to the
left; build up 3 in upper dial three; touch upper carriage shift key
to move carriage two places to the left so that the carriage position
indicator points to upper dial one; build up 9 in upper dial one.)
(b) Entering multipliers from the right
1. Depress lower green shift control key; put carriage in posi-
tion one.
2. Put 246 in keyboard dials three to one.
3. On the multiplier keyboard, touch keys 8 and 3 in that order.
The multiplication takes place; the product 9,348 appears in the
middle dials; the multiplier 38 appears in the upper dials; the multi-
plicand 246 remains on the keyboard and may be checked by read-
ing the keyboard dials.
(In D models touch X bar to build up 8 in upper dial one;
tap lower carriage shift key to move carriage one place to the right;
build up 3 in upper dial two.)
4. Clear the machine.
S
5. Set 6,789 in keyboard columns four to one.
6. Be sure carriage is in position one.
7. On the multiplier keyboard, touch keys 9, 0, 3, and 5 in that
order. The product 36,042,801 appears in the middle dials; the
multiplier 5,309 in the upper dials; and the multiplicand 6,789
remains in the keyboard and may be checked by reading the key-
board dials.
Multiplication
55
(In D models touch X bar to build up 9 in upper dial one; tap
lower carriage shift key to move carriage two places to the right
so that carriage position indicator points to upper dial three;
build up 3 in upper dial three; touch lower carriage shift key to
move carriage one place to the right; build up 5 in upper dial four.)
PROBLEMS: Multiplication of whole numbers. Check by multiplying
again, reversing the multipliers and multiplicands.
(1) 234 × 506
(2) 1,290 × 1,060
(3) 607 × 432
(4) 207 × 891
(5) 2,199 × 1,999
(6) 6,219 × 4,866
(7) 5,682 × 5,682
(8) 25 × 43,602 =
(9) 290 × 893,992
(10) 7,777 × 63,900
Monroe: Automatic
=
EXAMPLES: (a) 246 × 38
(b) 6,789 × 5,309
9,348
36,042,801
1. Put change lever at X. On CAA models be sure tab stop and
transfer slide are at 0. Carriage is in position one.
2. Set the multiplier 38 in extreme right of keyboard.
3. Depress multiplier setup bar (on some models this is marked
X; on CAA, SETUP). The keyboard clears on some machines, not
on others. On CAA machines, the multiplier appears in the multi-
plier setup dials.
=
4. Clear the keyboard if using a machine without automatic
keyboard clearance. Set the multiplicand 246 in extreme right of
keyboard.
5. Depress multiply key (marked = on some machines, × on
CAA machines).
6. The multiplication takes place. The product 9,348 appears
in the right lower dials and the multiplier 38 in the upper dials.
Disregard the number in left upper dials.
7. Clear dials and keyboard. Be sure carriage is in position one.
Set the next multiplier 5,309 in extreme right of keyboard.
.
56
Machine Computation of Elementary Statistics
8. Depress multiplier setup bar. Clear keyboard if necessary.
9. Set the multiplicand 6,789 in extreme right of keyboard and
depress multiply key.
10. Read the product 36,042,801 in the lower dials and the
multiplier 5,309 in the upper dials.
PROBLEMS: Multiplication of whole numbers. Check by multiplying
again, reversing the multipliers and multiplicands.
(1) 234 × 506
(2) 1,290 X 1,060
(3) 607 × 432
(4) 207 × 891
(5) 2,199 X 1,999
(6) 6,219 × 4,866
(7) 5,682 × 5,682
(8) 25 × 43,602 =
(9) 290 × 893,992
(10) 7,777 X 63,900
=
Monroe: Semiautomatic
On semiautomatic machines, multiplication is performed as on
the electric hand-operated machines, by building up the multiplier
in the upper dials with the plus bar. (See directions on pages 56,
57.) Note that on the semiautomatic machines, the plus bar is
used in positive multiplication, the minus bar in negative multipli-
cation; and that the carriage is moved by depressing the carriage
shift keys.
Monroe: Semiautomatic and Hand-operated
(a) Multiplication from the right
EXAMPLES: (a) 246 X 38 9,348
×
=
(b) 6,789 × 5,309 = 36,042,801
1. Depress repeat key. Be sure carriage is in position one.
2. Set 246 (the multiplicand of the first example) in the right
of the keyboard.
3. The order of multiplication is the same as that used when
performing the operation with pencil and paper; i.e., multiply first
by the units digit of the multiplier, then by the tens, and so forth.
Multiplication
57
Here the units digit is 8; turn the operating crank away from the
front of the machine eight times. Do not count the turns of the
crank; watch the first upper dial until the 8 is reached. (Hold the
plus bar down until 8 appears in the first upper dial.)
4. The 8 in the first upper dial indicates that the multiplicand
in the keyboard has been added eight times, or has been multiplied
by 8. The product of this multiplication 1968 appears in the lower
dials.
5. The tens digit of the multiplier is 3. Move the carriage one
place to the right by turning the carriage shift lever once. On most
hand-operated machines, turning of the carriage shift lever can be
done with the thumb of the left hand, resting the other fingers on
the machine and/or the table to the left of the lever.
6. Turn the operating crank away from the front of the machine
three times. (Hold the plus bar down until 3 appears in the second
upper dial.) The 3 in the second upper dial indicates that the multi-
plicand in the keyboard has been added thirty times, or multiplied
by 30. The 38 in the second and first upper dials indicate that the
number in the keyboard has been added thirty-eight times, or
multiplied by 38. The product 9,348 appears in the lower dials.
7. Clear the machine, return the carriage to position one.
8. Enter the new multiplicand 6,789 in the extreme right of the
keyboard.
9. The units digit of the multiplier is 9; therefore turn the op-
erating crank away from the front of the machine nine times.
(Hold plus bar down until 9 appears in the first upper dial.) Move
carriage one place to the right.
10. The tens figure is 0. Move the carriage another place to the
right, so that the carriage position indicator arrow points to the
third lower dial. Multiply by the hundreds digit 3. Move the
carriage one place to the right so that the carriage position indi-
cator points to the fourth lower dial.
11. Multiply by the thousands digit 5.
12. The multiplier 5309 appears in the upper dials; the product
36,042,801 appears in the lower dials.
PROBLEMS: Multiplication of whole numbers. Check by multiplying
again, reversing the multipliers and multiplicands. Be sure the
machine is cleared and the carriage returned to position one before
each new multiplication. If the multiplicand is added too many
58
Machine Computation of Elementary Statistics
*
times, simply subtract it in the same position until the desired
digit appears in the upper dials.
(1) 234 X 506 =
(2) 1,290 X 1,060
(3) 607 × 432
(4) 207 X 891
(5) 2,199 × 1,999
(6) 6,219 x 4,866
(7) 5,682 X 5,682
(8) 25 × 43,602 =
(9) 290 X 893,992 =
(10) 7,777 × 63,900 ·
(b) Multiplication from the left
EXAMPLES: (a) 246 X 38 = 9,348
(b) 6,789 × 5,309 = 36,042,801
Some operators prefer to perform the multiplication as it is
written, from left to right.
1. Depress repeat key.
2. Set the multiplicand of the first example (246) in the ex-
treme right of the keyboard.
3. Inspect the multiplier to determine the number of digits it
contains. Here there are two (38). Move the carriage to the right
so that the carriage position indicator points to the second lower
dial.
4. Multiply by 3; move carriage one place to the left; multiply
by 8.
5. The multiplier 38 appears in the upper dials; the product
9,348 in the lower dials.
6. Clear the machine.
7. Enter the second multiplicand 6,789 in the extreme right of
the keyboard.
8. Inspect the multiplier to determine the number of digits it
contains. Here there are four (5309). Move the carriage to the
right so that the carriage position indicator points to the fourth
lower dial; this may be done either with the left hand by turning the
carriage shift lever, or with the right by grasping the knob on the
Multiplication
59
right end of the carriage, lifting the carriage slightly, and sliding
it to the desired position.
9. Multiply by 5; move carriage one place to the left; multiply
by 3; move carriage two places to the left; multiply by 9.
10. The multiplier 5,309 appears in the upper dials; the product
36,042,801 appears in the lower dials.
PROBLEMS: Redo problems 1 to 5 on page 58, multiplying from
the left.
(c) Multiplication by multipliers with digits 6, 7, 8, 9
When a multiplier is a number containing 6, 7, 8, or 9 (e.g., 39,
198, 997), the multiplication may be accomplished by a combined
use of addition and subtraction. This reduces considerably the
number of revolutions of the crank required by the regular method.
=
There is one word of caution, however, which should be said.
On machines which do not have the carry-over feature in the upper
dials (e.g., the Educator), subtraction from zero is indicated by
red numbers in the upper dials. For example, 200 2 198 on
machines with carry-over; on machines which do not have this
feature 200 – 2 202, with the subtracted figure showing in red
in the first upper dial. On certain K models, the upper dials are
split in the middle; the left bank does not have the carry-over fea-
ture, the right bank does. In this case, the left bank would show
202, with the 2 in red, and the right bank 198 in black.
—
=
EXAMPLE: 2,146 X 198
424,908
1. Depress repeat key; be sure carriage is in position one.
2. Enter the multiplicand 2,146 in the extreme right of the key-
board.
3. The multiplier 198 is equal to 200 minus 2. Therefore shift
the carriage two places to the right and multiply by 200.
4. Return the carriage to position one, and subtract twice.
The product 424,908 appears in the lower dials. The multiplier
appears in the upper dials as follows: left bank of split upper dials
and Educator as 202, with the 2 in red; right bank of split upper
dials as 198 in black.
5. This method involves only four addition-subtraction opera-
tions, whereas the long way would have involved eighteen addi-
tion operations.
C
60
Machine Computation of Elementary Statistics
PROBLEMS: Multiplication by multipliers with digits 6, 7, 8, 9
(1) 234 X 88 = 234 X (100
× — 12)
(2) 4,629 × 99 = 4,629 X (100 — 1)
(3) 724 X 688 724 X (1,000 312)
×
–
(4) 879 X 987 = 879 X (1,000 — 13)
(5) 57,920 × 396 = 57,920 X (400 — 4)
=
B. MULTIPLICATION OF NUMBERS WITH DECIMALS
The number of decimal places in the product of two numbers is
equal to the sum of the number of decimal places in the multiplier
and the number of places in the multiplicand. For example, in the
problem 438.26 X 207.39, the first number, the multiplicand, con-
tains two decimal places, the second number, the multiplier, also
two. Therefore, their product will contain two plus two, or four,
places: 438.26 × 207.39 = 90,890.7414
Friden
(a) Multiplication of numbers with decimals
EXAMPLE: 438.26 × 207.39 = 90,890.7414
1. Inspect the multiplier and determine that there are two deci-
mal places; set the pointer beneath the result register of the multi-
plier over 2 and set the decimal marker for the lower dials between
dials two and three.
2. Inspect the multiplicand and determine that there are two
decimal places; turn up the keyboard decimal between columns
two and three.
3. Since the product will have two plus two, or four, decimal
places, set the upper dial decimal between dials four and five.
4. Perform the multiplication, being sure to set the multi-
plicand in the keyboard around the decimal point. Read the
product 90,890.7414 correctly pointed off in the upper dials.
PROBLEMS: Multiplication of numbers with decimals. Check by
multiplying again, reversing the multipliers and multiplicands.
Remember that the number of decimal places differs with each
problem; and that the number of places in the upper dials (product)
equals the number of decimal places in the lower dials (multiplier)
plus the number in the keyboard (multiplicand).
4
Multiplication
61
(1) 476.89 X 832
(2) 324.67 X 4.165
(3) 45.326 X 86.1
(4) 61.0 X 307
(5) 90.2 X 0.663
=
(6) 7.913 × 78.92
(7) 0.7580 X 0.0411
(8) 0.0164 × 0.0061
(9) 93,366 × 0.0048 =
(10) 458.459 X 0.000394
(11) (0.25)2
(12) (0.40)²
(13) (0.04)² =
(14) (0.00396)2
(b) Preset decimals
EXAMPLES: 35.24 X 63.511
320.6 X 1.6124
0.032 × 121.17
=
2,238.12764
516.93544
3.87744
1. When a series of multiplications of numbers involving deci-
mals is to be done, it is possible to preset the decimal points, and
perform all the multiplications around these preset decimals.
2. Inspect the series to determine the largest number of places
needed in the multiplicand. Here it is three (0.032). Turn up the
keyboard decimal between columns three and four.
3. Inspect the multipliers to determine the largest number of
decimal places needed. Here it is four (1.6124). Set the lower dial
decimal between dials four and five, and the multiplier unit decimal
over digit 4.
4. Assume that all products will have seven (four plus three)
decimal places, and set the upper dial decimal between dials seven
and eight.
5. Set 35.24 in the keyboard around the decimal point. To do
this, 35240 must be entered in columns five through one, giving
the maximum number of decimal places.
6. Read the multiplier as 63.5110, giving it four (the maximum
number) decimal places. Enter in the multiplier unit, noting that
the decimal point falls correctly between the 3 and the 5. Multiply.
62
Machine Computation of Elementary Statistics
7. The product 2,238.1276400 appears in the upper dials cor-
rectly pointed off. In recording, the two final O's to the right of the
decimal should be dropped. The multiplier 63.5110 appears in the
lower dials correctly pointed off.
8. Set the next multiplicand 320.6 around the decimal in the
keyboard (320600 in columns six through one). Set the multiplier
1.6124 in the multiplier unit. Note that since this has the maximum
number of decimal places, no zeros need be added.
9. Multiply. The product 516.9354400 appears in the upper
dials correctly pointed off; the multiplier 1.6124 appears in the
lower dials correctly pointed off.
10. Set the next multiplicand 0.032 in the keyboard around the
decimal, i.e., depress 3 in column two, and 2 in column one.
11. Enter the multiplier 121.1700 in the multiplier unit. Check
the proof register to be sure the decimal falls between the two
1's.
12. Multiply. The product 3.8774400 appears in the upper dials
correctly pointed off; the multiplier 121.1700 appears in the lower
dials correctly pointed off.
PROBLEMS: Multiplication with preset decimals. Redo problems
1 to 10 on page 61.
Marchant
Certain Marchant models are equipped with two sets of decimals,
white and orange. The white decimals parallel those on other makes
of machines; the orange decimals are for a special set of operations
which will be described on pages 65, 66.
(a) Multiplication of numbers with decimals; white decimals
EXAMPLE: 438.26 X 207.39
90,890.7414
1. Inspect the multiplier and determine that there are two
decimal places; set the upper dial decimal between dials two and
three.
2. Inspect the multiplicand and determine that there are two
decimal places; turn up the keyboard decimal between columns two
and three.
3. Since the product will have two plus two, or four, decimal
places, set the middle dial decimal between dials four and five.
Multiplication
63
4. Perform the multiplication, being sure to set the multi-
plicand in the keyboard around the decimal point. The product
90,890.7414 appears in the middle dials correctly pointed off.
5. It is possible to determine automatically the number of
decimal places in the product.*
(a) Set upper dial decimal between dials two and three (two
places in the multiplier); turn up keyboard decimal be-
tween columns two and three (two places in the multi-
plicand).
(b) Perform the multiplication.
(c) Position the carriage so that the red carriage position
indicator points to the first dial immediately to the left
of the decimal in the upper dials (dial three); set the
middle dial decimal directly above the keyboard dial
decimal. Note that it comes between middle dials four
and five, pointing off the product correctly (four decimal
places).
(1) 476.89 X 832
(2) 324.67 × 4.165 =
(3) 45.326 × 86.1
(4) 61.0 X 307
(5) 90.2 × 0.663
(6) 7.913 × 78.92
(7) 0.7580 X 0.0411
(8) 0.0164 X 0.0061 =
(9) 93,366 × 0.0048 =
(10) 458.459 × 0.000394
(11) (0.25)²
(12) (0.40)2
(13) (0.04)² =
(14) (0.00396)2
* See reference 4, p. 20.
PROBLEMS: Multiplication of numbers with decimals. Check by
multiplying again, reversing the multipliers and multiplicands.
Remember that the number of decimal places differs with each
problem, and that the number of decimals in the middle dials
(product) equals the number of decimals in the upper dials (multi-
plier) plus the number in the keyboard (multiplicand).
=
64
Machine Computation of Elementary Statistics
(b) Preset decimals; white decimals
EXAMPLES: (a) 35.24 X 63.511
(b) 320.6 X 1.6124
0.032 X 121.17
(c)
2,238.12764
516.93544
3.87744
1. When a series of multiplications of numbers involving deci-
mals is to be done, it is possible to preset the decimal points, and
perform all the multiplications around these preset decimals.
2. Inspect the series to determine the largest number of places
needed in the multiplicand. Here it is three (0.032). Turn up the
keyboard decimal between columns three and four.
3. Inspect the multipliers to determine the largest number of
decimal places needed. Here it is four (1.6124). Set the upper dial
decimal between dials four and five.
4. Assume that all products will have seven (four plus three)
decimal places, and set the middle dial decimal between dials
seven and eight.
5. Set 35.24 in the keyboard around the decimal point. To do
this, 35240 must be entered in columns five through one, giving the
maximum number of decimal places.
6. Read the multiplier as 635110, giving it four (the maximum)
decimal places. This means that there are six digits in the multi-
plier; touch selective tab key six and multiply. Or, if multiplying
from the right, enter a O in the multiplier keyboard first, then 1, 1,
5, 3, and 6.
(In D models use lower carriage shift bar to put carriage in
position six if multiplying from the left; if multiplying from the
right, touch upper carriage shift bar to put carriage in position two,
and build up the multiplier from that position.)
7. The product 2,238.1276400 appears in the middle dials cor-
rectly pointed off. In recording, the two final O's to the right of the
decimal should be dropped. The multiplier 63.5110 appears in the
upper dials correctly pointed off.
8. Clear the machine.
9. Set the next multiplicand 320.6 around the decimal in the
keyboard (320600 in columns six through one).
10. Since the multiplier 1.6124 has the maximum number of
decimal places, it is not necessary to add any zeros to it. Touch
selective tab key five and multiply.
11. The product 516.935440 appears in the middle dials correctly
Multiplication
65
pointed off; the multiplier 1.6124 appears in the lower dials cor-
rectly pointed off.
12. Clear the machine.
13. Set the next multiplicand 0.032 in the keyboard around the
decimal, i.e., depress 3 in column two, and 2 in column one.
14. Read the multiplier as 121.1700; touch selective tab key
seven, and multiply.
15. The product 3.8774400 appears in the middle dials correctly
pointed off; the multiplier 121.1700 appears in the lower dials cor-
rectly pointed off.
PROBLEMS: Multiplication with preset decimals. Redo problems 1 to
10 on page 63.
(c) Preset decimals; orange decimals*
Certain models have supplementary orange decimals for the
middle and upper dials. These remain fixed regardless of the posi-
tion of the carriage, and are useful in a series of single multiplica-
tions with multipliers and multiplicands containing approximately
the same number of decimal places. They are not designed for
accumulative multiplication (see pages 75 to 79).
EXAMPLES: (a) 35.24 X 63.511
(b) 320.6 X 1.6124
(c) 0.032 × 121.17
-
=
2,238.12764
516.93544
3.87744
1. Inspect the series to determine the largest number of places
needed in the multiplicand. Here it is three (0.032). Turn up the
keyboard decimal between columns three and four.
2. Inspect the multipliers to determine the largest number of
decimal places needed. Here it is four (1.6124). Set the upper dial
orange decimal window over 4.
3. Depress selective tab key ten.
4. Set the lower dial orange decimal so that the number on its
arm, corresponding to the number in the upper dial orange decimal
window (here 4), is directly above the keyboard decimal point.
(If the arm has only 0 to 3 on it, slide the 3 one place to the left
of the keyboard decimal, so that if a 4 were there, it would be over
the keyboard decimal.)
* See reference 4, p. 21-22; and personal communication from Miss Ann
Curran of the New York office of the Marchant Calculating Machine Company.
66
Machine Computation of Elementary Statistics
5. Set the first multiplicand 35.24 in the keyboard around the
decimal point.
6. Multiply by 635110. Note that it is necessary to allow for
the largest number of multiplier decimal places. Check by being
sure that the upper dial orange decimal points off the multiplier
correctly. Instead of depressing the multiplier keyboard 0, it is
possible to adjust the carriage to the correct position by touching
the carriage shift key. The product 2,238.12764 appears in the
middle dials, correctly pointed off by the orange decimal.
7. Clear the machine. Be sure the carriage is moved all the way
to the right.
8. Set the second multiplicand 320.6 in the keyboard around the
decimal.
9. Multiply by 1.6124. Since this multiplier contains the maxi-
mum number of decimal places, it is not necessary to adjust the
carriage to point off the multiplier correctly in the upper dials. The
product 516.93544 appears in the middle dials correctly pointed off
by the middle dial orange decimal.
10. Clear the machine.
11. Set the third multiplicand 0.032 in the keyboard around the
decimal.
12. Multiply by 121.17. It is necessary to adjust the carriage so
that the multiplier is correctly pointed off in the upper dials by
the upper dial orange decimal.
13. The product 3.87744 appears in the middle dials correctly
pointed off by the middle dial orange decimal.
PROBLEMS: Multiplication with preset decimals. Redo problems 1 to
5 on page 63.
Monroe: Automatic
(a) Multiplication of numbers with decimals
EXAMPLE: 438.26 X 207.39 = 90,890.7414
1. Carriage is in position one; change lever at X.
2. Inspect the multiplier and determine that there are two
decimal places; set the right upper dial decimal marker between
upper dials two and three.
3. Inspect the multiplicand and determine that there are two
decimal places; set the keyboard decimal marker between columns
two and three.
Multiplication
67
4. Since the product will have two plus two, or four, decimal
places, set the lower dial decimal marker between dials four and
five.
5. Perform the multiplication, being sure to set the multi-
plicand in the keyboard around the decimal point. Read the
product 90,890.7414 correctly pointed off in the upper dials.
PROBLEMS: Multiplication of numbers with decimals. Check by
multiplying again, reversing the multipliers and multiplicands.
Remember that the number of decimal places differs with each
problem, and that the number of places in the lower dials (product)
equals the number of decimal places in the upper dials (multiplier)
plus the number in the keyboard (multiplicand).
(1) 476.89 × 832
(2) 324.67 X 4.165
(3) 45.326 × 86.1
(4) 61.0 × 307
(5) 90.2 × 0.663
(6) 7.913 × 78.92
(7) 0.7580 X 0.0411
(8) 0.0164 × 0.0061
(9) 93,366 × 0.0048
(10) 458.459 X 0.000394
(11) (0.25)² =
(12) (0.40)2
2
(13) (0.04)² =
(14) (0.00396)²
=
=
(b) Preset decimals
EXAMPLES: (a) 35.24 X 63.511
(b) 320.6 X 1.6124
(c) 0.032 X 121.17
= 2,238.12764
516.93544
3.87744
1. When a series of multiplications of numbers involving deci-
mals is to be done, it is possible to preset the decimal points and
perform all the multiplications around these preset decimals.
2. Inspect the series to determine the largest number of places
needed in the multiplicand. Here it is three (0.032). Turn up the
keyboard decimal between columns three and four.
3. Inspect the multipliers to determine the largest number of
68
Machine Computation of Elementary Statistics
decimal places needed. Here it is four (1.6124). Set the upper dial
decimal marker between dials four and five.
4. Assume that all products will have seven (four plus three)
decimal places, and set the lower dial decimal marker between
dials seven and eight.
5. Read the first multiplier as 64.5110, giving it four (the maxi-
mum number) decimal places. Disregard decimal marker in the
keyboard. Set up.
6. Set the first multiplicand 35.24 in the keyboard around the
decimal point. To do this, 35240 must be entered in columns five
through one, giving the maximum number of decimal places.
7. Multiply. The product 2,238.1276400 appears in the lower
dials correctly pointed off. In recording, the two final O's to the
right of the decimal should be dropped. The multiplier 63.5110
appears in the upper dials correctly pointed off.
8. Clear the machine and set up the next multiplier 1.6124.
Clear keyboard and set the multiplicand 320.600 around the deci-
mal in the keyboard.
9. Multiply. The product 516.9354400 appears in the lower
dials correctly pointed off; the multiplier 1.6124 appears in the
upper dials correctly pointed off.
10. Clear machine and set up the next multiplier 121.1700. Set
the multiplicand 0.032 in keyboard columns two and one.
11. Multiply. The product 3.8774400 appears in the lower dials
correctly pointed off; the multiplier 121.1700 appears in the upper
dials correctly pointed off.
PROBLEMS: Multiplication with preset decimals. Redo problems 1 to
10 on page 67.
;
Monroe: Semiautomatic and Hand-operated
(a) Multiplication of numbers with decimals
EXAMPLE: 438.26 X 207.39 90,890.7414
1. Inspect the multiplier and determine that there are two
decimal places; set the upper dial decimal marker between upper
dials two and three.
=
2. Inspect the multiplicand and determine that there are two
decimal places; turn up the keyboard decimal marker between
columns two and three.
Multiplication
69
3. Since the product will have two plus two, or four, decimal
places, set the lower dial decimal marker between lower dials four
and five.
4. Perform the multiplication, being sure to set the multi-
plicand in the keyboard around the decimal point. Read the product
90,890.7414 correctly pointed off in the lower dials.
PROBLEMS: Multiplication of numbers with decimals. Check by
multiplying again, reversing the multipliers and multiplicands.
Remember that the number of decimal places differs with each
problem, and that the number of places in the lower dials (product)
equals the number of decimal places in the upper dials (multiplier)
plus the number in the keyboard (multiplicand).
(1) 476.89 × 832
(2) 324.67 × 4.165
(3) 45.326 X 86.1
(4) 61.0 X 307
(5) 90.2 X 0.663 =
(6) 7.913 × 78.92
(7) 0.7580 X 0.0411
(8) 0.0164 × 0.0061
(9) 93,366 × 0.0048
(10) 458.459 × 0.000394
(11) (0.25)2
(12) (0.40)²
(13) (0.04)² =
(14) (0.00396)² =
=
(b) Preset decimals
EXAMPLES: (a) 35.24 × 63.511
(b) 320.6 X 1.6124
(c) 0.032 X 21.17
=
2,238.12764
516.93544
0.67744
1. When a series of multiplications of numbers involving deci-
mals is to be done, it is possible to preset the decimal points, and
perform all the multiplications around these preset decimals.
2. Inspect the series to determine the largest number of places
needed in the multiplicand. Here it is three (0.032). Turn up the
keyboard decimal marker between columns three and four.
3. Inspect the multipliers to determine the largest number of
70
Machine Computation of Elementary Statistics
decimal places needed. Here it is four (1.6124). Set the upper dial
decimal marker between upper dials four and five.
4. Assume that all products will have seven (four plus three)
decimal places, and set the lower dial decimal marker between lower
dials seven and eight.
5. Set 35.24 in the keyboard around the decimal point. To do
this, 35240 must be entered in columns five through one, giving the
maximum number of decimal places.
6. Read the multiplier as 63.5110, giving it four (the maxi-
mum number) decimal places. Multiply.
7. The product 2,238.1276400 appears in the lower dials cor-
rectly pointed off. In recording, the two final 0's to the right of the
decimal should be dropped. The multiplier 63.5110 appears in the
upper dials correctly pointed off.
8. Clear the machine, return carriage to position one. Set the
next multiplicand 320.6 around the decimal in the keyboard
(320600 in columns six through one). Multiply by 1.6124.
9. The product 516.9354400 appears in the lower dials correctly
pointed off; the multiplier 1.6124 appears in the upper dials cor-
rectly pointed off.
10. Clear the machine, return carriage to position one. Set the
next multiplicand 0.032 around the decimal, i.e., set 3 in column
two, and 2 in column one.
11. Multiply by 21.1700. The product 0.67744 appears in the
lower dials correctly pointed off; the multiplier 21.1700 appears in
the upper dials correctly pointed off.
PROBLEMS: Multiplication with preset decimals. Redo problems 1 to
9 on page 69.
C. ACCUMULATIVE AND NEGATIVE MULTIPLICATION
It is often necessary in statistics to sum the cross products of
two columns of numbers, as in computing the Pearson product-
moment coefficient of correlation (r) and in performing matrix
multiplication; or to subtract one product from another, as in the
evaluation of determinants.
When the cross products are all positive, the process of multiply-
ing and adding the products is termed "accumulative multiplica-
tion." When some of the cross products are negative, as they would
Multiplication
71
be if one of the two numbers concerned were negative, the process
of multiplying and subtracting the product is termed “subtractive”
or "negative multiplication."
Friden
(a) Accumulative multiplication
EXAMPLE: (21 × 565) + (31 × 122)
15,647
1. Multiply 21 by 565. The result 11,865 appears in the upper
dials.
2. Set 31 in the keyboard and 122 in the multiplier. Touch the
ACCUM MULT key. The carriage returns to position one and multipli-
cation is made without clearing the dials; therefore, the second
product is added to the first which is already in the upper dials,
giving a total of 15,647. The sum of the two multipliers 687 ap-
pears in the lower dials.
=
(b) Negative multiplication
EXAMPLE: (21 X 565) - (31 X 122)
×
8,083
1. Multiply 21 by 565. The result 11,865 appears in the upper
dials.
2. Set 31 in the keyboard and 122 in the multiplier. Touch the
NEG MULT key. The carriage returns to position one and multipli-
cation is made without clearing the dials; the product of the second
pair is subtracted from the product of the first pair which is already
in the upper dials. The difference of the two products 8,083 ap-
pears in the upper dials; the difference of the two multipliers 443
appears in the lower dials.
488482
=
(c) Accumulative and negative multiplication
EXAMPLE: Find the sum of cross products of scores on two tests,
mathematics and intelligence. Let X stand for the scores on the
mathematics test, and Y for the scores on the intelligence test.
The scores are:
Y
59
48
83 64
45 50
52
29
37
72
Machine Computation of Elementary Statistics
The term "sum of the cross products" is written as ΣXY. The
operation, in detail, is as follows:
ΣΧΥ
=
When negative scores are included in either of the variates
(X or Y), negative multiplication is involved.
(59 × 48) + (83 × 64) + (45 × 50) + (63 × 52)
+ (29 × 37) = 14,743.
ΣΧΥ
(1) X Y
31
30
29
28
27
=
Here, ΣXY = (12 × 48) — (7 × 30) – (10 × 36) + (25 × 59)
+ (6 × 33) = 1,679.
It is not necessary to write out these operations in detail. By
using a guide, each pair can be read directly from the raw data into
the machine.
(4) X
16
17
18
19
20
X
12
- 7
PROBLEMS: Sum of the cross products of two variates, X and Y.
Check by multiplying again, reversing multipliers and multipli-
cands. If an answer is negative, be sure to return the carriage to the
first position before finding the complement.
24
16
20
15
13
1 1
-10
Y
25
6
ΣΧΥ
(2) X
Y
48
30
36
59
33
138
149
120
110
139
Y
46
57
30
19
49
(5)
X
45
Ta
36
19
- 136
57
269
— 14
- 16
49
398
(3) X
Y
-59
3
80
25
86
498
537
465
500
698
ΣΧΥ
=
(6) X
0.34
0.53
0.38
-076
-0.86
Y
56.09
68.87
49.30
53.09
74.78
Y
0.98
-0.86
-0.20
-0.43
0.89
Multiplication
73
ΣΧΥ
8882F
ΣΧ
71 EX²
(8) X
133
49
200
954
ΣΧ
112 EX2
ΣΧΥ
(d) Sum and sum of squares
PROBLEMS: Sum and sum of squares of a variate X. To do this,
set each value in the keyboard and multiplier, and accumulate
multiply. Read EX in the lower dials and ZX2 in the upper dials.
(7) X
=
ΣΧΥ
X
0.97
0.03
0.62
-0.31 EX
0.58 ΣΧ2
=
In problems 9 to 11, negative quantities are involved. In order
to obtain the algebraic sum of the numbers, the negative sign must
be taken into consideration.
After setting the negative quantity (e.g., -0.31 in problem 9),
move down the counter control key located in the upper right of
the keyboard. When the accumulative multiplication operation is
performed, 31 will be subtracted from the quantity in the lower
dials, and the square of 31 will be added into the upper dials.
That is, when the counter control key is down, it reverses the di-
rection of travel of the lower dials.
Be sure it is returned to its normal position for the next multipli-
cation, which involves a positive number as multiplier.
(9)
74
Machine Computation of Elementary Statistics
(10)* X
-397
830
-669
(11)* X
39
50
1.34
5.90
-3.00
-8.66 ZX
4.00 ΣΧ2
ΣΧ
ΣΧ2
Problems 12 to 15 are for students who are familiar with matrix
algebra.
|| ||
PROBLEMS: Matrix multiplication and evaluation of determinants.
If the final result is negative, be sure to return carriage to position
one before finding the complement; also see note on problems 10
and 11 of this section. Check by recomputing.
Note that when a negative multiplication operation is the first
of a series, the machine must be cleared first (cf. problem 13).
(12)
56 197
13 64
= [66
11.11
(13) [1082
X
103 -6
32 45
1[183
(56 X 13)+(19 X 19)
×
.(13 × 13) + (64 X 19)
13 46
X
167
19 12
[198
103 -6
32 45
(56 X 46) + (19 × 12)
(13 X 46) + (64 × 12)]
12)]
Check:
(103 X 103)
-(1036)
(6 × 45)˜¯
[(323x1093)+(4632) -
(32 × 103) + (45 × 32)
―
− (32 × 6) + (45 × 45)
[
Check:
* Note that EX is negative. After recording ΣX², write the amount from the
lower dials in the keyboard, put add key up, clear dials, and subtract. The
complement appears in the upper dials.
Multiplication
75
(14)
(15)
56
13
19
64
103 -6
32 45
=
| = (56 × 64) – (13 × 19)
= (103 X 45) + (32 × 6)
Marchant
(a) Accumulative multiplication
EXAMPLE: (21 × 565) + (31 X 122)
15,647
1. Since there are three digits in the first multiplier, depress
selective tab key three. Multiply 21 by 565. The result 11,865
appears in the middle dials.
2. Without clearing the upper and middle dials, depress CLEAR
TAB and KEYBOARD DIAL keys to clear the keyboard and position
the carriage for the next multiplication.
3. Multiply 31 by 122. Since the carriage dials were not cleared,
the product of the second multiplication is added to the first
product to give a total of 15,647 in the middle dials. The sum of the
two multipliers 687 appears in the upper dials.
=
EXAMPLE: (35 × 46) + (98 × 164) + (112 × 59) = 24,290
1. Since there are two digits in the first multiplier, depress
selective tab key two. Multiply 35 by 46.
2. Note that the second multiplier 164 has three digits. With-
out clearing the upper and middle dials, depress KEYBOARD DIAL
and selective tab key three simultaneously. This clears the key-
board and positions the carriage for the next multiplication. Multi-
ply 98 by 164.
3. Note that the third multiplier has two digits. Depress KEY-
BOARD DIAL and selective tab key two simultaneously. This clears
the keyboard and positions the carriage for the next multiplica-
tion. Multiply 112 by 59.
4. The result 24,290 appears in the middle dials, the sum of the
multipliers 269 in the upper dials.
—>
(b) Negative multiplication
EXAMPLE: (21 × 565) (31 X 122)
8,083
1. Multiply 21 by 565. The result 11,865 appears in the middle
dials.
=
76
Machine Computation of Elementary Statistics
2. Without clearing the upper and middle dials, depress CLEAR
TAB and KEYBOARD DIAL keys to clear the keyboard and position
the carriage for the next multiplication.
3. Set 31 in the keyboard. Using the left hand, depress the
REVERSE bar, and hold it there while using the right hand to enter
the multiplier 122 in the multiplier keyboard. (In D models, use
SHORT CUT bar instead of X bar.)
4. Since the carriage dials were not cleared, the product of the
second multiplication is subtracted from the first to give the
difference 8,083.
5. The difference between the two multipliers 443 appears in
the upper dials.
6. If the REVERSE bar sticks, release it by depressing the di-
vision STOP key.
(c) Accumulative and negative multiplication
EXAMPLE: Find the sum of cross products of scores on two tests,
mathematics and intelligence. Let X stand for the scores on the
mathematics test, and Y for the scores on the intelligence test.
The scores are:
X
59
83
45
63
29
ΣΧΥ =
The term "sum of the cross products" is written as ZXY. The
operation in detail is as follows:
(59 × 48) + (83 × 64) + (45 × 50) + (63 × 52)
+ (29 × 37) = 14,743.
When negative scores are included in either of the variates (X or
Y), negative multiplication is involved.
X
Y
48
64
50
52
37
Y
12 48
30
- 7
| |
-10
25
6
8888
36
33
A
Multiplication
77
Here, ΣΧΥ
(1) X
31
30
29
It is not necessary to write out these operations in detail. By
using a guide, each pair can be read directly from the raw data into
the machine.
28
27
PROBLEMS: Sum of the cross products of two variates, X and Y.
Check by multiplying again, reversing multipliers and multipli-
cands. If an answer is negative, be sure to return the carriage to the
first position before finding the complement.
ΣΧΥ
=
(4) X
16
17
18
19
20
ΣΧΥ
=
=
(7) X
36
60
89
22
71
× —
(12 X 48) (7 X 30) (10 × 36) + (25 × 59)
+ (6 × 33) = 1,679
Y
24
16
20
15
13
Y
45
36
19
-14
- 16
-
ΣΧ
EX² =
(2) X
(5)
138
149
120
110
139
ΣΧΥ
X
- 136
57
269
49
398
ΣΧΥ
Y
46
57
30
19
49
Y
- 59
3
80
25
86
(3) X
498
537
465
500
698
ΣΧΥ
(6)
X
0.34
0.53
0.38
-0.76
-0.86
ΣΧΥ
=
Y
56.09
68.87
49.30
53.09
74.78
(d) Sum and sum of squares
PROBLEMS: Sum and sum of squares of a variate X. To do this, set
each value in the keyboard and multiplier, and accumulate multi-
ply. Read EX in the upper dials and 2X2 in the middle dials.
Y
0.98
-0.86
-0.20
-0.43
0.89
78
Machine Computation of Elementary Statistics
(8) X
133
49
200
954
ΣΧ
112 ΣΧ2
In problems 9 to 11, negative quantities are involved. In order
to obtain the algebraic sum of numbers, the negative sign must be
taken into consideration.
After setting the negative quantity (e.g., -0.31 in problem 9),
move down the counter control key located in the upper right of
the keyboard. When the accumulative multiplication operation is
performed, 31 will be subtracted from the quantity in the upper
dials, and the square of 31 will be added into the middle dials.
That is, when the counter control key is down, it reverses the direc-
tion of travel of the upper dials.
Be sure it is returned to its normal position for the next multipli-
cation, which involves a positive number as multiplier.
(9) X
12
0.97
0.03
0.62
-0.31
0.58
(10)* X
(11)*
ཻ།
-397
830
-669
39 ΣΧ
- 50 EX²
X
1.34
5.90
ΣΧ
ΣΧ
-3.00
-8.66
4.00
|| ||
ΣΧ
ΣΧ2
* Note that ZX is negative. After recording ZX2, return the carriage to position
one, write the amount from the upper dials in the keyboard, clear the carriage
dials, depress REVERSE and 1 multiplier key simultaneously. The comple-
ment appears in the middle dials.
Multiplication
79
Problems 12 to 15 are for students who are familiar with matrix
algebra.
PROBLEMS: Matrix multiplication and evaluation of determinants.
If the final result is negative, be sure to return carriage to position
one before finding the complement; also see note on problems 10
and 11 of this section. Check by recomputing.
(12) [56
[56 19
(14)
(15) |
13 64
||_||
19]
- [168
103
32
= [(193
(13) [192
[103 78] × [183
-6
X
32 45
56 19
13 64
X
´(56 × 13) + (19 × 19)
[183
13 46
18]
(13 × 13) + (64 X 19)
S
-6
45
19 12
-67
【(103 × 103) — (6 × 32)
(32 × 103) + (45 × 32)
]
(56 × 64) – (13 × 19)
32 45
(56 × 46) + (19 × 12)
(13 X 46) + (64 × 12).
Check:
(103 × 6) –
– (103 × 6) – (6 × 45)
− (32 × 6) + (45 × 45)
Check:
(103 X 45) + (32 × 6) =
Monroe: Automatic
(a) Accumulative multiplication
EXAMPLE: (21 × 565) + (31 × 122) 15,647
1. Carriage in position one, change lever at X.
2. Multiply 21 by 565. The result 11,865 appears in the lower
dials.
=
3. Without clearing the dials, clear the keyboard. Be sure
carriage is in position one.
4. Multiply 31 by 122. The second product is added to the first
which is already in the lower dials, giving a total of 15,647. The
sum of the two multipliers 687 appears in the upper dials.
80
Machine Computation of Elementary Statistics
(b) Negative multiplication
EXAMPLE: (21 X 565) - (31 × 122)
8,083
1. Multiply 21 by 565. The result 11,865 appears in the lower
dials.
2. Clear keyboard. Set up 122. Set 31 in the keyboard. On CAA
machines, put change lever at X and depress negative multiply
key (X). On other models, put change lever at÷, push up counter
control key, and multiply.
3. The product of the second pair of numbers is subtracted
from the product of the first pair which is already in the lower
dials. The difference of the two products 8,083 appears in the
lower dials; the difference of the two multipliers 443 appears in
the upper dials.
X
59
(c) Accumulative and negative multiplication
EXAMPLE: Find the sum of cross products of scores on two tests,
mathematics and intelligence. Let X stand for the scores on the
mathematics test, and Y for the scores on the intelligence test.
The scores are:
Y
48
64
45 50
888888
83
63
29
=
X
52
37
The term "sum of the cross products" is written as ZXY. The
operation in detail is as follows:
12
- 7
A
EXY = (59 X 48) + (83 X 64) + (45 X 50) + (63 × 52)
ΣΧΥ
=
+ (29 × 37) 14,743
When negative scores are included in either of the variates
(X or Y), negative multiplication is involved.
-10
25
6
Y
48
30
36
59
33
Here, ΣXY = (12 × 48) — (7 × 30) — (10 × 36) + (25 × 59)
+(6 × 33) = 1,679
Multiplication
81
It is not necessary to write out these operations in detail. By
using a guide, each pair can be read directly from the raw data into
the machine.
PROBLEMS: Sum of the cross products of two variates, X and Y.
Check by multiplying again, reversing multipliers and multipli-
cands. If an answer is negative, be sure to return the carriage to the
first position before finding the complement.
(1) X
31
24
30 16
20
15
13
22227
29
28
ΣΧΥ
(4) X
16
17
18
19
20
ΣΧΥ
(7) X
36
88827
60
89
Y.
Y
71
——
399895
36
19
- 14
- 16
22 ΣΧ
ΣΧ
45
(2) X Y
138
149
120
110
139
ΣΧΥ
(5) X
=
- 136
57
269
49
398
ΣΧΥ
46
57
30
19
49
Y
-59
3
80
8988
25
86
(3) X
498
537
465
500
698
ΣΧΥ
(6)
=
X
0.34
0.53
0.38
-0.76
-0.86
ΣΧΥ
(d) Sum and sum of squares
PROBLEMS: Sum and sum of squares of a variate X. To do this,
set up each value, and accumulate multiply. Read EX in the upper
dials and EX² in the lower dials. In CAÂ machines, the setup key
should be held down until the amount in the keyboard is trans-
ferred into the machine; in this case, the keyboard does not clear,
and squaring is completed when the multiply key is depressed. On
machines with squaring lock, turn lock to "on" position.
=
Y
56.09
68.87
49.30
53.09
74.78
Y
0.98
-0.86
-0.20
-0.43
0.89
82
Machine Computation of Elementary Statistics
(8) X
133
49
200
954
ΣΧ
112 ΣΧ2
In problems 9 to 11, negative quantities are involved. In order
to obtain the algebraic sum of the numbers, the negative sign
must be taken into consideration.
After setting the negative quantity (e.g., -0.31 in problem 9),
push change lever to÷position. When the accumulative multipli-
cation operation is performed, 31 will be subtracted from the quan-
tity in the upper dials, and the square of 31 will be added into the
lower dials. That is, when the change lever is at ÷ position, it
reverses the direction of travel of the upper dials.
Be sure it is returned to its normal position for the next multipli-
cation, which involves a positive number as multiplier.
(9) X
0.97
0.03
0.62
-0.31
0.58
(10)* X
-397
830
-669
1
39 ΣΧ
50 ΣΧ
(11)* X
1.34
5.90
EX
ΣΧ2
-3.00
-8.66
4.00
ΣΧ
ΣΧ2
* Note that EX is negative. After recording ZX2, write the amount from the
upper dials in the keyboard, clear dials, and subtract. The complement appears
in the upper dials.
Multiplication
83
;
Problems 12 to 15 are for students who are familiar with matrix
algebra.
PROBLEMS: Matrix multiplication and evaluation of determinants.
If the final result is negative, be sure to return carriage to position
one before finding the complement; also see note on problems 10
and 11 of this section. Check by recomputing.
56
(12) [58 19]
13 64
(14)
(13) [108
(15)
11
(56 × 13) + (19 × 19)
(56 × 46) + (19 × 12)
= (13 × 13) + (64 × 19) (13 × 46) + (64 × 12)_
11
X
•
||
32 45
-6
[133
56
19
13 64
103 - 6
32
45
X
13 46
19
19 12
**
103 -67
(103 × 103) – (6 × 32)
(32 × 103) + (45 × 32)
]
[103
32 45
(56 X 64)
Check:
A
− (103 × 6) – (6 × 45)7
- (32 × 6) + (45 × 45).
Check:
(13 X 19)
(103 X 45) (32 X 6)
Monroe: Semiautomatic and Hand-operated
(a) Accumulative multiplication
EXAMPLE: (21 × 565) + (31 × 122)
15,647
1. Multiply 21 by 565. The result 11,865 appears in the lower
dials.
=
2. Return carriage to position one.
3. Clear upper dials and keyboard.
4. Multiply 31 by 122.
84
Machine Computation of Elementary Statistics
5. Since the second multiplication was made without clearing
the first product from the lower dials, the second product was
added to the first to give the sum of the two 15,647 in the lower
dials.
(b) Negative multiplication
EXAMPLE: (21 X 565) (31 × 122)
8,083
1. Multiply 21 by 565. The result 11,865 appears in the lower
dials.
2. Return carriage to position one.
3. Clear upper dials and keyboard.
4. Set the second multiplicand 31 in the keyboard. Negative
multiply by subtracting the multiplier 122.
5. Since the second multiplication was made without clearing
the first product from the lower dials, the second product was sub-
tracted from the first to give the difference 8,083 in the lower dials.
6. Note that non-carry-over dials will show the multiplier 122
in red; carry-over dials will show the complement of 122 (878)
preceded by several 9's.
7. On models with a counter control lever (marked X÷) put
the lever at position for the negative multiplication; the upper
dials will count forward in subtraction and show 122 in black.
(c) Accumulative and negative multiplication
EXAMPLE: Sum of cross products of scores on two tests, mathematics
and intelligence. Let X stand for the scores on the mathematics
test, and Y for the scores on the intelligence test. The scores are:
X
59
83
=
28482
=
45
63
29
Y
48
64
50
52
37
The term "sum of the cross products" is written as EXY. The
operation in detail is as follows:
ΣΧΥ
(59 X 48) + (83 × 64) + (45 × 50) + (63 × 52)
+ (29 × 37) = 14,743
Multiplication
85
When negative scores are included in either of the variates (X or
Y), negative multiplication is involved.
(1) X Y
31
30
29
28
27
ΣΧΥ =
(4) X
16
17
18
Here, ΣXY = (12 × 48) – (7 × 30) – (10 × 36) + (25 × 59)
+ (6 × 33) 1,679
=
19
20
It is not necessary to write out these operations in detail. By
using a guide, each pair can be read directly from the raw data into
the machine.
ΣΧΥ
PROBLEMS: Sum of the cross products of two variates, X and Y.
Check by multiplying again, reversing multipliers and multipli-
cands. If an answer is negative, be sure to return the carriage to the
first position before finding the complemen
=
24
16
20
15
13
Y
X
12
- 7
- 10
25
6
45
36
19
– 14
- 16
ΣΧΥ
(5)
Y
48
30
888888
X
36
(2) X Y
138
46
149
57
120
30
110
19
139
49
136
57
269
49
398
59
33
ΣΧΥ
Stag
Y
- 59
- 3
80
25
86
(3) X
498
537
465
500
698
ΣΧΥ
(6)
=
X
0.34
0.53
0.38
-0.76
-0.86
ΣΧΥ
Y
56.09
68.87
49.30
53.09
74.78
Y
0.98
-0.86
-0.20
-0.43
0.89
86
Machine Computation of Elementary Statistics
(d) Sum and sum of squares (can be carried out on machines with
split upper dials only; this cannot be done on the Educator model).
EXAMPLE: Find the sum and sum of squares of a variate X.
(7) X
36
60
89
22
71
(8) X
133
49
200
954
112
ΣΧ
ΣΧ2
X
1. Clear the machine, put carriage in position one.
2. Push upper dial lock lever to the left. This locks the right
bank of the upper dials.
3. Square 25. Clear the upper dials. Note that only the left
bank clears. The number 25 appears in the right bank of the upper
dials, and 625 (the square of 25) in the lower dials.
4. Square the next value 60. The number 60 appears in the
left bank of the upper dials, 85 (25 +60) in the right bank, and
the sum of the two squares 4,225 in the lower dials.
5. Clear the upper dials, and square the next value 68. Con-
tinue through all five values, being sure to clear only the upper
dials and keyboard each time.
6. Read EX (330) in the right upper dials; EX2 (28,838) in the
lower dials.
ΣΧ
25
60
68
42
135
PROBLEMS: Sum and sum of squares of a variate X. Check by re-
computing.
ΣΧ
ΣΧ
ΣΧ2
=
=
330
28838
Multiplication
87
(e) Sum and sum of squares, negative quantities involved
This operation can be performed best on machines with split
upper dials and a counter control lever.
EXAMPLE: Find EX and EX²
X
43
29
12
-16
48
X
0.97
0.03
0.62
-0.31
C
1. Clear machine; put carriage in position one.
2. Push upper dial lock lever to the left.
3. Square 43, 29, 12 as before, clearing only the upper dials
and keyboard after each operation.
4. Before squaring the negative quantities (-16 and -48),
push counter control lever to the position. In semiautomatic
machines, push change lever to divide. In this way, the squares of
16 and 48 are added into the lower dials; the quantities 16 and 48
are subtracted from the right upper dials. Note that 16 and 48
both appear in red in the left upper dials, indicating the subtrac-
tion operation.
ΣΧ =
ΣΧ
20
ΣΧ2 = 5394
5. Be sure to return the counter control lever to the X position
for squaring of positive quantities.
6. On machines with split upper dials and no counter control
lever, sum, square, and sum the squares of the positive numbers.
Then record ZX and EX². Clear machine, and sum, square, and
sum the squares of the negative numbers, disregarding the negative
signs. Record. Add the two sums of squares to get the total sum of
squares; subtract the sum of the negative numbers from the sum
of the positive numbers to get ZX.
0.58 ΣΧ2
PROBLEMS: Sum and sum of squares with negative quantities in-
volved. Remember to unlock and clear the right upper dials after
completion of each problem.
(9)
88
Machine Computation of Elementary Statistics
(10)*
(11)*
X
-397
830
-669
39
50
X
1.34
5.90
-3.00
-8.66
4.00
ΣΧ
ΣΧ2
56 19
(12) [18 64
13
ΣΧ
ΣΧ2
Problems 12 to 15 are for students who are familiar with matrix
algebra.
PROBLEMS: Matrix multiplication and evaluation of determinants.
If the final result is negative, be sure to return carriage to position
one before finding the complement; also see note on problems 10
and 11 of this section. Check by recomputing.
|| ||
X
= [(193
13 467
[133 16 ]
19 12
(56 × 13) + (19 × 19)
(13 X 13) + (64 X 19)
-6
(13) [103
[103 78] × [198
[103
X
32 45
(103 × 103)
-67
45
(6 X 32)
(32 × 103) + (45 × 32)
(56 × 46) + (19 × 12)
(13 X 46) + (64 × 12)]
12]
Check:
(103 × 6) – (6 × 45)'
– (32 × 6) + (45 × 45).
Check:
[
* Note that ZX is negative. After recording ZX2, clear the keyboard and lower
dials. Write the amount from the upper dials in the keyboard, retaining only
two 9's, and subtract. The complement appears in the lower dials.
Multiplication
89
(14)
(15)
56
13
103
32
19
64
-6
45
D. MULTIPLICATION BY A CONSTANT
| = (56 × 64) — (13 × 19)
Month
January
February
March
April
May
June
July
August
= (103 × 45) + (32 × 6)
Friden
When each of a series of numbers is to be multiplied by the same
number (a constant), the constant can be locked in the keyboard
by pushing up the positive keyboard lock. The add key should be
up.
September
October
EXAMPLE: Prepare a table showing the decimal equivalent of a
whole year, of the last day of each month. Use a 365-day year;
report to five places.
November
December
Ordinal Number
of Last Day
31
59
90
120
151
181
212
243
273
304
334
365
Decimal Equivalent
of Last Day
0.08493
0.16164
0.24657
0.32876
0.41369
0.49589
0.58082
0.66575
0.74794
0.83287
0.91506
1.00000
1. Add key up.
2. The decimal equivalent of one day (15)
365) is 0.0027397.
3. Enter 0.0027397 in the keyboard. Lock by pushing up the
positive keyboard lock.
4. Since the multiplicands are all whole numbers, and there are
seven decimal places in the constant multiplier, there will be
0 + 7 = 7 decimal places in the product. Put the upper dial
decimal between dials seven and eight.
90
Machine Computation of Elementary Statistics
5. Multiply by 31; then by 60, and so on.
6. When doing an extensive series of multiplications by a deci-
mal fraction, it is well to carry two more decimal places than are
desired for the final answer.
Marchant
When each of a series of numbers is to be multiplied by the same
number (a constant), the constant can be entered in the keyboard
and the keyboard dial clearance key locked by the keyboard lock.
Since on the Marchant calculator, the keyboard does not clear in
the multiplication operation, it is not necessary to lock the con-
stant in. However, use of the keyboard lock guards against acci-
dental clearance by striking the keyboard dial clearance key.
On D models, if successive multipliers are similar, it is sometimes
possible to go from one multiplication to the next without clearing
the machine by building up or down the necessary digits in the
upper dials.
EXAMPLE: Prepare a table showing the decimal equivalent of a
whole year, of the last day of each month. Use a 365-day year;
report to five decimal places.
Month
January
February
March
April
May
June
July
August
September
October
November
December
Ordinal Number
of Last Day
31
59
90
120
151
181
212
243
273
304
334
365
Decimal Equivalent
of Last Day
0.08493
0.16164
0.24657
0.32876
0.41369
0.49589
0.58082
0.66575
0.74794
0.83287
0.91506
1.00000
1. The decimal equivalent of one day (15)
365
2. Enter 0.0027397 in the keyboard. Lock keyboard dial clear-
ance key.
is 0.0027397.
Multiplication
91
3. Since the multiplicands are all whole numbers, and there are
seven decimal places in the constant multiplier, there will be
0 + 7 = 7 decimal places in the product. Put the middle dial
decimal between dials seven and eight.
4. Multiply by 31; record product and clear machine; multiply
by 60; record product and clear machine; and so on.
5. When doing an extensive series of multiplications by a
decimal equivalent, it is well to carry two more decimal places
than are desired for the final answer.
Monroe: Automatic
When each of a series of numbers is to be multiplied by the same
number (a constant), the constant multiplier can be locked in the
machine by a constant lever. The change lever must be at X.
EXAMPLE: Prepare a table showing the decimal equivalent of a
whole year, of the last day of each month. Use a 365-day year;
report to five places.
Month
January
February
March
April
May
June
July
August
September
October
November
December
Ordinal Number
of Last Day
31
59
90
120
151
181
212
243
273
304
334
365
Decimal Equivalent
of Last Day
0.08493
0.16164
0.24657
0.32876
0.41369
0.49589
0.58082
0.66575
0.74794
0.83287
0.91506
1.00000
1. The decimal equivalent of one day (35)
2. Enter 0.0027397 in extreme right of keyboard. Set up.
3. Lock constant multiplier lever (C on CAA machines) or
push up lower automatic constant lever (A models).
4. Since the multiplicands are all whole numbers, and there
are seven decimal places in the constant multiplier, there will be
is 0.0027397.
92
Machine Computation of Elementary Statistics
0+ 7 = 7 decimal places in the product. Put the lower dial
decimal marker between dials seven and eight. Put upper dial
decimal marker between seven and eight.
5. Multiply by 31. Record product 0.08493 from lower dials.
Disregard number in left upper dials; right upper dials show
0.0027397, the constant multiplier. Clear dials and keyboard.
6. Set up and multiply by 60. Record product; clear dials and
keyboard.
7. Continue in this fashion throughout the series.
8. To clear the constant multiplier out of the machine: On CAA
machines, release constant multiplier lock by moving C lever up
before touching the multiply key in the final multiplication. On A
models, push lower constant lever and return upper constant lever
after clearing the dials and keyboard after the last multiplication.
9. When doing an extensive series of multiplications by a
decimal fraction, it is well to carry two more decimal places than
are desired for the final answer.
Monroe: Semiautomatic and Hand-operated
When each of a series of numbers is to be multiplied by the same
number (a constant), the constant multiplier is entered in the key-
board and the repeat key depressed. Care must be taken not to
clear the keyboard after each multiplication operation.
EXAMPLE: Prepare a table showing the decimal equivalent of a
whole year, of the last day of each month. Use a 365-day year;
report to five places.
Month
January
February
March
April
May
June
July
August
September
October
November
December
Ordinal Number
of Last Day
31
59
90
120
151
181
212
243
273
304
334
365
Decimal Equivalent
of Last Day
0.08493
0.16164
0.24657
0.32876
0.41369
0.49589
0.58082
0.66575
0.74794
0.83287
0.91506
1.00000
Multiplication
93
1. Push down repeat key.
1
2. The decimal equivalent of one day (35)
3. Enter 0.0027397 in the keyboard.
4. Since the multiplicands are all whole numbers, and there
are seven decimal places in the constant multiplier, there will be
0+ 7 = 7 decimal places in the products. Place the lower dial
decimal marker between lower dials seven and eight.
is 0.0027397.
5. Multiply by 31. Record product, clear the dials, and multiply
by 60. Record the product, clear the dials, and continue through
the problem.
6. When doing an extensive series of multiplications by a
decimal fraction, it is well to carry two more decimal places than
are desired for the final answer.
E. USE OF TABLES OF PRODUCTS
The student will often have occasion to handle multiplication of
one- and two-digit numbers. A rapid method of determining the
products of such numbers is to read them from a table. Crelle's
tables (22)
may be used to find the products of any two numbers
from one to one thousand. Holzinger (25) provides two tables, one
(Table II, p. 8 to 27) of Products of Integers to 100 X 99 and the
other (Table IV, p. 48 to 49) of Products: (1-100) X 22, 32, ...,
(12)2. Tables of squares may be found in Barlow (20) for integers up
to 12,500; Holzinger (25) (Table I, p. 1 to 7) for integers up to
1,050; Walker (19) (Appendix D, p. 340-342) for numbers 1.0 to 9.9
and for integers up to 100; and in almost every text in statistics.
EXAMPLE: 46 X 29 = 1,334
(1) 56 × 28
(4) 49 X 47
(7) 96 X 144
(10) (139) 2
(13) 92 X 112
1. Turn to Holzinger's Table II. Find 46 in the first column
(in bold-face type), and read across the row to the column headed
29 (bold-face type at top of table). The number in the cell where the
46 row crosses the 29 column is the product of 46 and 29 (1,334).
2. Verify by multiplying on the machine.
PROBLEMS: Read products from the appropriate tables. Check by
multiplication on the machine.
(2) 31 × 94
(5) 69 × 83
(8) 39 X 62
(11) (5.5)2
(14) 82 X 144
(3) 98 X 49
(6) 38 X 121
(9) (325)2
(12) (1.94)2
(15) 99 X 99
{
CHAPTER 5
DIVISION
Division requires certain operations peculiar to itself as well as:
a mastery of subtraction and multiplication. Fundamentally,
division involves a series of subtractions of the divisor from the
dividend, just as in paper and pencil operations. On a hand-
operated computing machine, the subtractions and movement of
the carriage from left to right must be made manually; on an auto-
matic machine, the process is completely automatic. The dividend
is the number to be divided; the divisor is the number by which
the dividend is divided; the quotient is the number resulting
from the division of dividend by divisor. For example, 6 ÷ 3 = 2;
6 is the dividend, 3 the divisor, and 2 the quotient.
A. DIVISION OF WHOLE NUMBERS
Friden
(a) Quotient with no remainder
EXAMPLE: 39,483 ÷ 321
123
1. Put add key down and carriage in position one.
2. Set the dividend 39,483 in the extreme right of the keyboard.
Add into the machine; it appears in the upper dials.
3. Clear the 1 from the lower dials manually by moving the
lower clearing lever to the right.
4. Set the divisor 321 in keyboard columns three through one.
5. In order to line up the first digits of dividend and divisor,
move the carriage to the right until the first digit of the dividend
in the upper dials is directly above the first digit of the divisor in
the keyboard. Since there are three digits in the divisor, the car-
riage position indicator points to the third lower dial.
6. Touch both divide keys. Division is performed automatically.
The quotient 123 appears in lower dials three through one.
=
94
Division
95
PROBLEMS: Division of whole numbers. Be sure to touch both
divide keys. Check by multiplying divisor by quotient to get the
dividend as product. Clear and return the machine after each check.
237
573
347
(1) 50,481
(2) 75,636
(3) 56,561
(4) 43,605
(5) 95,953
(6) 98,010990
459
793
(b) Quotient with remainder
EXAMPLE: 56,733 ÷ 888 63 and 789/888
1 to 6. Same as in section (a).
7. The remainder 789 appears in the upper dials.
PROBLEMS: Division of whole numbers; quotients with remainders.
Check by multiplying divisor by quotient and adding remainder
to get the dividend. Be sure to return carriage to first position
before adding remainder.
(7) 67,837 ÷ 865 =
(8) 59,632 ÷ 795 =
(9) 52,138 173 =
(10) 37,153 ÷ 269 =
Marchant
(a) Quotient with no remainder
EXAMPLE: 39,483 321
-
123
1. Place carriage in position one.
2. Set the dividend 39,483 in the extreme right of the keyboard
in columns five through one. Add into machine; the number ap-
pears in the middle dials.
3. Clear the 1 from the upper dial manually by touching the
upper dial clear key.
4. Set the divisor 321 in keyboard columns three through one.
5. With the right carriage shift key (lower key, arrow pointing
to the right) move the carriage until the first digit of the dividend
in the middle dials is directly above the first digit of the divisor in
the keyboard, i.e., the first 3 of 39,483 is above the 3 of 321.
96
Machine Computation of Elementary Statistics
6. Depress the automatic divide key. Division is performed
automatically. The quotient 123 appears in upper dials three
through one.
PROBLEMS: Division of whole numbers. Check by multiplying divisor
by quotient to get dividend as product. Be sure to clear the ma-
chine after each check.
(1) 50,481 ÷ 237
(2) 75,636 ÷ 573 =
(3) 56,561 ÷ 347
(4) 43,605
459
793 =
(5) 95,953
(6) 98,010 990 =
(b) Quotient with remainder
EXAMPLE: 56,733 ÷ 888 63 and 789/888
1. Push up division clear-return key. This causes the remainder
to appear in the middle dials.
2 to 6. Proceed as on page 95, steps 2 to 6.
7. The remainder 789 appears in the middle dials.
=
PROBLEMS: Division of whole numbers; quotients with remainders.
Check by multiplying divisor by quotient and adding remainder
to get the dividend. Be sure to return the carriage to first position
before adding remainder.
(7) 67,837 865
(8) 59,632 ÷ 795 =
(9) 52,138 173 =
(10) 37,153 ÷ 269 =
Monroe: Automatic and Semiautomatic
(a) Quotient with no remainder
EXAMPLE: 39,483 ÷ 321 = 123
1. Carriage is in position one; nonrepeat key is depressed.
2. Set the dividend 39,483 in the extreme right of the keyboard.
3. Clear the 1 from the upper dials.
4. Set the divisor 321 in keyboard columns three through one.
5. In order to line up the first digits of dividend and divisor,
move the carriage to the right until the first digit of the dividend
Division
97
in the lower dials is directly above the first digit of the divisor in
the keyboard. Since there are three digits in the divisor, the car-
riage position indicator points to the third lower dial.
6. Push divide lever up (in CAA models, depress divide key).
Division is performed automatically.
7. The quotient 123 appears in red in the left upper dials, its
complement in the right upper dials. In CAA machines, the quo-
tient appears in the upper dials.
PROBLEMS: Division of whole numbers. Check by multiplying divisor
by quotient to get the dividend as product.
(1) 50,481 ÷ 237
(2) 75,636
573 =
347 =
(3) 56,561
(4) 43,605 ÷ 459 =
(5) 95,953 ÷ 793
(6) 98,010 ÷ 990 =
(b) Quotient with remainder
EXAMPLE: 56,733 ÷ 888 = 63 and 789/888
1 to 6. Same as in section (a).
7. The quotient 63 appears in the upper dials, and the remainder
789 in the lower dials.
PROBLEMS: Division of whole numbers; quotients with remainders.
Check by multiplying divisor by quotient and adding remainder
to get the dividend.
(7) 67,837 ÷ 865
(8) 59,632 795
(9) 52,138 ÷ 173
(10) 37,153 269 =
Monroe: Hand-operated
(a) Quotient with no remainder
EXAMPLE: 39,483 321
123
1. Put repeat key down and carriage in position one.
2. On machines with counter control lever, set lever to DIVIDE
position.
=
98
Machine Computation of Elementary Statistics
3. Set the dividend 39,483 in the extreme right of the keyboard.
Add into the machine; the dividend appears in the lower dials.
Number 1 appears in the left upper dials, and in the right bank of
upper dials of machines without a counter control lever; a row of
9's appears in the right upper dials of machines with a counter
control lever.
4. Clear the keyboard, and touch the minus bar once to take
the 1 out of the upper dials.
5. Set the divisor 321 in keyboard columns three through one.
6. In order to line up the first digits of dividend and divisor,
move the carriage to the right until the first digit of the dividend
in the lower dials is directly above the first digit of the divisor in
the keyboard. Since there are three digits in the divisor, the carriage
position indicator points to the third lower dial.
7. Subtract until a negative number appears in the lower dials.
This is indicated by the ringing of the bell and the appearance of a
row of 9's preceding the numbers in the lower dials. In electric K
models, the bell need not be used; the machine automatically stops
subtracting when the first oversubtraction takes place.
8. Add the divisor in again to remove the negative number,
and move the carriage one place to the left.
9. Subtract until a negative number appears in the lower dials,
add once to make the number positive. Note that the upper dials
register the number of times the divisor is subtracted in each po-
sition.
10. Move the carriage one place to the left. Subtract until the
bell rings, add in once. The lower dials register 0 and the left upper
dials show 123, the quotient. The quotient 123 appears also in the
right upper dials of machines with a counter control key.
PROBLEMS: Division of whole numbers. Check by multiplying di-
visor by quotient to get the dividend as product. Be sure to clear
the machine and return carriage to position one after completing
each problem.
(1) 50,481 237 =
(2) 75,636 ÷ 573 =
(3) 56,561
347 =
(4) 43,605
459 =
793 =
(5) 95,953
(6) 98,010 ÷ 990 =
Division
99
(b) Quotient with remainder
EXAMPLE: 56,733 ÷ 888 = 63 and 789/888
1 to 6. Same as in section (a). Note that the first subtraction
results in a negative number (576 – 888). If this is noted in time,
the carriage can be moved immediately, and the subtraction begun
in the second position.
7. The remainder 789 appears in the lower dials.
(7) 67,837 ÷ 865 =
795 =
(8) 59,632
(9) 52,138 173 =
(10) 37,153 ÷ 269 =
PROBLEMS: Division of whole numbers; quotients with remainders.
Check by multiplying divisor by quotient and adding remainder
to get the dividend. Be sure to return carriage to position one
before adding remainder.
B. DIVISION OF NUMBERS WITH DECIMALS
There are two methods of performing division with decimals:
(a) setting both dividend and divisor in the extreme left of the
keyboard, and
(b) using a constant or preset decimal setup.
Friden
(a) Numbers in extreme left of keyboard
EXAMPLE: 7,659.96 44.44 172.3663366+
÷
1. Put add key down.
2. Depress tab key ten.
3. Enter 765996 in extreme left of keyboard in columns ten
through five.
4. Touch DIV'D TAB key. This automatically clears the dials,
positions the carriage to the extreme right, enters the dividend in
the upper dials, clears the 1 from the lower dials, and clears the
keyboard.
(In models not having the DIV'D TAB key, position carriage to
extreme right and add dividend into the machine. If there is not
an automatic clearance of the 1 in the lower dials, this must be done
manually.)
=
100
Machine Computation of Elementary Statistics
5. Put the upper dial decimal marker in its place between the
two 9's (upper dials fifteen and sixteen). This means that there are
fifteen places to the right of the decimal in the upper dials.
6. Set the divisor 4444 in the extreme left of the keyboard in
columns ten through seven.
7. Turn up the keyboard decimal marker between columns
eight and nine where it actually appears in the divisor.
8. To determine the number of decimal places in the quotient,
subtract the number of places to the right of the decimal in the
divisor in the keyboard (here 8) from the number to the right of
the decimal in the dividend in the upper dials (here 15), and set
the lower dial decimal marker accordingly (here 15 8 = 7; i.e.,.
set decimal marker between lower dials seven and eight).
9. Touch both divide keys. Division is performed automatically.
10. The quotient 172.3663366+ appears in the lower dials
correctly pointed off.
REMEMBER: The number of places to the right of the decimal in
the upper dials is equal to the number to the right of the decimal
in the lower dials plus the number to the right in the keyboard.
(Note that this is the same as in multiplication.)
PROBLEMS: Division of numbers with decimals; working from extreme
left of keyboard. Check by multiplying divisor by quotient to get
the dividend as product. Remember that the number of decimal
places changes with each problem.
(1) 67,893.221 ÷ 569.003
(2) 20,589.02 ÷ 76.5992 =
(3) 4.211199 ÷ 9.91124
(4) 0.112390090.11290909
(5) 0.055468 ÷ 3.45482
(b) Constant or preset decimal setup
The preset decimal setup is particularly useful for a series of
divisions where a fixed number of decimal places is required in the
quotients.
EXAMPLES: Given the following series of divisions to be performed;
in each problem the quotient is required to be correct to four deci-
mal places. Check by multiplying divisor by quotient; note that the
Division
101
product in the check will agree in all but the last digit of the original
dividend.
a. 38.1022 ÷ 19.779 =
1.92639 rounded to 1.9264
b. 94.56 ÷ 12.3 = 7.68780 rounded to 7.6878
c. 1.038.93 0.11534 rounded to 0.1153
d. 8,756 ÷ 847.9
847.9 = 10.32668 rounded to 10.3267
e. 0.0365 ÷ 1.12 0.03258 rounded to 0.0326
=
1. Put add key down.
2. Since four decimal places are required in the quotient, the
division should be carried through five places, and the quotient
rounded to four. Set the lower dial decimal marker between dials
five and six, and press selective tabulator key six. REMEMBER: Press
the tab key which is directly to the left of the decimal marker.
3. Examine the several dividends and divisors to determine the
maximum number of decimals in any number. In the example, this
is four. Turn up the keyboard decimal marker between columns
four and five. This determines the position of the decimal point for
all parts of the example.
4. Set the first dividend in the keyboard, being sure that the
number is so placed that the decimal marker falls between the
correct numbers. Here 38.1022 should be set in columns six through
one, with the decimal marker between the 8 and the 1.
5. Touch the DIV'D TAB key.
6. Set the upper dial decimal between the same figures as it was
in the keyboard (here between the 8 and the 1, dials ten and nine).
The number of decimal places in the upper dials must be equal to
the sum of the number of places in the lower dials and the number
in the keyboard. Here there are nine places to the right of the
decimal in the upper dials, which is the sum of five (lower dials)
and four (keyboard).
7. Set the divisor 19.779 in keyboard columns six through two
with the decimal between the first 9 and 7.
8. Touch divide keys.
9. The quotient 1.92639 appears in the lower dials. Since only
four place accuracy is desired, round this quotient and record as
1.9264.
10. Repeat the operations for parts b. and c. of the series.
11. If the dividend in the upper dials has a greater number of
102
Machine Computation of Elementary Statistics
digits to the left of the decimal point than the divisor set in the
keyboard, as in part d. of the series, move the carriage to the right
until the first digits of the dividend and the divisor line up. Then
touch the divide keys.
12. In part e. of this series, the first digit of the divisor is to the
left of the first digit of the dividend. Since the difference here is
small (only two places) it is not necessary to line up the first digits
as was done in step 11. However, if the difference were greater,
moving the carriage to line up the first two digits would make the
division faster. This is not necessary, however.
PROBLEMS: Division of a series of problems with preset decimals.
Report quotients correct to six decimal places. Check by multi-
plying quotient by divisor.
(1) 547.60036 ÷ 1.62
(2) 5128 ÷ 29.04
(3) 416.637 ÷ 29.04
(4) 197.882 ÷ 32.445
(5) 55.021 ÷ 99.119
(6) 789.43009 ÷ 0.98
(7) 0.48596 ÷ 80.76
(8) 0.0925 ÷ 0.682
(9) 847.000898,253.09
=
rounded to
=
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
Marchant
(a) Numbers in extreme left of keyboard
EXAMPLE: 7,659.96 44.44 172.3663366+
÷
1. Be sure division clear-return key is down (normal position).
2. Depress selective tabulator key ten.
Division
103
3. Enter 765996 in extreme left of the keyboard in columns ten
through five.
4. Add.
5. Put the middle dial decimal marker in its place between the
two 9's (middle dials fifteen and sixteen).
6. Set the divisor 4444 in the extreme left of the keyboard in
columns ten through seven.
7. Turn up the keyboard decimal marker between columns
eight and nine where it actually appears in the divisor.
8. To determine the number of decimal places in the quotient,
subtract the number of places to the right of the decimal in the
divisor in the keyboard (here 8) from the number to the right
of the decimal in the dividend in the middle dials (here 15), and
set the upper dial decimal marker accordingly (here 15 -8= 7;
i.e., set decimal between upper dials seven and eight).
9. Touch automatic divide key. The number 1 clears out of the
upper dials and division is performed automatically.
10. The quotient 172.3663366 appears in the upper dials cor-
rectly pointed off; the middle dials and keyboard clear and the
carriage returns to the extreme right.
REMEMBER: The number of places to the right of the decimal in the
middle dials is equal to the number to the right of the decimal in
the upper dials plus the number to the right in the keyboard.
(Note that this is the same as multiplication.)
PROBLEMS: Division of numbers with decimals, working from extreme
left of keyboard. Check by multiplying divisor by quotient to get
the dividend as product. Remember that the number of decimal
places changes with each problem.
(1) 67,893.221 ÷ 569.003
(2) 20,589.02 ÷ 76.5992
(3) 4.211199 ÷ 9.91124
(4) 0.11239009 ÷ 0.11290909
(5) 0.055468 ÷ 3.45482
=
=
(b) Constant or preset decimal setup
The preset decimal setup is particularly useful for a series of
divisions where a fixed number of decimal places is required in the
quotient.
104
Machine Computation of Elementary Statistics
EXAMPLES: Given the following series of divisions to be performed;
in each problem the quotient is required correct to four decimal
places. Check by multiplying divisor by quotient; note that the
product in the check will agree in all but the last digit of the original
dividend.
a. 38.1022 ÷ 19.779
b. 94.56 ÷ 12.3 7.68780 rounded to 7.6878
c. 1.03 ÷ 8.93 = 0.11534 rounded to 0.1153
d. 8,756 ÷ 847.9
e. 0.0365 ÷ 1.12
=
=
********
1.92639 rounded to 1.9264
10.32668 rounded to 10.3267
0.03258 rounded to 0.0326
1. Be sure division clear-return key is down.
2. Since four decimal places are required in the quotient, the
division should be carried through five places and the quotient
rounded to four. Set the upper dial decimal marker between dials
five and six, and press selective tabulator key six. REMEMBER: press
the tab key which is the same number (here six) as the upper dial
immediately to the left of the upper dial decimal marker.
3. Examine the several dividends and divisors to determine the
maximum number of decimals in any number. In the example,
this is four. Turn up the keyboard decimal between columns four
and five. This determines the position of the decimal point for all
the examples.
4. Set the first dividend in the keyboard around the decimal
point. Here 38.1022 should be set in columns six through one,
with the decimal between the 8 and the 1.
5. Add.
6. Set the middle dial decimal marker between the same figures
as it was in the keyboard (here between the 8 and the 1, dials ten
and nine). The number of decimal places in the middle dials must
be equal to the number of places in the upper dials plus the number
in the keyboard. Here there are nine places to the right of the
decimal in the middle dials, which is the sum of five (upper dials)
and four (keyboard).
7. Set the divisor 19.779 in keyboard columns six through two
around the decimal.
8. Touch automatic divide key.
9. The quotient 1.92639 appears in the upper dials. Since only
four place accuracy is desired, round this quotient and record as
1.9264.
10. Repeat the operations for parts b. and c. of the series.
Division
105
11. If the dividend in the middle dials has a greater number of
digits to the left of the decimal point than the divisor set in the
keyboard, as in part d. of the example, move the carriage to the
right until the first digits of the dividend and the divisor line up.
Then touch the automatic divide key.
12. In part e. of this series, the first digit of the divisor is to the
left of the first digit of the dividend. Since the difference here is
small (only two places) it is not necessary to line up the first
digits as was done in step 11. However, if the difference were
greater, moving the carriage to line up the first two digits would
make the division faster. This is not necessary.
PROBLEMS: Division of a series of problems with preset decimals.
Report quotients correct to six decimal places. Check by multiply-
ing quotient by divisor.
(1) 547.60036 ÷ 1.62
(2) 5128 ÷ 29.04
(3) 416.637 ÷ 29.04
(4) 197.882 ÷ 32.445
(5) 55.021 99.119
(6) 789.43009 ÷ 0.98 =
(7) 0.48596 ÷ 80.76
(8) 0.0925 ÷ 0.682
(9) 847.0008 ÷ 98,253.09 =
=
=
=
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
Monroe: Automatic and Semiautomatic
(a) Numbers in extreme left of keyboard
EXAMPLE: 7,659.96 44.44
÷
172.3663366+
1. Carriage in position ten, to extreme right. On CAA machines,
depress tab stop ten.
106
Machine Computation of Elementary Statistics
2. Enter the dividend 765996 in extreme left of keyboard in
columns ten through five.
3. Add into machine, clear 1 from upper dials. (In CAA ma-
chines, depress DIVD TAB and the keys below it with the fore- and
middle fingers; this automatically positions the carriage to the
extreme right, enters the dividend in the lower dials, clears the 1
from the upper dials, and clears the keyboard.)
4. Put lower dial decimal marker in its place between the two
9's (lower dials fifteen and sixteen). This means that there are
fifteen decimal places to the right of the decimal in the lower dials.
5. Set the divisor 4444 in the extreme left of the keyboard
in columns ten through seven.
6. Turn up the keyboard decimal marker between columns
eight and nine where it actually appears in the divisor.
7. To determine the number of decimal places in the quotient,
subtract the number of places to the right of the decimal in the
divisor in the keyboard (here 8) from the number to the right of
the decimal in the dividend in the lower dials (here 15), and set the
left upper dial decimal marker accordingly (here 1587; i.e.,
set decimal marker between upper dials seven and eight).
8. Push divide lever. Division is performed automatically.
9. The quotient 172.3663366+ appears in the (left) upper dials.
REMEMBER: The number of places to the right of the decimal in
the lower dials is equal to the number to the right of the decimal
in the upper dials plus the number to the right in the keyboard.
(Note that this is the same as multiplication.)
PROBLEMS: Division of numbers with decimals, working from extreme
left of keyboard. Check by multiplying divisor by quotient to get
the dividend as product. Remember that the number of decimal
places changes with each problem.
(1) 67,893.221 ÷ 569.003 =
(2) 20,589.02 ÷ 76.5992
(3) 4.211199 ÷ 9.91124
(4) 0.11239009 ÷ 0.11290909
(5) 0.055468 ÷ 3.45482
←
(b) Constant or preset decimal setup
EXAMPLES: Given the following series of divisions to perform; in
each problem the quotient is required correct to four decimal
Division
107
places. Check by multiplying divisor by quotient; note that the
product in the check will agree in all but the last digit of the
original dividend.
a. 38.1022 ÷ 19.779 = 1.92639 rounded to 1.9264
b. 94.56 ÷ 12.3
7.68780 rounded to 7.6878
0.11534 rounded to 0.1153
8.93
*********
c. 1.03
d. 8,756
847.9
10.32668 rounded to 10.3267
e. 0.0365 ÷ 1.12 = 0.03258 rounded to 0.0326
=**
1. Put repeat key down.
2. Since four decimal places are required in the quotient, the
division should be carried through five places, and the quotient
rounded to four. Set the (left) upper dial decimal marker between
dials five and six. On CAA machines, press tab stop six; on other
models, move carriage to position six. NOTE: the number of the
position to which the carriage is moved is one more than the number
of decimal places set off for the quotient.
3. Examine the several dividends and divisors to determine
the maximum number of decimals in any number. In the example
this is four. Turn up the keyboard decimal marker between columns
four and five. This determines the position of the decimal point for
all parts of the example.
4. Set the first dividend in the keyboard, being sure that the
number is so placed that the decimal marker falls between the cor-
rect numbers. Here 38.1022 should be set in columns six through
one, with the decimal marker between the 8 and the 1.
5. Add. (In CAA machines, depress DIVD TAB.)
6. Set the lower dial decimal marker between the same figures
as it was in the keyboard (here between the 8 and the 1, dials ten
and nine). The number of decimal places in the lower dials must
be equal to the sum of the number of places in the upper dials and
the number in the keyboard. Here there are nine places to the
right of the decimal in the lower dials, which is the sum of five
(upper dials) and four (keyboard).
7. Set the divisor 19.779 in keyboard columns six through two
with the decimal between the first 9 and 7.
8. Divide.
9. The quotient 1.92639 appears in the upper dials. Since only
four place accuracy is desired, round this quotient and record as
1.9264.
108
Machine Computation of Elementary Statistics
10. Repeat the operations for parts b. and c. of the example.
Remember to move the carriage to position six before beginning
each part of the example.
11. If the dividend in the lower dials has a greater number of
digits to the left of the decimal point than the divisor set in the
keyboard, as in part d. of the example, move the carriage to the
right until the first digits of the dividend and the divisor line up.
Then divide.
12. In part e. of this series, the first digit of the divisor is to the
left of the first digit of the dividend. Time may be saved in division
by lining up the first digits of the dividend and divisor, even
though it is not absolutely necessary.
PROBLEMS: Division of a series of problems with preset decimals.
Report quotients correct to six decimal places. Check by multiply-
ing quotient by divisor.
(1) 547.60036 ÷ 1.62
(2) 5128 ÷ 29.04
(3) 416.637 ÷ 29.04
(4) 197.882 ÷ 32.445
(5) 55.021 ÷ 99.119
(6) 789.43009 ÷ 0.98
(7) 0.48596 ÷ 80.76
(8) 0.0925 0.682
(9) 847.0008 ÷ 98,253.09
=
=
rounded to
-
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
Monroe: Hand-operated
(a) Numbers in extreme left of keyboard
EXAMPLE: 7,659.96 44.44 172.3663366+
÷
1. Put repeat key down and counter control key at divide.
2. Move carriage to extreme right.
+
Division
109
3. Set 765996 in extreme left of keyboard, in columns ten
through six. (In Educator model, seven through two.)
4. Add; clear upper dials and keyboard. Set lower dial decimal
marker between the two 9's of the dividend, between dials fifteen
and sixteen. This means there are fifteen places to the right of the
dividend in the lower dials. (Educator: put decimal marker between
lower dials nine and eight; this means there are eight places to the
right of the decimal.)
5. Set the divisor 4444 in the extreme left of the keyboard in
columns ten through seven. (Educator: seven through four.)
6. Turn up the keyboard decimal marker between columns
eight and nine where it actually appears in the divisor. (Educator:
five and four.)
7. To determine the number of places in the quotient, subtract
the number of places to the right of the divisor in the keyboard
(here 8; Educator: 5) from the number to the right of the decimal
in the dividend (here 15; Educator: 8), and set the upper dial deci-
mal marker accordingly (here 1587; Educator, 8 - 5 = 3).
Set the upper dial decimal marker (either left or right bank, or
both) between dials seven and eight (Educator: three and four).
8. Divide. The quotient 172.3663366+ appears in the upper
dials correctly pointed off (Educator: 172.366+).
REMEMBER: The number of places to the right of the decimal in
the lower dials is equal to the number to the right of the decimal
in the upper dials plus the number to the right in the keyboard.
(Note that this is the same as multiplication.)
PROBLEMS: Division of numbers with decimals, working from extreme
left of keyboard. Check by multiplying divisor by quotient to get the
dividend as product. Remember that the number of decimal places
changes with each problem.
(1) 67,893.221 ÷ 569.003
(2) 20,589.02 76.5992 =
(3) 4.211199 ÷ 9.91124
(4) 0.11239009 0.11290909
(5) 0.055468 ÷ 3.45482
(b) Constant or preset decimal setup
The preset decimal setup is particularly useful for a series of
divisions where a fixed number of decimal places is required in the
quotients.
110
Machine Computation of Elementary Statistics
EXAMPLES: Given the following series of divisions to be performed;
in each problem the quotient is required correct to four decimal
places. Check by multiplying divisor by quotient; note that the
product in the check will agree in all but the last digit of the original
dividend.
a. 38.1022 ÷ 19.779 = 1.92639 rounded to 1.9264
÷
c. 1.03 ÷ 8.93
d. 8,756 ÷ 847.9
e. 0.0365 ÷ 1.12
b. 94.56 12.3 = 7.68780 rounded to 7.6878
0.11534 rounded to 0.1153
10.32668 rounded to 10.3267
0.03258 rounded to 0.0326
=
=
1. Put repeat key down and counter control key at divide.
2. Since four decimal places are required in the quotient, the
division should be carried through five places, and the quotient
rounded to four. Set the upper dial decimal marker between dials
five and six.
3. Examine the several dividends and divisors to determine the
maximum number of decimals in any number. In the example this
is four. Turn up the keyboard decimal marker between columns
four and five. This determines the position of the decimal point for
all parts of the example.
4. Set the first dividend in the keyboard, being sure that the
number is so placed that the decimal marker falls between the
correct digits. Here 38.1022 should be set in columns six through
one, with the decimal marker between the 8 and the 1.
5. Move the carriage to the right so that the carriage position
indicator points to the lower dial which is one immediately to the
left of the decimal marker in the upper dials. (Here, the pointer
would indicate lower dial six.)
6. Add; clear upper dials. Set the lower dial decimal marker
between the same figures as it was in the keyboard (here 8 and 1,
dials ten and nine). Note that the lower dial decimal marker is
directly above the keyboard decimal marker. The number of places
in the lower dials must be equal to the sum of the number of places
in the upper dials and the number in the keyboard. Here there are
nine places to the right of the decimal in the lower dials, which is
the sum of five (upper dials) and four (keyboard).
7. Set the divisor 19.779 in keyboard columns six through two
with the decimal between the first 9 and 7.
Division
111
8. Divide. The quotient 1.92639 appears in the upper dials.
Since only four place accuracy is desired, round this quotient and
record as 1.9264.
9. Repeat the operations for parts b. and c. of the series.
10. If the dividend in the lower dials has a greater number of
digits to the left of the decimal point than the divisor set in the
keyboard, as in part d. of the series, move the carriage to the right
until the first digits of the dividend and divisor line up. Then divide.
11. In part e. of this series, the first digit of the divisor is to the
left of the first digit of the dividend. Move the carriage to the
left until the first digits of dividend and divisor line up. Then
divide.
PROBLEMS: Division of a series of problems with preset_decimals.
Report quotients correct to six decimal places. Check by multi-
plying quotient by divisor.
(1) 547.60036 ÷ 1.62
(2) 5128 ÷ 29.04
=
=
(3) 416.637 29.04
(4) 197.882 ÷ 32.445
(5) 55.021 ÷ 99.119
(6) 789.43009 ÷ 0.98
(7) 0.48596 ÷ 80.76 =
=
(8) 0.0925 ÷ 0.682
(9) 847.0008 ÷ 98,253.09 =
=
=
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
rounded to
C. DIVISION BY A CONSTANT
Division by a constant is an operation used in many statistical
computations, e.g., percentages, birth rates, death rates, index
numbers. On the Friden machine it is possible to lock a constant
112
Machine Computation of Elementary Statistics
divisor in the keyboard, and enter the variable dividends in the
upper dials manually by the upper dial twirlers.
However, the quotient of a number X divided by another num-
ber Y is equivalent to the product of the first number X multiplied
by 1/1
Y'
X
11/17 - 11/x
X
Y
When dividing by a constant divisor, it is possible to utilize this
relationship by dividing 1 by the constant divisor to get a new con-
stant, which may be used as a constant multiplier.
This quotient of 1 divided by a number is called the reciprocal of
the number. (See next section.)
D. RECIPROCALS
The reciprocal of any number is 1 divided by that number. For
example, 1 divided by 5 is 0.2; therefore the reciprocal of 5 is 0.2.
Friden
(a) Reciprocal of a number greater than 1
The reciprocal of a number greater than 1 is a decimal.
EXAMPLE: Find the reciprocal of 1,319; i.e., divide 1 by 1,319.
1 ÷ 1,319 0.000758150114
=
1. Depress add key.
2. Depress tab key ten.
3. Depress 1 in column ten.
4. Touch DIV’D TAB key.
5. Set 1,319 in keyboard columns nine through six. Touch both
divide keys.
6. The digits 7581501137 appear in lower dials ten through one.
7. To determine the place of the decimal point of the reciprocal,
remember first that the reciprocal of a number greater than 1 is a
decimal. There is a relationship between the number of digits to the
left of the decimal of the original number and the number of zeros
to the right of the decimal of the reciprocal, expressed by the fol-
lowing rule:
Prefix to the reciprocal one less zero than there are digits to
the left of the decimal in the original number; e.g., 1,319 has
Division
113
four digits to the left of the decimal; the reciprocal has three
(4 1 = 3) zeros to the right of the decimal point
0.000758150114. Check by multiplying the original number
1,319 by the reciprocal. The product is approximately 1 (ten
9's followed by other digits).
▬▬▬▬▬
PROBLEMS: Reciprocals. Note that the first digit of the number
must always be entered on the keyboard in the column immediately
to the right of the 1 in the upper dials in order to obtain the maxi-
mum number of significant digits in the reciprocal. If the maximum
number is not needed, depress the division stop key at the desired
point. Check by multiplying the original number by the reciprocal
to get 1 (approximately).
(1) 367
(2) 1,462
(3) 6,312
(4) 6,252
(5) 63.77
(b) Reciprocal of a number less than 1
The reciprocal of a number less than 1 is greater than 1; 1 di-
vided by a decimal fraction gives a quotient greater than 1; e.g.,
1 ÷ 0.5 = 2.0. There is a relationship between the number of
O's to the right of the decimal of the original number and the num-
ber of digits to the left of the decimal point of the reciprocal. This
is expressed by the following rule:
Point off one more digit to the left of the decimal point in the
reciprocal than there are O's to the right of the decimal in the
original number; e.g., the reciprocal of 0.1319 is 7.58150114;
the reciprocal of 0.001319 is 758.150114.
PROBLEMS: Find reciprocals to nine significant digits. Check by
multiplying the original number by the reciprocal.
(6) 49.3604
(7) 0.0896
(8) 99.9
(9) 109
(10) 0.0109
(11) 3.439
(12) 343.9
114
Machine Computation of Elementary Statistics
(13) Inspect your answers for (9) and (10), and for (11) and (12).
Formulate another rule for reciprocals.
(14) Compute the cumulative percentage distribution of the fol-
lowing group data (fictitious) by finding the percentage of
cases with scores less than the upper limit of each interval of
the frequency distribution. To find each cumulative per-
centage value, it is necessary to divide each cumulative fre-
quency by the total number of cases in the distribution.
Score
150-159
140-149
130-139
120-129
110-119
100-109
90-99
80-89
70-79
60-69
50-59
40-49
30-39
20- 29
10- 19
0 9
Frequency
2378
7
5
6
15
19
17
8
6
8
4
6
2
Cumulative
Frequency
123
121
118
111
103
96
91
85
70
51
34
26
1. Depress selective tab key ten.
2. Set 1 in column ten.
20
12
NOOR
2
Cumulative
Percentage
100.0
74.0
16.3
(a) Find the reciprocal of 123 (the total number of cases in the
distribution).
(b) Lock the reciprocal in the keyboard.
(c) Multiply the reciprocal by each cumulative frequency.
(d) Enter values in table in cumulative percentage column, round-
ing to one decimal place.
Marchant
(a) Reciprocal of a number greater than 1
The reciprocal of a number greater than 1 is a decimal.
EXAMPLE: Find the reciprocal of 1,319; i.e., divide 1 by 1,319.
1 ÷ 1,319 0.000758150114
Division
115
3. Add into machine.
4. Set 1,319 in keyboard columns nine through six.
5. Divide.
6. The digits 7581501137 appear in upper dials ten through one.
7. To determine the place of the decimal point of the reciprocal,
remember first that the reciprocal of a number greater than 1 is a
decimal. There is a relationship between the number of digits to
the left of the decimal of the original number and the number of O's
to the right of the decimal of the reciprocal. This is expressed by
the following rule:
Prefix to the reciprocal one less 0 than there are digits to
the left of the decimal in the original number; e.g., 1,319
has four digits to the left of the decimal; the reciprocal has
three (4 – 1 = 3) 0's to the right of the decimal point
0.000758150114. Check by multiplying the original number
1,319 by the reciprocal. The product is approximately 1
(ten 9's followed by other digits).
(1) 367
(2) 1,462
(3) 6,312
PROBLEMS: Reciprocals. Note that the first digit of the number
must always be entered on the keyboard in the column immediately
to the right of the 1 in the middle dials in order to obtain the maxi-
mum number of significant digits in the reciprocal. If the maximum
number is not needed, depress the division stop key at the desired
point. Check by multiplying the original number by the reciprocal
to get 1 (approximately).
(4) 6,252
5) 63.77
(b) Reciprocal of a number less than 1
The reciprocal of a number less than 1 is greater than 1; 1 di-
vided by a decimal fraction gives a quotient greater than 1; e.g.,
10.5 = 2.0. There is a relationship between the number of O's
to the right of the decimal of the original number and the number
of digits to the left of the decimal point of the reciprocal, expressed
by the following rule:
Point off one more digit to the left of the decimal point in the
reciprocal than there are O's to the right of the decimal in the
original number; e.g., the reciprocal of 0.1319 is 7.58150114;
the reciprocal of 0.001319 is 758.150114.
116
Machine Computation of Elementary Statistics
1
PROBLEMS: Reciprocals to nine significant digits. Check by multi-
plying the original number by the reciprocal.
(6) 49.3604
(7) 0.0896
(8) 99.9
(9) 109
(10) 0.0109
(11) 3.439
(12) 343.9
(13) Inspect your answers for (9) and (10), and for (11) and (12).
Formulate another rule for reciprocals.
(14) Compute the cumulative percentage distribution of the
following group data (fictitious) by finding the percentage of
cases with scores less than the upper limit of each interval of
the frequency distribution. To find each cumulative per-
centage value it is necessary to divide each cumulative fre-
quency by the total number of cases in the distribution.
Score
150-159
140-149
130-139
120-129
110-119
100-109
90-99
80-89
1
70-79
60-69
50- 59
40-49
30-39
20-29
10- 19
0 9
Frequency
237876OL OF
5
15
19
17
8
6
8
4
+62
Cumulative
Frequency
123
121
118
111
103
96
91
85
25*220*2
34
12
Cumulative
Percentage
100.0
74.0
16.3
(a) Find the reciprocal of 123 (the total number of cases in the
distribution).
(b) Enter the reciprocal in the keyboard, and lock the keyboard
dial clearance key.
Division
117
(c) Multiply the reciprocal by each cumulative frequency.
(d) Enter values in table in the cumulative percentage column,
rounding to one decimal place.
Monroe: Automatic and Semiautomatic
(a) Reciprocal of a number greater than 1
The reciprocal of a number greater than 1 is a decimal.
EXAMPLE: Find the reciprocal of 1,319; i.e., divide 1 by 1,319.
1 ÷ 1.319 0.000758150114
=
1. Put carriage in position ten (in CAA machines: depress tab
stop ten).
2. Depress 1 in keyboard column ten. Add (in CAA machines:
depress DIVD TAB).
3. Clear keyboard and upper dials.
4. Set 1,319 in keyboard columns nine through six. Divide.
5. The digits 7581501137 appear in upper dials nine through
one.
6. To determine the place of the decimal point of the reciprocal,
remember that the reciprocal of a number greater than 1 is a deci-
mal. There is a relationship between the number of digits to the
left of the decimal of the original number and the number of O's
to the right of the decimal of the reciprocal. This is expressed by
the following rule:
(1) 367
(2) 1,462
Prefix to the reciprocal one less 0 than there are digits to
the left of the decimal in the original number; e.g., 1,319 has
four digits to the left of the decimal; the reciprocal has
three (413) 0's to the right of the decimal point
0.000758150114. Check by multiplying the original number
1,319 by the reciprocal. The product is approximately 1 (ten
9's followed by other digits).
PROBLEMS: Reciprocals. Note that the first digit of the number must
always be entered on the keyboard in the column immediately to
the right of the 1 in the lower dials in order to obtain the maximum
number of significant digits in the reciprocal. Check by multiply-
ing the original number to get 1 (approximately).
+
118
Machine Computation of Elementary Statistics
(3) 6,312
(4) 6,252
(5) 63.77
(b) Reciprocal of a number less than 1
The reciprocal of a number less than 1 is greater than 1; 1 di-
vided by a decimal fraction gives a quotient greater than 1; e.g.,
10.5 2.0. There is a relationship between the number of O's
to the right of the decimal of the original number and the number
of digits to the left of the decimal point of the reciprocal, expressed
by the following rule:
=
Point off one more digit to the left of the decimal point in
the reciprocal than there are O's to the right of the decimal
in the original number; e.g., the reciprocal of 0.1319 is
7.58150114; the reciprocal of 0.001319 is 758.150114.
PROBLEMS: Reciprocals to nine significant digits. Check by multi-
plying the original number by the reciprocal.
(6) 49.3604
(7) 0.0896
(8) 99.9
(9) 109
(10) 0.0109
(11) 3.439
(12) 343.9
(13) Inspect your answers for (9) and (10), and for (11) and (12).
Formulate another rule for reciprocals.
(14) Compute the cumulative percentage distribution of the fol-
lowing group data (fictitious) by finding the percentage of
cases with scores less than the upper limit of each interval of
the frequency distribution. To find each cumulative per-
centage value it is necessary to divide each cumulative fre-
quency by the total number of cases in the distribution.
Division
119
Score
150-159
140-149
130-139
120-129
110-119
100-109
90-99
80-89
70-79
60- 69
50-59
40- 49
30-39
20-29
10- 19
0- 9
Frequency
2378760-7∞∞∞∞
5
←
15
19
17
4
Cumulative
Frequency
123
121
118
111
103
96
91
85
70
51
34
26
20
12
8
2
Cumulative
Percentage
100.0
Monroe: Hand-operated
74.0
16.3
(a) Find the reciprocal of 123 (the total number of cases in the
distribution).
(b) Lock the reciprocal in the machine as a constant multiplier.
(c) Multiply each cumulative frequency by the reciprocal.
(d) Enter values in table in cumulative percentage column, round-
ing to one decimal place.
(a) Reciprocal of a number greater than 1
The reciprocal of a number greater than 1 is a decimal.
EXAMPLE: Find the reciprocal of 1,319; i.e., divide 1 by 1,319.
1 ÷ 1,319 0.000758150114
1. Repeat key down, counter control key at divide.
2. Move carriage to extreme right.
3. Set 1 in keyboard column ten (Educator: seven); then in key-
board columns nine through six (Educator: six through three),
set the number whose reciprocal is being found (1,319).
4. Add. Clear the first 1 in column ten (seven) out of the key-
board by depressing the zero key in that column.
120
Machine Computation of Elementary Statistics
5. Divide. Note that the first subtraction clears the 1 from the
upper dials and the 1,319 from the lower dials, leaving 1 followed
by O's as the dividend in the lower dials. The number 1,319, the
divisor, is still in the keyboard.
6. The digits 7581501137 appear in upper dials ten through one
(Educator: 758150 in upper dials six through one).
7. To determine the place of the decimal point of the reciprocal,
remember first that the reciprocal of a number greater than one is a
decimal. There is a relationship between the number of digits to
the left of the decimal of the original number and the number of O's
to the right of the decimal of the reciprocal. This is expressed by
the following rule:
Prefix to the reciprocal one less 0 than there are digits to
the left of the decimal in the original number; e.g., 1,319
has four digits to the left of the decimal; the reciprocal has
three (413) 0's to the right of the decimal point
0.000758150114. Check by multiplying the original number
1,319 by the reciprocal. The product is approximately 1 (ten
9's followed by other digits).
PROBLEMS: Reciprocals. Check by multiplying the original number
by the reciprocal to get 1 (approximately).
(1) 367
(2) 1,462
(3) 6,312
(4) 6,252
(5) 63.77
(b) Reciprocal of a number less than 1
The reciprocal of a number less than 1 is greater than 1; 1 di-
vided by a decimal fraction gives a quotient greater than 1; e.g.,
1 ÷ 0.5 = 2.0. There is a relationship between the number of O's
to the right of the decimal of the original number and the number
of digits to the left of the decimal point of the reciprocal. This is
expressed by the following rule:
Point off one more digit to the left of the decimal point in the
reciprocal than there are O's to the right of the decimal in the
original number; e.g., the reciprocal of 0.1319 is 7.58150114;
the reciprocal of 0.001319 is 758.150114.
Division
121
PROBLEMS: Reciprocals to nine significant digits. Check by multi-
plying the original number by the reciprocal.
(6) 49.3604
(7) 0.0896
(8) 99.9
(9) 109
(10) 0.0109
(11) 3.439
(12) 343.9
(13) Inspect your answers for (9) and (10), and for (11) and (12).
Formulate another rule for reciprocals.
(14) Compute the cumulative percentage distribution of the fol-
lowing group data (fictitious) by finding the percentage of
cases with scores less than the upper limit of each interval of
the frequency distribution. To find each cumulative percent-
age value it is necessary to divide each cumulative frequency
by the total number of cases in the distribution.
Score
150-159
140-149
130-139
120-129
110-119
100-109
90-99
80-89
70-79
60-69
50- 59
40-49
30-39
20- 29
10- 19
0 9
Frequency
237875
6
15
19
17
8
6
8462
Cumulative
Frequency
123
121
118
111
103
96
3873 REH2 22*~
Cumulative
Percentage
100.0
74.0
16.3
(a) Find the reciprocal of 123 (the total number of cases in the
distribution).
(b) Lock the reciprocal in the keyboard.
122
Machine Computation of Elementary Statistics
(c) Multiply the reciprocal by each cumulative frequency.
(d) Enter values in table in cumulative percentage column, round-
ing to one decimal place.
E. USE OF TABLES OF RECIPROCALS
Consult Barlow's Tables (21) and verify the values of the re-
ciprocals of problems 1 to 5 and 7 to 12 on page 120. Enter the
table at the number whose reciprocal is being sought, and read the
entry in the
1
column.
n
Note that it will be necessary to apply the rules for determining
the decimal point of the reciprocal.
F. USE OF TABLES OF QUOTIENTS
The student will often have occasion to handle division of one-
and two-digit numbers. A rapid method of determining the quo-
tients of such numbers to four decimal places is to read them from a
table. Holzinger (25) includes such a table (Table III, p. 28-47) of
Quotients of Integers to 100 ÷ 99.
EXAMPLE: 68 ÷ 56 = 1.214
1. Find the column headed 68 as the dividend.
2. Read down the divisor column to 56.
3. Read the quotient 1.214 in the cell where the column and row
cross.
PROBLEMS: Check by multiplying quotient by divisor to get divi-
dend as product.
(1) 47 ÷ 29
(2) 50 ÷ 91
(3) 62 ÷ 79
(4) 80 ÷ 81
(5) 2 ÷ 83
-
CHAPTER 6
SQUARE ROOT
There are several methods by which the root of a number may
be extracted. Two of the machine methods will be outlined here:
subtraction of successive odd numbers, and approximation by
division.
A. EXTRACTING THE SQUARE ROOT OF A NUMBER BY
SUBTRACTION OF SUCCESSIVE ODD NUMBERS
Friden
(a) Roots of whole numbers
EXAMPLE: (2,401)
49
First estimate the magnitude of the root by pointing off groups
of two digits to the left of the decimal. Here there are two such
groups 24'01. The square root of 24 would be greater than 4
(4 X 4 16) but less than 5 (5 × 5 25). The square root of
2,400 would be greater than 40 (40 X 40 1,600) but less than 50
×
(50 × 50 = 2,500); therefore it will be of a magnitude between 40
and 50, much nearer to 50. Having determined this, go on with the
machine computation by the method of subtraction of successive
odd numbers.
1. Depress tabulator key ten.
2. Put add key up.
3. Put counter control key down. This causes the lower dials
to count forward in subtraction.
4. Set 2,401 in extreme left of keyboard and touch DIV'D TAB
key.
5. Clear keyboard.
6. Starting with the decimal point, point off the figures in the
upper dials to the left in groups of two, 24'01. There will be as
many figures in the root before the decimal place as there are
groups of figures in the original number. Since there are two groups
123
124
Machine Computation of Elementary Statistics
in this example, start with the lower dial indicated by the carriage
position indicator arrow as one, count two, and set the lower dial
decimal marker to the right of the second dial, i.e., between dials
nine and eight.
7. Directly below the decimal marker in the upper dials,
indicating the position of the first group (here between 24 and 01),
turn over a keyboard marker. Using the left hand, in the column
directly to the left of the keyboard decimal marker, depress the
1 key, i.e., the first odd number to be subtracted.
There is a general rule for starting the subtraction: When the
number of digits to the left of the decimal point is even, start in
the next to the last column on the left of the keyboard; when the
number of digits is odd, start in the last column on the left.
8. Subtract the 1. Be sure to subtract only once.
9. The next odd number is 3; so depress 3 in the same column,
and subtract once.
10. Depress the next odd number 5 and subtract once; then de-
press 7 and subtract. Note that the value 801 appears in the upper
dials and 4 in the lower dials.
11. The next odd number is 9. If this were subtracted, a negative
number would result, as 9 is greater than the 8 which appears in
the dial directly above the column in which the numbers are being
depressed; and it would be necessary to add the 9 back in again.
If noted in time, it is not necessary to do this; if not, a bell will ring
when the number is subtracted, and a row of 9's will appear in the
upper dials, indicating that the result of the subtraction is negative.
Whenever the number in the keyboard is about to become greater
than the value directly above it in the dials, make the next number
even. Thus, in this case, change the 7 to 8, i.e., increase the value
in the keyboard by one, but do not subtract. Note that now the
value in the keyboard is twice the root already found (8 = 2 × 4).
X
12. Shift the carriage one space to the left by touching the upper
carriage shift key (arrow points to the left).
13. Set 1 in the column to the right of the column in which the
8 is depressed. The keyboard register reads 81.
14. Subtract the 81. The next odd number is 83; so depress 3 in
the column where the 1 was depressed.
15. Subtract successively 83, 85, 87, 89. The next odd number
is 91.
Square Root
125
16. Set 91 in the keyboard in the same columns in which 89 was
set.
17. Subtract successively 91, 93, 95, 97.
18. With the subtraction of 97, the upper dials all read 0. The
lower dials read 49.000000000, the exact square root of 2401.
Note that the root is one-half the value in the keyboard plus 1
(97 + 1 = 98 = 49 × 2).
EXAMPLE: (5,929)*
1. Depress tabulator key ten; be sure add key is up, counter
control key down.
2. Set 5,929 in extreme left of the keyboard, and touch DIV'D
TAB key.
=
77
3. Clear keyboard.
4. Point off 5,929 in groups of two, starting from the decimal
point and set the lower dial decimal marker for two digits to the
left of the decimal in the root, i.e., between dials nine and eight.
5. Using the general rule for starting the subtraction of suc-
cessive odd numbers, set the 1 in the keyboard column nine.
6. Subtract successively 1, 3, 5, 7, 9.
7. The next odd number is 11. Set the tens 1 in the column to
the left of that in which the 9 is depressed (column ten) and the
units 1 in the same column as that in which 9 is depressed (column
nine).
8. Subtract successively 11, 13.
9. With subtraction of 13, the numbers in the two upper dials
directly above the keyboard columns in which the 13 is set become
10, less than 15, the next successive odd number.
10. Change the 13 to 14. Note that the value in the keyboard
14 is twice that in the lower dials.
11. Shift the carriage one space to the left and depress 1 in the
column to the right of the column in which the 4 is set. The key-
board register reads 141.
12. Subtract successively 141, 143, 145, 147, 149.
13. With the subtraction of 149, the number in the three upper
dials above the keyboard columns in which 149 is depressed is 304,
i.e., larger than the next successive odd number, 151. So set 5 and 1
in the columns in which the 4 and 9 were set.
14. Subtract 151 and 153. With the subtraction of 153, the upper
126
Machine Computation of Elementary Statistics
dials register O's. The lower dials read 77 000000000, the exact
root of 5,929.
15. As a check, 77 is one-half the last number subtracted plus
1 (153 + 1 = 154 = 2 X 77).
16. Some computers prefer to listen for the bell to indicate
oversubtraction. However, there are two points to be kept in mind
if this method is preferred: (1) the bell does not always ring, and
(2) there are some machines which do not have a bell (e.g., Mar-
chant).
PROBLEMS: Extraction of square roots, having first estimated the
magnitude of each one. Remember this is a process of subtraction of
successive odd numbers. Check by multiplication.
(1) 8,836
(3) 6,724
(5) 7,921
(2) 2,704
(4) 9,801
(b) Roots of decimals
EXAMPLE: (0.49815364)
0.7058
1. Depress tabulator key ten; be sure add key is up, counter
control key down.
2. Set 49815364 in extreme left of keyboard; touch DIV'D TAB
key.
3. Place an upper dial decimal marker to the left of the first
4 to indicate the decimal point of the number. Point off periods of
two digits to the right of the decimal, and approximate the root.
It will be estimated as a little greater than 0.7000 [(0.7000)2
0.49000000].
-
=
4. Mark the decimal place in the root by putting the lower dial
decimal marker directly to the left of the dial indicated by the
carriage position indicator, here to the left of dial ten.
5. Perform the first series of subtractions: 1, 3, 5, 7, 9, then 11,
and 13. In setting in the 11, the units 1 is in the column where the
9 was, and the tens 1 is in the column to the left.
6. After the subtraction of 13, there are two zeros in the upper
dials above the two columns in which 13 is set. Move up one key
to 4 in the column where the 3 is depressed. Check to see that the
number in the keyboard register is 14, twice the root in the lower
dials 7.
Square Root
127
7. Move carriage one place to the left, and set 1 in the next
column of the keyboard. The register now reads 141. The three
dials above the three columns in which the 141 is set read 081.
Since 141 is greater than 081, it cannot be subtracted from it with-
out obtaining a negative number. Therefore, take out the 1 just
set in, move the carriage another space to the left, and set a 1 in
the next column. The keyboard register now reads 1401, which can
be subtracted from the number 8153 in the upper dials directly
above.
8. Subtract 1401, 1403, 1405, 1407, 1409. The next odd number
is 1411, which is greater than 1128 in the dials directly above.
9. Therefore, take out the last 1; note that the number in the
keyboard 1410 is twice the root in the lower dials 705.
10. Move the carriage one place to the left, and set a 1 in the
next column. Subtract successively 14101, 14103, 14105, 14107,
14109, 14111, 14113, 14115. With the subtraction of 14115, the
upper dials all read 0, and the lower dials 7058, with the decimal
marker to the left of the 7; note that the number in the keyboard
plus 1 (14116) is twice the root 7058. Read the exact root of
0.49815364 as 0.7058.
PROBLEMS: Extraction of square roots. Remember that the decimal
points change from problem to problem. Check by multiplication;
in some cases the square will only approximate the original num-
ber. Be sure to clear the machine before each new problem.
(6) 228.01
(7) 0.38416
(8) 442,500
(9) 8,482.84
(10) 848.284
(11) 84.8284
(12) Inspect the answers to (9), (10), and (11). What conclusions
can you draw from these answers? (At least two.)
(13) 3.4641016
(14) 6.9282032
(15) 6.3245553
(16) 12.6491106
(17) Inspect the answers to (13) and (14), then (15) and (16),
and compare them with the relationship between the original num-
bers. What conclusion can you draw from the comparison?
128
Machine Computation of Elementary Statistics
Marchant*
(a) Roots of whole numbers
EXAMPLE: (2,401)}
= 49
First estimate the magnitude of the root by pointing off groups
of two digits to the left of the decimal. Here there are two such
groups 24'01. The square root of 24 would be greater than 4
(4 X 4 = 16) but less than 5 (5 × 5 = 25). The square root of
2,400 would be greater than 40 (40 X 40 1,600) but less than
50 (50 X 50 = 2,500); therefore, it will be of a magnitude between
40 and 50, much nearer to 50. Having determined this, go on with
the machine computation by the method of subtraction of suc-
cessive odd numbers.
=
1. Put nonshift key down.
2. Depress selective tab key ten.
3. Put counter control key down. This causes the upper dials
to count forward in subtraction.
4. Set 2,401 in extreme left of keyboard, and add into machine.
5. Clear upper dials.
6. Starting with the decimal point, point off the figures in the
middle dials to the left in groups of two, 24'01. There will be as
many figures in the root before the decimal place as there are
groups of figures in the original number. Since there are two groups
in this example, start with the upper dial indicated by the carriage
position indicator arrow as one, count two, and set the upper dial
decimal marker to the right of the second dial, i.e., between dials
nine and eight.
7. Directly below the decimal marker in the middle dials indi-
cating the position of the first group, between lower dials eighteen
and seventeen, turn over a keyboard marker. Using the left hand,
in the column directly to the left of the keyboard decimal marker,
depress the 1 key, i.e., the first odd number to be subtracted.
There is a general rule for starting the subtraction: When the
number of digits to the left of the decimal point is even, start in
the next to the last column on the left of the keyboard; when the
number of digits is odd, start in the last column on the left.
8. Subtract the 1 by touching the REVERSE key and the 1 of
the multiplier keyboard simultaneously.
* See reference 5, methods MM-302, MM-95, tables 56, 57.
Square Root
129
9. The next odd number is 3; so depress 3 in the same column,
and subtract once in the same manner. It is necessary to use the
reverse-multiplier 1 combination so that the keyboard will not
clear.
10. Depress the next odd number 5 and subtract once; then de-
press 7 and subtract. Note that the value 801 appears in the middle
dials and 4 in the upper dials.
11. The next odd number is 9. If this were subtracted, a negative
number would result, as 9 is greater than the 8 which appears in
the dial directly above the column in which the numbers are being
depressed; and it would be necessary to add the 9 back in again.
If noted in time, it is not necessary to do this; if not, a row of 9's
will appear in the middle dials, indicating that the result of the
subtraction is negative. If oversubtraction takes place, add the
amount back into the machine by depressing the multiplier 1 key.
Whenever the number in the keyboard is about to become greater
than the value directly above it in the dials, make the next number
even. Thus, in this case, change the 7 to 8, i.e., increase the value in
the keyboard by one, but do not subtract. Note that now the value
in the keyboard is twice the root already found (8 = 2 × 4).
X
12. Shift the carriage one space to the left by touching the upper
carriage shift key (arrow points to the left).
13. Set 1 in the column to the right of the column in which the
8 is depressed. The keyboard register reads 81.
14. Subtract the 81. The next odd number is 83; so depress 3 in
the column where the 1 was depressed.
15. Subtract successively 83, 85, 87, 89. The next odd number is
91.
16. Set 91 in the keyboard in the same columns in which 89
was set.
17. Subtract successively 91, 93, 95, 97.
18. With the subtraction of 97, the middle dials all read 0. The
upper dials read 49.000000000, the exact root of 2,401. Note
that the root is one-half the value in the keyboard plus 1
(97 + 1 = 98 = 49 × 2).
EXAMPLE: (5,929) = 77
1. Be sure that nonshift key is down; depress selective tab key
ten; put counter control key down.
130
Machine Computation of Elementary Statistics
2. Set 5,929 in extreme left of the keyboard, and add into the
machine.
3. Clear upper dials.
4. Starting with the decimal point, point off the figures in the
middle dials in groups of two, and set the upper dial decimal point
for two digits to the left of the decimal in the root, i.e., between
dials nine and eight.
5. Using the general rule for starting the subtraction of suc-
cessive odd numbers, set the 1 in keyboard column nine.
6. Using the REVERSE and multiplier keyboard 1 key, subtract
successively 1, 3, 5, 7, 9.
7. The next odd number is 11. Set the tens 1 in the column to
the left of that in which the 9 is depressed (column ten), and the
units 1 in the same column as that in which 9 is depressed (nine).
8. Subtract successively 11, 13.
9. With the subtraction of 13, the numbers in the two middle
dials, directly above the keyboard columns in which the 13 is set,
become 10, i.e., less than 15, the next successive odd number.
10. Change the 13 to 14. Note that the value in the keyboard 14
is twice that in the upper dials, 7.
11. Shift the carriage one place to the left and depress the 1 in
the column to the right of the column in which the 4 is set. The
keyboard register reads 141.
12. Subtract successively 141, 143, 145, 147, 149.
13. With the subtraction of 149, the number in the three middle
dials above the keyboard columns in which 149 is depressed is 304,
i.e., larger than the next successive odd number 151. So set 5
and 1 in the columns in which the 4 and 9 were set.
14. Subtract 151 and 153. With the subtraction of 153, the
middle dials register O's. The upper dials read 77.000000000, the
exact square root of 5,929.
15. As a check, 77 is one-half the last number subtracted plus
1 (153 + 1 = 154
2 X 77).
=
PROBLEMS: Extraction of square roots, having first estimated the
magnitude of each one. Check by multiplication. Remember this is a
process of subtraction of successive odd numbers.
(1) 8,836
(3) 6,724
(5) 7,921
(2) 2,704
(4) 9,801
Square Root
131
-
(b) Roots of decimals
EXAMPLE: (0.49815364)
0.7058
1. Depress selective tab key ten; be sure nonshift key and
counter control key are down.
2. Set 49815364 in extreme left of keyboard; add. Clear 1 from
upper dials.
3. Place a decimal marker to the left of the first 4 to indicate
the decimal point of the number. Point off periods of two digits to
the right of the decimal, and approximate the root. It will be a
little greater than 0.7000 [(0.7000)² = 0.49000000].
4. Mark the decimal place in the root by putting the upper dial
decimal marker directly to the left of the dial indicated by the
carriage position indicator, here to the left of dial ten. Turn over
the keyboard decimal marker between columns nine and eight to
indicate the place of the first group.
5. Perform the first series of subtractions: 1, 3, 5, 7, 9, then 11,
and 13. In setting in the 11, the units 1 is in the column where the
9 was,
and the tens 1 is in the column to the left.
6. After subtraction of 13, there are two O's in the middle dials
above the two columns in which 13 is set. Move up one key to 4
in the column where the 3 is depressed. Check to see that the num-
ber in the keyboard register is 14, twice the root in the upper dials 7.
7. Move carriage one place to the left, and set 1 in the next
column of the keyboard. The register now reads 141. The three
dials above the three columns in which the 141 is set read 081.
Since 141 is greater than 081, it cannot be subtracted from it with-
out obtaining a negative number. Therefore, take out the 1 just
set in, move the carriage another space to the left, and set a 1 in
the next column. The keyboard register now reads 1401, which
can be subtracted from the number 8153 in the middle dials directly
above.
8. Subtract 1401, 1403, 1405, 1407, 1409. The next odd number
is 1411, which is greater than 1128 in the dials directly above.
9. Therefore, take out the last 1; note that the number in the
keyboard 1410 is twice the root in the upper dials 705.
10. Move the carriage one place to the left, and set a 1 in the
next column. Subtract successively 14101, 14103, 14105, 14107,
14109, 14111, 14113, 14115. With the subtraction of 14115 the
132
Machine Computation of Elementary Statistics
middle dials all read 0; and the upper dials 7058, with the decimal
marker to the left of the 7; note that the number in the keyboard
plus 1, 14116, is twice the root 7058. Read the exact root of
0.49815364 as 0.7058.
PROBLEMS: Extraction of square roots. Remember that the decimal
points change from problem to problem. Check by multiplication;
in some cases, the square will only approximate the original
number. Be sure to clear the machine before each new problem.
(6) 228.01
(7) 0.38416
(8) 442,500
(9) 8,482.84
(10) 848.284
(11) 84.8284
(12) Inspect the answers to (9), (10), and (11). What conclusions
can you draw from these answers? (At least two.)
(13) 3.4641016
(14) 6.9282032
(15) 6.3245553
(16) 12.6491106
(17) Inspect the answers to (13) and (14), then (15) and (16), and
compare them with the relationship between the original
numbers. What conclusion can you draw from the com-
parison?
(a) Roots of whole numbers
EXAMPLE: (2,401)
49
=
First estimate the magnitude of the root by pointing off groups of
two digits to the left of the decimal. Here there are two such
groups 24'01. The square root of 24 would be greater than 4
(4 X 4 16) but less than 5 (5 × 5 25). The square root of
2,400 would be greater than 40 (40 × 40 1,600) but less than 50
(50 X 50 = 2,500); therefore, it will be of a magnitude between 40
and 50, much nearer to 50. Having determined this, go on with the
machine computation by the method of subtraction of successive
odd numbers.
=
-
Monroe: Automatic
=
1. Put carriage in position ten (in CAA machines: depress tab
stop ten).
2. Depress repeat key.
3. Change lever at divide.
Square Root
133
4. Set 2,401 in extreme left of keyboard. Add (in CAA ma-
chines: touch DIVD TAB).
5. Clear keyboard and upper dials.
6. Starting with the decimal point, point off the figures in the
lower dials to the left in groups of two, 24'01. There will be as many
figures in the root before the decimal place as there are groups of
figures in the original number. Since there are two groups, start
with (left) upper dial ten as one, count two, and set the (left) upper
dial decimal marker to the right of the second dial counted, i.e.,
between dials nine and eight.
7. Directly below the decimal marker in the lower dials, indi-
cating the position of the first group (here between 24 and 01),
turn over a keyboard marker. Using the left hand, in the column
directly to the left of the keyboard decimal marker, depress the
1 key, i.e., the first odd number to be subtracted.
There is a general rule for starting the subtraction: When the
number of digits to the left of the decimal point is even, start in
the next to the last column on the left of the keyboard; when the
number of digits is odd, start in the last column to the left.
8. Subtract the 1. Be sure to subtract only once.
9. The next odd number is 3; so depress 3 in the same column,
and subtract once.
10. Depress the next odd number 5 and subtract once; then de-
press 7 and subtract. Note that the value 801 appears in the lower
dials and 4 in the upper dials.
11. The next odd number is 9. If this were subtracted, a negative
number would result, as 9 is greater than the 8 which appears in
the dial directly above the column in which the numbers are being
depressed, and it would be necessary to add the 9 back in again.
If noted in time, it is not necessary to do this. Oversubtraction is
indicated by a row of 9's in the lower dials.
Whenever the number in the keyboard is about to become
greater than the value directly above it in the dials, make the next
number even. Thus, in this case, change the 7 to 8, i.e., increase the
value in the keyboard by one, but do not subtract. Note now
that the value in the keyboard is twice the root already found
(8 = 2 × 4).
X
12. Shift the carriage one space to the left by touching the
carriage shift key on which the arrow points to the left.
134
Machine Computation of Elementary Statistics
13. Set 1 in the column to the right of the column in which the
8 is depressed. The keyboard register reads 81.
14. Subtract the 81. The next odd number is 83; so depress 3 in
the column where the 1 was depressed.
15. Subtract successively 83, 85, 87, 89. The next odd number is
91.
16. Set 91 in the keyboard in the same columns in which 89
was set.
17. Subtract successively 91, 93, 95, 97.
18. With the subtraction of 97, the lower dials all read 0.
The upper dials read 49.00000000, the exact square root of 2,401.
Note that the root is one-half the value in the keyboard plus 1
(97 + 1 = 98 49 × 2).
=
EXAMPLE: (5,929)} = 77
1. Carriage in position ten (in CAA machines: tab stop ten).
2. Depress repeat key.
3. Change lever at divide.
4. Point off 5,929 in groups of two, starting from the decimal
point, and set the upper dial decimal marker for two digits to the
left of the decimal in the root, i.e., between dials nine and eight.
5. Using the general rule for starting the subtraction of suc-
cessive odd numbers, set the 1 in keyboard column nine.
6. Subtract successively 1, 3, 5, 7, 9.
7. The next odd number is 11. Set the tens 1 in the column to
the left of that in which the 9 is depressed (column ten) and the
units 1 in the same column as that in which the 9 is depressed
(column nine).
8. Subtract successively 11, 13.
9. With the subtraction of 13, the numbers in the two lower
dials, directly above the keyboard columns in which the 13 is set,
become 10, i.e., less than 15, the next successive odd number.
10. Change the 13 to 14. Note that the value in the keyboard, 14,
is twice that in the upper dials, 7.
11. Shift the carriage one place to the left and depress 1 in the
column to the right of the column in which the 4 is set. The key-
board register reads 141.
12. Subtract successively 141, 143, 145, 147, 149.
13. With the subtraction of 149, the number in the three lower
Square Root
135
dials above the keyboard columns in which 149 is depressed is 304,
i.e., larger than the next successive odd number, 151. So set 5 and
1 in the columns in which the 4 and 9 were set.
14. Subtract 151 and 153. With the subtraction of 153, the lower
dials register O's. The upper dials read 77.00000000, the exact root
of 5,929.
15. As a check, 77 is one-half the last number subtracted plus 1
(153 + 1 = 154 = 2 × 77).
PROBLEMS: Extraction of square roots, having first estimated the
magnitude of each one. Remember this is a process of subtraction
of successive odd numbers. Check by multiplying.
(1) 8,836
(3) 6,724
(5) 7,921
(2) 2,704
(4) 9,801
(b) Roots of decimals
Example: (0.49815364) = 0.7058
1. Put carriage in position ten (in CAA machines: depress tab
stop ten).
2. Depress repeat key.
3. Change lever at divide.
4. Set 49815364 in extreme left of keyboard. Add (Automatic:
touch DIVD TAB). Clear keyboard and upper dials.
5. Place a lower dial decimal marker to the left of the first 4
to indicate the decimal point of the number. Point off periods of
two digits to the right of the decimal, and approximate the root.
It will be a little greater than 0.7000 [(0.7000)² = 0.49000000].
6. Mark the decimal place in the root by putting the upper
dial decimal marker directly to the left of the dial numbered the
same as the lower dial indicated by the carriage position indicator,
here to the left of dial ten.
7. Perform the first series of subtractions: 1, 3, 5, 7, 9, then 11,
and 13. In setting in the 11, the units 1 is in the column where the
9 was, and the tens 1 is in the column to the left.
8. After the subtraction of 13, there are two zeros in the lower
dials above the two columns in which 13 is set. Move up one key to
14. Check to see that the number in the keyboard register is 14,
twice the root in the upper dials 7.
136
Machine Computation of Elementary Statistics
9. Move the carriage one place to the left, and set 1 in the next
column of the keyboard. The register now reads 141. The three
lower dials above the three columns in which 141 is set read 081.
Since 141 is greater than 081, it cannot be subtracted from it with-
out obtaining a negative number. Therefore, take out the 1 just set
in, move the carriage another space to the left, and set a 1 in the
next column. The keyboard register now reads 1401, which can be
subtracted from the number 8153 in the lower dials directly above.
10. Subtract 1401, 1403, 1405, 1407, 1409. The next odd num-
ber is 1411, which is greater than 1128 in the dials directly above.
11. Therefore, take out the last 1; note that the number in the
keyboard, 1410, is twice the root in the upper dials, 705.
12. Move the carriage one place to the left, and set a 1 in the
next column. Subtract successively 14101, 14103, 14105, 14107,
14109, 14111, 14113, 14115. With the subtraction of 14115 the
lower dials all read 0, and the upper dials 7508, with the decimal
marker to the left of the 7; note that the number in the keyboard
plus 1 is twice the root (14115 + 1 = 14116 2 X 7508). Read
the exact root of 0.49815364 as 0.7058.
=
PROBLEMS: Extraction of square roots. Remember that the decimal
points change from problem to problem. Check by multiplication;
in some cases, the square will only approximate the original num-
ber. Be sure to clear the machine before each new problem.
(6) 228.01
(7) 0.38416
(8) 442,500
(9) 8,482.84
(10) 848.284
(11) 84.8284
(12) Inspect the answers to (9), (10) and (11). What conclusions
can you draw from these answers? (At least two.)
(13) 3.4641016
(14) 6.9282032
(15) 6.3245553
(16) 12.6491106
(17) Inspect the answers to (13) and (14), then (15) and (16),
and compare them with the relationship between the original
numbers. What conclusion can you draw from the com-
parison?
Square Root
137
Monroe: Hand-operated
(K Models)
(a) Roots of whole numbers
EXAMPLE: (2,401)}
49
First estimate the magnitude of the root by pointing off groups
of two digits to the left of the decimal. Here there are two such
groups 24'01. The square root of 24 would be greater than 4
(4 × 4 16) but less than 5 (5 × 5 25). The square root of
2,400 would be greater than 40 (40 X 40 1,600) but less than 50
×
(50 × 50 = 2,500); therefore, it will be of a magnitude between
40 and 50, much nearer to 50. Having determined this, go on with
the machine computation by the method of subtraction of successive
odd numbers.
=
=
--
1. Put repeat key down and carriage at extreme right.
2. Put counter control lever at divide. This causes the right
upper dials to count forward in subtraction.
3. Set 2,401 in extreme left of keyboard. Add.
4. Clear upper dials and keyboard.
5. Starting with the decimal point, point off the figures in the
lower dials to the left in groups of two, 24 and 01. There will be as
many digits in the root before the decimal place as there are groups
of digits in the original number. Since there are two groups in this
example, start with right upper dial ten as one, count two, and set
the upper dial decimal to the right of the second dial, i.e., between
dials nine and eight.
6. Directly below the decimal marker in the lower dials, indi-
cating the position of the first group (here between 24 and 01),
turn over a keyboard marker. Using the left hand, in the column
directly to the left of the keyboard decimal marker, depress the
1 key, i.e., the first odd number to be subtracted.
There is a general rule for starting the subtraction: When the
number of digits to the left of the decimal point is even, start
in the next to the last column on the left of the keyboard; when
the number of digits is odd, start in the last column on the
left.
7. Subtract the 1. Be sure to subtract only once.
8. The next odd number is 3; so depress 3 in the same column,
and subtract once.
138
Machine Computation of Elementary Statistics
9. Depress the next odd number 5 and subtract once; then 7
and subtract. Note that the value 801 appears in the lower dials and
4 in the upper dials.
10. The next odd number is 9. If this were subtracted, a negative
number would result, as 9 is greater than the 8 which appears in
the dial directly above the column in which the numbers are being
depressed, and it would be necessary to add the 9 back in again.
If noted in time, it is not necessary to do this; if not, a bell will
ring when the number is subtracted, and a row of 9's will appear in
the lower dials, indicating that the result of the subtraction is
negative.
When the number in the keyboard is about to become greater
than the value directly above it in the dials, make the next number
even. Thus in this case, change the 7 to 8, i.e., increase the value in
the keyboard by one, but do not subtract. Note now that the value
in the keyboard is twice the root already found (8 = 2 × 4).
11. Shift the carriage one space to the left.
12. Set 1 in the column to the right of the column in which the
8 is depressed. The keyboard register reads 81.
13. Subtract the 81. The next odd number is 83; so depress 3 in
the column where the 1 was depressed.
14. Subtract successively 83, 85, 87, 89. The next odd number
is 91.
15. Set 91 in the keyboard in the same columns in which 89
was set.
16. Subtract successively 91, 93, 95, 97.
17. With the subtraction of 97, the lower dials all read 0. The
upper dials read 49.000000000, the exact square root of 2,401.
Note that the root is one-half the value in the keyboard plus 1
(97 + 1 = 98
98 = 49 × 2).
EXAMPLE: (5,929)
1. Put repeat key down and counter control key at divide.
2. Set 5,929 in extreme left of the keyboard. Add.
=
77
3. Clear upper dials and keyboard.
4. Point off 5,929 in groups of two, starting from the decimal
point, and set the upper dial decimal marker for two places to the
left of the decimal in the root, i.e., between upper dials nine and
eight.
Square Root
139
5. Using the general rule for starting the subtraction of suc-
cessive odd numbers, set the 1 in keyboard column nine.
6. Subtract successively 1, 3, 5, 7, 9.
7. The next odd number is 11. Set the tens 1 in the column to
the left of that in which the 9 is depressed (column ten), and the
units 1 in the same column as that in which 9 is depressed (column
nine).
8. Subtract successively 11, 13.
9. With subtraction of 13, the numbers in the two lower dials,
directly above the keyboard columns in which the 13 is set, become
10, i.e., less than 15, the next successive odd number.
10. Change the 13 to 14. Note that the value in the keyboard
14 is twice that in the upper dials.
11. Shift the carriage one space to the left and depress 1 in the
column to the right of the column in which the 4 is set. The key-
board register reads 141.
12. Subtract successively 141, 143, 145, 147, 149.
13. With the subtraction of 149, the number in the three lower
dials above the keyboard columns in which 149 is depressed, be-
comes 304, i.e., larger than the next successive odd number, 151.
So set 5 and 1 in the columns in which the 4 and 9 were set. The
keyboard register reads 151.
14. Subtract 151 and 153. With the subtraction of 153, the lower
dials register O's. The upper dials read 77.000000000, the exact
root of 5,929.
15. As a check, 77 is one-half the last number subtracted plus
1 (153 +1 154 = 2 × 77).
-
PROBLEMS: Extraction of square roots, having first estimated the
magnitude of each one. Check by multiplication Remember this
is a process of subtraction of successive odd numbers.
(1) 8,836
(3) 6,724
(5) 7,921
(2) 2,704
(4) 9,801
(b) Roots of decimals
EXAMPLE: (0.49815364)
= 0.7058
1. Put repeat key down, counter control lever at divide,
carriage at extreme right.
140
Machine Computation of Elementary Statistics
t
2. Set 49815364 in extreme left of keyboard. Add. Clear upper
dials and keyboard.
3. Place lower dial decimal marker to the left of the first 4 to
indicate the decimal point of the number. Point off periods of two
digits to the right of the decimal, and approximate the root. It
will be a little greater than 0.7000 [(0.7000)2 = 0.49000000].
4. Mark the decimal place in the root by putting the upper
dial decimal marker directly to the left of the last right upper dial
(ten).
5. Perform the first series of subtractions, setting them in the
next to the last column on the left since there is an even number of
digits in the first group: 1, 3, 5, 7, 9, then 11, and 13. In setting in
the 11, the units 1 is in the column where the 9 was, and the tens
1 is in the column to the left.
6. After the subtraction of 13, there are two zeros in the lower
dials above the two columns in which 13 is set. Move up one key
to 4 in the column where the 3 is depressed. Check to see that the
number in the keyboard register is 14, twice the root in the upper
dials, 7.
7. Move carriage one place to the left, and set 1 in the next
keyboard column. The register now reads 141. The three dials
above the three columns in which the 141 is set read 081. Since 141
is greater than 081, it cannot be subtracted from it without obtain-
ing a negative number. Therefore, take out the 1 just set in, move
the carriage another space to the left, and set a 1 in the next
column. The keyboard register now reads 1401, which can be sub-
tracted from the number 8153 in the lower dials directly above.
8. Subtract successively 1401, 1403, 1405, 1407, 1409. The next
odd number is 1411, but it is greater than 1128 in the dials directly
above.
9. Therefore, take out the last 1; note that the number in the
keyboard, 1410, is twice the root in the upper dials, 705.
10. Move the carriage one place to the left, and set a 1 in the
next column. Subtract successively 14101, 14103, 14105, 14107,
14109, 14111, 14113, 14115. With the subtraction of 14115 the
lower dials all read 0, and the upper dials 7058, with the decimal
marker to the left of the 7; note that the number in the keyboard
plus 1, i.e., 14116, is twice the root, 7058. Read the exact root of
0.49815364 as 0.7058.
Square Root
141
PROBLEMS: Extraction of square roots. Remember that the decimal
points change from problem to problem. Check by multiplication;
in some cases, the square will only approximate the original num-
ber. Be sure to clear the machine before each new problem.
(6) 228.01
(7) 0.38416
(8) 442,500
(9) 8,482.84
(10) 848.284
(11) 84.8284
(12) Inspect the answers to (9), (10), and (11). What conclusions
can you draw from these answers? (At least two.)
(13) 3.4641016
(14) 6.9282032
(15) 6.3245553
(16) 12.6491106
(17) Inspect the answers to (13) and (14), then (15) and (16), and
compare them with the relationship between the original
numbers. What conclusion can you draw from the com-
parison?
B. EXTRACTING THE SQUARE ROOT OF A NUMBER BY
APPROXIMATION BY DIVISION*
Friden
=
EXAMPLE: (13,125)}
114.564392
1. Depress tab key ten, put add key up and counter control key
up (normal position).
2. Set 13,125 in the extreme left of the keyboard and touch
DIV'D TAB key.
3. Determine the number of digits to the left of the decimal
in the root-three. Set the lower dial decimal between lower dials
eight and seven.
4. Approximate the root by inspection. Here, say 114+, about
115. Call 115.0000000 the first trial root. Set up a table as follows,
and enter values at each step.
* The basic principles of this set of operations were suggested by Mr. Herman
Hovemann of the New York office of the Marchant Calculating Machine
Company for the Marchant machine. The author utilized Mr. Hovemann's
suggestions in setting up procedures for the other machines.
142
Machine Computation of Elementary Statistics
First:
Second:
Third:
Final:
Approximations
(Trial
Roots)
115.0000000
114.5652173
114.5643923
114.564392
Quotient
114.1304347
114.5635674
key.
(b) Unlock the upper dials.
114.5643924
Difference
(Approx.-
Quotient)
0.8695653
0.0016499
0.0000001
5. Clear the keyboard. Divide 13,125 by 115, to get a quotient
of 114.1304347. Since this quantity is less than the first trial root,
that first approximation is too large; but 114.1304347 is too small.
Therefore, it is necessary to take the average of the two as the
second trial root. This may be done on the machine as follows:
(a) Without clearing machine, put add key up to facilitate
checking numbers placed in keyboard, and put carriage
in position one.
(b) To get the sum of the two numbers, copy the amount
from the keyboard (the trial divisor) 115.0000000 in the
multiplier unit. Accumulate multiply. The sum
229.1304347 appears in the lower dials. The number in
the upper dials is of no consequence in this operation
and will be cleared out automatically with the next
operation.
(c) Write this sum 229.1304347 in the keyboard.
(d) Multiply by the reciprocal of 2 (0.5) to get 114.56521735,
which is the new (second) trial root. Record in table.
(Note that the upper dial decimal must be moved one
place to the left to get the answer correctly pointed off.)
Do NOT clear the machine.
6. Divide the number whose root is being found by the second
approximation as follows:
(a) Lock upper dials, clear the keyboard, and set the original
dividend 13,125 in the machine by using the DIVD TAB
Square Root
143
(c) Move carriage to position two so that the first digit
in the lower dials is directly above the extreme left
keyboard column. Write the new approximation
114.5652173 in the keyboard, copying directly from the
upper dials.
(d) Subtract. Take the 5 in the first upper dial out by using
the dial twirler. The O's in the upper dials prove that
the new divisor was correctly copied.
(e) Clear lower dials manually, and return carriage to ex-
treme right.
(f) Divide. The quotient is 114.5635674. Record in table.
7. Following the procedure outlined in step 5, starting with
copying of the previous trial root 114.5652173 from the keyboard
into the multiplier unit, compute the average of this quotient
114.5635674 and the second trial root 114.5652173 to get a third
approximation 114.5643923. Note that the new quotient and the
previous approximation are closer together than the first trial root
and first quotient.
8. Following the procedure outlined in step 6, divide 13,125 by
the third trial root 114.5643923; the quotient is 114.5643924.
9. Since the difference between this quotient and the previous
(third) approximation is only 0.0000001, square the fourth approxi-
mation 114.5643924. The result is 13,125 followed by several O's.
With rounding, 114.564392 is considered the root correct to six
decimal places.
PROBLEMS: Redo problems 6 to 11 on page 141 by approximation
by division.
Marchant
EXAMPLE: (13,125)}
114.564392
1. Depress selective tab key ten, put counter control key up
(normal position).
2. Set 13,125 in the extreme left of the keyboard and add.
3. Determine the number of digits to the left of the decimal in
the root-three. Set the upper dial decimal between dials eight
and seven.
4. Approximate the root by inspection. Here, say 114+, about
115. Call 115.0000000 the first trial root. Set up a table as follows,
and enter values at each step.
144
Machine Computation of Elementary Statistics
First:
Second:
Third:
Final:
Approximations
(Trial
Roots)
115.0000000
114.5652173
114.5643923
114.564392
Quotient
114.1304347
114.5635674
114.5643924
Difference
(Approx.
Quotient)
0.8695653
0.0016499
0.0000001
5. Divide 13,125 by 115.0000000 to get a quotient of
114.1304347. Since this quantity is less than the first trial root,
the first approximation is too large, but 114.1304347 is too small.
Therefore, it is necessary to take the average of the two as the
second trial root. This may be done on the machine as follows:
(a) Put carriage in position ten.
(b) Add the first divisor 115.0000000 to the quotient in the
upper dials 114.1304347 by putting 115 in the multiplier
keyboard.
(c) Write the sum 229.1304347 in the keyboard.
(d) Clear the dials and multiply by the reciprocal of 2 (0.5)
to get 114.56521735, which is the new (second) trial root.
(Note that the middle dial decimal must be moved one
place to the left to get the answer correctly pointed off.)
6. Divide the number whose root is being found by the second
approximation as follows:
(a) Without clearing, move the carriage to the extreme right
and add in the original number 13,125.
(b) Move the carriage so that the first digit of the trial
divisor in the upper dials is directly above the keyboard
column on the extreme left, and write the new trial
divisor 114.5652173 from the dials in the keyboard.
(c) Take the divisor out of the dials by pressing REVERSE
and 1 key of the multiplier keyboard.
(d) Tab carriage to position ten and divide. The quotient is
114.5635674. Record in table.
Square Root
145
7. Following the procedure outlined in step 5, starting with
multiplication by the number in the keyboard to add it to the
quotient in the upper dials, compute the average of this quotient
114.5635674 and the second trial root 114.5652173 to get a third
approximation 114.5643923. Note that the new quotient and the
previous approximation are closer together than the first trial root
and first approximation.
8. Following the procedure outlined in step 6, divide 13,125 by
the third trial root 114.5643923; the quotient is 114.5643924.
9. Since the difference between this quotient and the previous
(third) approximation is only 0.0000001, square the fourth approxi-
mation 114.5643924. The result is 13,125 followed by several O's.
Therefore, 114.564392 is considered the root correct to six decimal
places.
PROBLEMS: Redo problems 6 to 11 on page 141 by approximation
by division.
Monroe: Automatic and Semiautomatic
EXAMPLE: (13,125)*
= 114.564392
1. Put carriage in position ten and repeat key down; put change
lever to multiply.
2. Set 13,125 in extreme left of keyboard. Add (in CAA ma-
chines: touch DIVD TAB), clear keyboard and upper dials.
3. Determine the number of digits to the left of the decimal
in the root-three. Set the upper dial decimal between upper dials
eight and seven.
4. Approximate the root by inspection. Here, say 114+, about
115. Call 115.0000000 the first trial root. Set up a table as follows,
and enter values at each step.
First:
Second:
Third:
Final:
Approximations
(Trial
Roots)
115.0000000
114.5652173
114.5643923
114.564392
Quotient
114.1304347
114.5635674
114.5643924
Difference
(Approx.-
Quotient)
0.8695653
0.0016499
0.0000001
146
Machine Computation of Elementary Statistics
5. Divide 13,125 by 115.0000000 to get a quotient of
114.1304347. Since this quantity is less than the first trial root,
the first approximation is too large, but 114.1304347 is too small.
Therefore, it is necessary to take the average of the two as the
second trial root. This may be done on the machine as follows:
(a) Put carriage in position one.
(b) To get the sum of the two numbers, put change lever
to multiply, clear lower dials, set up, multiply by 1. The
sum 229.1304347 appears in the right upper dials. Clear
lower dials and keyboard.
(c) Set up reciprocal of 2 (0.5). Write the sum from the
upper dials 229.1304347 in the keyboard, and multiply
to get 114.56521735, which is the new (second) trial
root. Record in table. Do NOT clear the machine.
6. Divide the number whose root is being found (13,125) by
the second approximation as follows:
(a) Lock lower dials, clear keyboard, move carriage to posi-
tion ten, enter original number 13,125 in extreme left
of keyboard. Add.
(b) Unlock lower dials.
(c) Move carriage to position two so that the first digit in
the upper dials is directly above the extreme left
keyboard column. Write the
Write the new approximation
114.5652173 in the keyboard, copying directly from the
lower dials.
(d) Subtract.
(e) Clear upper dials and return carriage to position ten.
Put change lever at divide.
(f) Divide. The quotient is 114.5635674. Record in table.
7. Following the procedure outlined in step 5, starting with
setting up the previous trial root 114.5652173, compute the average
of this quotient 114.5635674 and the second trial root 114.5652173
to get a third approximation 114.5643923. Note that the new
quotient and the previous approximation are closer together than
the first trial root and first quotient.
8. Following the procedure outlined in step 6, divide 13,125
by the third trial root 114.5643923; the quotient is 114.5643924.
9. Since the difference between this quotient and the previous
(third) approximation is only 0.0000001, square the fourth (final)
Square Root
147
approximation. The result is 13,125 followed by several O's. There-
fore, 114.564392 is considered the root correct to six decimal
places.
PROBLEMS: Redo problems 6 to 11 on page 141 by approximation
by division.
EXAMPLE: (13,125) = 114.564392
1. Put repeat key down, counter control lever at divide, and
carriage at extreme right.
2. Set 13,125 in extreme left of keyboard. Add. Clear upper
dials and keyboard.
3. Determine the number of digits to the left of the decimal in
the root-three. Set the right upper dial decimal marker between
dials seven and eight, and do the same for the left upper dial
decimal marker.
4. Approximate the root by inspection. Here, say 114+, about
115. Call 115.0000000 the first trial root. Set up a table as follows,
and enter values at each step.
First:
Monroe: Hand-operated
Second:
Third:
Final:
Approximations
(Trial
Roots)
115.0000000
114.5652173
114.5643923
114.564392
Quotient
114.1304347
114.5635674
114.5643924
Difference
(Approx.
Quotient)
0.8695653
0.0016499
0.0000001
5. Divide 13,125 by 115.0000000 to get a quotient of
114.1304347. Since this quantity is less than the first trial root,
the first approximation is too large, but 114.1304347 is too small.
Therefore, it is necessary to take the average of the two as the
second trial root. This may be done on the machine as follows:
(a) Put carriage in position one, lock right upper dial.
(b) Clear upper dials, unlock.
148
Machine Computation of Elementary Statistics
(c) To get the sum of the two numbers, multiply by the
previous trial root which is in the keyboard 115.0000000.
Since the counter control lever is at divide, negative
multiply to cause the right upper dials to count forward,
thus adding the multiplier to the number already in the
dials. Use left upper dial to check multiplier digits and
decimal point. The sum 229.1304347 appears in the
right upper dials. The numbers in the lower dials are of
no consequence.
(d) Write the sum 229.1304347 in the keyboard, clear both
sets of dials.
(e) Put carriage in position one. Multiply by the reciprocal
of 2 (0.5) to get 114.5652173, which is the new (second)
trial root. Record in table. Do not clear machine.
6. Divide the number whose root is being found by the second
approximation as follows:
(a) Put carriage to extreme right.
(b) Set 13,125 in extreme left of keyboard. Add.
(c) Move carriage so that the first digit of the new trial
divisor, still in the lower dials, is directly above the key-
board column on the extreme left.
(d) Clear keyboard, and copy the new divisor 114.5652173
from the lower dials directly to the keyboard.
(e) Subtract. The O's in the lower dials (except the first one)
prove that the divisor was correctly copied.
(f) Clear upper dials, move carriage to extreme right.
(g) Divide. The quotient is 114.5635674. Record in table.
7. Following the procedure outlined in step 5, starting with
locking of right upper dials, clearance of left upper dials, and multi-
plication by the number in the keyboard, compute the average of
this quotient 114.5635674 and the second trial root 114.5652173 to
get a third approximation 114.5643923. Note that the new quotient
and the previous approximation are closer together than the first
trial root and the first approximation.
8. Following the procedure outlined in step 6, divide 13,125 by
the third trial root 114.5643923; the quotient is 114.5643924.
9. Since the difference between this quotient and the previous
(third) approximation is only 0.0000001, square the fourth approxi-
mation 114.5643924. The result is 13,125 followed by several O's.
Square Root
149
Rounding, 114.564392 is considered the root correct to six decimal
places.
PROBLEMS: Redo problems 6 to 11 on page 141 by approximation
by division.
C. USE OF TABLES OF SQUARES AND SQUARE ROOTS
Whenever numbers are small, the use of a table of square roots
is the most satisfactory method of finding a root. In addition to
the Barlow (21) and Holzinger (25) tables and those in the Walker (19)
and Garrett (17) texts, the Dunlap and Kurtz tables (23) are recom-
mended. The student who plans to study in advanced statistics
courses should also know the Fisher and Yates (24) and Kelley (26)
tables.
EXAMPLE: (11,943)}
109.284034
1. Use Barlow's Tables. Find 11,943 in the first column (n).
2. Read across to the column headed √n. The number in the
cell, which marks the crossing of the row and column, is the square
root of 11,943 correct to six decimal places.
=
EXAMPLE: (119,430)} = 344.586458
1. Use Barlow's Tables. Find 11943 in the n column as before.
2. Read across the row to the column headed √10n. The num-
ber in the cell, which marks the crossing of the row and column, is
the square root of 10 × 11,943, correct to six decimal places.
EXAMPLE: (956,484)
978.0
1. This number is too large to be read from the n column of
any of the published tables. However, the square column of a table
of squares goes to 10,000 for numbers to 100; and to 1,000,000 for
numbers to 1,000. Barlow's Tables go to 156,250,000 in the squares
column.
=
2. Read the table in the squares column (n²), and find 95 64 84.
Read to the left along the row to the n column. The number 978
(n) is the square root of 956,484 (√n²
n).
=
EXAMPLE: (447,598) = 669.0276520
1. The number is greater than 12,500. Therefore, the table must
be read in the squares column.
150
Machine Computation of Elementary Statistics
2. The number 44 75 61 is the square of 669; the next number
in the column 44 89 00 is the square of 670. The number 44 75 98
falls between these two numbers; therefore, the square root of
447,598 must fall between 669 and 670,
3. If there is no machine available, then the decimal portion of
the root may be found by interpolation. If there is a machine, the
table will serve as a means to approximate the root extracted on
the machine.
PROBLEMS: Estimation of the magnitude of the root. Find its ap-
proximate value from a table; then compute on the machine.
Check by multiplication.
(1) 535,824
(2) 869.4673
(3) 0.495938
(4) 1,004.372
(5) 0.80656
(6) 134,369
(7) 962,614,400
(8) 0.056892
PART TWO
HOW TO OBTAIN CERTAIN COMMON STATISTICS
"
1
CHAPTER 7
THE MEAN AND STANDARD DEVIATION FROM A
GROUPED FREQUENCY DISTRIBUTION USING
AN ARBITRARY ORIGIN*
Since with a machine, large numbers are less inconvenient than
negative signs, the arbitrary origin should be placed at the mid-
point of the lowest step interval.** The quantities for substitution
in the formulas for the mean can be secured at the same time as
those for the standard deviation. The same checks, therefore, serve
for both statistics.
Computational formulas are as follows.***
(1) mean:
(2) standard deviation:
X = A +
S
N
·
(Σfx')
i
√ √N Zƒ(x')² — (Zƒx')²
N
Check formulas are as follows†:
(3) Σf(x' + 1) = Σfx' + N
(4) Zƒ(x' + 1)² = Σƒ(x')² + 2 Σƒx' + N
where
x = m
= mean of the distribution of X scores
A = the arbitrary origin placed at the midpoint of the
lowest step interval
width of the step interval
i
f a frequency
x' = deviation from the arbitrary origin
N = number of cases in the distribution
S s = standard deviation of the observed sample
* Patterned after Dunlap (reference 16, pp. 36-40).
** See reference 19, p. 119.
*** See reference 16, pp. 32, 37; reference 19, pp. 116, 118, 119.
† See reference 19, p. 118.
153
154
Machine Computation of Elementary Statistics
Set up a computing sheet. (See p. 195). Enter values for N, i,
i
N'
~, and A. Then compute the quantities as follows and enter on
computing sheet. The student should know the squares of all
integers from 1 to 25.
EXAMPLE: Find the mean (X) and standard deviation (s) of the
following frequency distribution of scores of male college freshmen
on a physics achievement test (unpublished data):
Score
63-65
60-62
57-59
54-56
51-53
48-50
45-47
42-44
39-41
36-38
33-35
30-32
Total
f
1OLSO OŊHMN MN
5
6
6
7
4
3
2
3
2
40
Friden*
x'
11
REGOOD OBHBQ –
OBH♡ 21
10
9
8
7
6
5
4
3
2
1
0
x' + 1
I
12
11
10
9
8
7
6
5
4
3
I
1. Put add key up.
2. Set decimal markers on both upper dials and keyboard be-
tween 8 and 9.
3. The problem is essentially one of sum, sum of squares, and
sum of cross products. The ƒ values are entered in the multiplier
unit; the x' values in the left of the keyboard, using column nine
as the units place, and the (x')² values in the right of the keyboard,
using column one as the units place.
4. Tap the lowest ƒ value 2 in the multiplier unit; set the lowest
x' value 0 in keyboard column nine, and its (x')2 value 0 in column
one.
* For EXAMPLE, refer to p. 154.
Mean and Standard Deviation
155
5. Multiply. The lowest f value 2 appears in lower dial one;
all other dials register 0.
6. Tap the next ƒ value 3 in the multiplier; set the next x' value
1 in column one and its (x')2 value 1 in column one; accumulate
multiply.
7. Continue in this fashion for all entries in the table, being
sure to accumulate multiply each time. In the dials read the follow-
ing values:
upper dials 11 to 9: 205
4 to 1: 1,309
lower dials 2 to 1:
40
Record on computing sheet in proper spaces.
8. Repeat for (x' + 1) and (x' + 1)2 values. In the dials read
these values:
upper dials 11 to 9: 245 =
4 to 1: 1,759
lower dials 2 to 1:
Σƒ(x' + 1)
Σƒ(x' + 1)²
Record on computing sheet in proper spaces.
9. Note that the computing sheet is set up to allow for sub-
stitution in check formulas as follows:
-
S =
****
=
=
Zfx'N 205 +40 245
(check)
Σf(x')² + 2 Σƒx' + N 1,309 +2 (205) + 40
= 1,759
(check)
10. If all quantities check, substitute as outlined on computing
sheet, and enter values found in the spaces provided.
11. As a check, repeat, using (x + 1) and (x + 1)2 values.
Remember that the arbitrary origin A is moved down one step
interval.
12. Final values should be:
=
=
=
3
40
Σfx'
Σf(x')²
Σf = N.
40 = Σf = N
Χ
X = 28+ (245) = 46.375
Σf(x' + 1)
Σf(x' + 1)²
X
3
= 31+ (205) 31+ 0.075 (205)
46.375 rounded to 46.4
=
·
(205)² = 0.075√‍10,335
3√40(1,309) — (205)²
0.075(101.7) = 7.6275 rounded to 7.6
13. Check, using (x' + 1) and (x' + 1)2 values:
****
=
3
s =
40 √40(1,759) — (245)² = 0.075√10,335
= 7.6275
(check)
(check)
156
Machine Computation of Elementary Statistics
Marchant*
1. Set decimal markers on both middle dials and keyboard be-
tween 8 and 9.
2. The problem is essentially one of sum, sum of squares, and
sum of cross products. The x' values are entered in the left of the
keyboard, using column nine as the units place; the (x')2 values are
entered in the right of the keyboard, using column one as units
place, and the ƒ values are entered in the multiplier keyboard.
3. Put carriage in position one, depress upper green carriage
shift key for multiplication from the left. Set the lowest x' value
O in keyboard column nine, its (x')² value 0 in column one; multiply
by the corresponding ƒ value 2.
4. The lowest f value 2 appears in upper dial one; all other dials
register 0.
5. Without clearing the keyboard dials, set the next x' value 1
in column nine and its (x')² value 1 in column one, and multiply
by the corresponding ƒ value 3.
6. Continue to accumulate multiply in this fashion for all
entries in the table. In the dials read these values:
40
Ef
Σf = N
upper dials 2 to 1:
middle dials 11 to 9: 205 Σfx'
4 to 1: 1,309 = Σf(x')²
=
=
Record on computing sheet in the proper spaces.
7. Clear the machine and repeat for (x + 1) and (x' + 1)²
values. In the dials read these values:
upper dials 2 to 1:
40 = Σƒ = N
middle dials 11 to 9: 245 Σf(x' + 1)
4 to 1: 1,759 = Σƒ(x' + 1)²
Record on computing sheet in proper spaces.
8. Note that the computing sheet is set up to allow for sub-
stitution in check formulas as follows:
=
Zfx' + N = 205 +40 245
Σf(x' + 1)
(check)
Σƒ(x' + 1)² = Σf(x')² + 2 Σfx' + N = 1,309 +2(205) + 40
= 1,759
(check)
* For EXAMPLE, refer to p. 154.
=
Mean and Standard Deviation
157
9. If all quantities check, substitute as outlined on computing
sheet, and enter values found in the spaces provided.
10. As check, repeat, using (x' + 1) and (x' + 1)² values. Re-
member that the arbitrary origin A is moved down one step
interval.
11. Final values should be:
X = 31 +(205) = 31 + 0.075(205)
= 46.375 rounded to 46.4
0.07510,335
40
3
s = √40(1,309) — (205)²
0.075(101.7) = 7.6275 rounded to 7.6
=
12. Check, using (x' + 1) and (x' + 1)² values:
X = 28+ (245)
3
46.375
40
=
=
-
3
40
7.6275
S = √40(1,759) — (245)²· ·
=
0.07510,335
(check)
(check)
Monroe: Automatic*
1. Put repeat key down and carriage in position one.
2. Set decimal markers both on lower dials and keyboard be-
tween 8 and 9.
3. The problem is essentially one of sum, sum of squares, and
sum of cross products. The ƒ values are set in the machine as multi-
pliers; the x' values in the left of the keyboard, using column nine
as the units place and the (x')² values in the right of the keyboard,
using column one as the units place.
4. Set the lowest ƒ value 2 in column one; set up; clear key-
board.
5. Enter the corresponding x' value 0 in column nine, and its
(x')² value 0 in column one.
6. Multiply. The lowest ƒ value 2 appears in upper dial one;
all other dials register 0.
7. Set the next ƒ value 3 in column one; set up.
8. Set the corresponding x' value 1 in column nine, and its
(x')² value 1 in column one; multiply.
9. Clear keyboard and continue in this fashion for all entries
* For EXAMPLE, refer to p. 154.
158
Machine Computation of Elementary Statistics
in the table, being sure to clear the keyboard after each operation.
In the dials read these values:
upper dials 2 to 1:
lower dials 11 to 9:
Record on computing sheet in proper spaces.
10. Clear the machine and repeat for (x + 1) and (x' + 1)²
values. In the dials read these values:
40
upper dials 2 to 1:
lower dials 11 to 9: 245 =
4 to 1: 1,759
40 = Σf = N
Zfx'
205
4 to 1: 1,309 = Σf(x′)²
Σƒ(x' + 1)
Σf(x' + 1)² =
=
=
Record on computing sheet in proper spaces.
11. Note that the computing sheet is set up to allow for substi-
tution in check formulas as follows:
=
X = 31 + % (205)
3
40
s = √40(1,309)
3
40
=
Σfx' + N = 205 + 40 = 245
(check)
Σƒ(x')² + 2 Σƒx' + N = 1,309 + 2(205) + 40
1,759
(check)
12. If all quantities check, substitute as outlined on computing
sheet, and enter values found in the spaces provided.
13. Final values should be:
46.375 rounded to 46.4
=
Σf = N
ΣΤ
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
f(x' + 1)
Σƒ(x' + 1)²
-
(205)2 = 0.07510,335
0.075(101.7) = 7.6275 rounded to 7.6
14. Check, using (x' + 1) and (x' + 1)2 values:
X
=28+ (245)
3
46.375
40
31 +0.075(205)
S = √40(1,759) — (245)² = 0.075√10,335
3
40
= 7.6275
Monroe: Semiautomatic and Hand-operated*
(check)
(check)
(a) Machines with split dials feature
1. Depress repeat key, put upper dial lock lever to left and put
carriage in position one.
* For EXAMPLE, refer to p. 154.
Mean and Standard Deviation
159
2. Set decimal markers on both lower dials and keyboard be-
tween 8 and 9.
3. The problem is essentially one of sum, sum of squares, and
sum of cross products. The x' values should be entered in the left
of the keyboard with column nine as units place; the corresponding
(x')2 values in the right of the keyboard with column one as units
place; and the ƒ values used as multipliers.
4. Set the lowest x' value 0 in keyboard column nine and its
(x')2 value 0 in column one.
5. Multiply by the ƒ value 2. The lowest f value 2 appears in
upper dial one; all other dials register 0.
6. Clear upper dials.
7. Set the next x' value 1 in column nine and its (x')² value
1 in column one. Multiply by the ƒ value 3.
8. Clear upper dials.
9. Continue in this fashion for all entries in the table, being
sure to clear the upper dials only between each accumulative
multiplication. In the dials read these values:
right upper dials 2 to 1:
lower dials 11 to 9:
right upper dials
lower dials
40
205
4 to 1: 1,309
=
Record on computing sheet in proper spaces.
10. Repeat for (x' + 1) and (x' + 1)2 values. In the dials read
these values:
=
=
-
=
2 to 1:
11 to 9:
Σƒ(x' + 1)
4 to 1: 1,759 = Σf(x' + 1)²
Record on computing sheet in proper spaces.
11. Note that the computing sheet is set up to allow for sub-
stitution in check formulas as follows:
Σf(x' + 1) Zfx' + N = 205 +40 245
Σƒ(x' + 1)² = Σƒ(x')² + 2 Σƒx' + N
1,759
=
Σf = N
Σξχ'
Σf(x)2
40 = Σf = N
245
=****
(check)
1,309 + 2(205) + 40
(check)
12. If all quantities check, substitute as outlined on computing
sheet, and enter values found in the spaces provided.
160
Machine Computation of Elementary Statistics
13. As check, repeat, using (x' + 1) and (x + 1)2 values. Re-
member that the arbitrary origin A is moved down one step
interval.
14. Final values should be:
X
31 + ³ (205) = 31 + 0.075(205)
3
40
= 46.375 rounded to 46.4
=
s = 3√40(1,309) — (205)²
(205)² = 0.075√10,335
0.075(101.7) = 7.6275 rounded to 7.6
=
15. Check, using (x' + 1) and (x' + 1)² values:
X
28+ (245)
46.375
s =
S √40(1,759) — (245)² = 0.075√10,335
√40(1,759)
= 7.6275
3
40
3
40
(check)
(check)
(b) Machines having no split dials feature; carry-over in upper dials
The process is the same as that outlined under section (a) with
this exception: it is necessary to count the turns of the operating
crank in multiplication; or to estimate quickly before each accumu-
lative multiplication what Σf value will next appear in the upper
dials. Since the ƒ values are usually small, this should not prove
difficult.
(c) Machines having no split dials feature; no carry-over in upper
dials
Here it is probably less confusing to the student to clear the
upper dials after each accumulative multiplication. In this case,
there will, of course, be no Σƒ = N value read in the upper dials.
On machines with only eight (or seven) banks in the keyboard,
it is necessary to set the lower dial and keyboard decimal markers
between six and seven (or between five and six for Educator
models); then the x' and (x + 1) units place changes to the
seventh (or sixth) column; and Ex' and 2(x' + 1) are read in lower
dials nine through seven (or eight through six).
If on these smaller machines, N is so large that Σƒx' and Σƒ(x')²
might overlap in the lower dials, the data may be divided into sub-
groups, and totals summed to give the grand total for the entire
group. This method is suggested for use in computing the correla-
tion coefficient and is described in the following section.
CHAPTER 8
THE PEARSON PRODUCT-MOMENT CORRELATION
COEFFICIENT; MEAN, AND STANDARD
DEVIATION FROM RAW SCORES*
Under no circumstances is the computation of the product-
moment correlation coefficient an easy task. When the number of
cases is greater than 150 to 200, the advantages of machine calcu-
lation are lost and greater speed is obtained through plotting the
paired values on a correlation chart.† This is true, of course, only
of data for which International Business Machine punch card
routines and resources are not available.
When the number of pairs of observations is less than 100 to 150,
computation for the quantities needed for r can be carried out di-
rectly on the machine. In this case, the raw scores of the variates
should not contain more than two digits. If they run into more
than two figures, they can be reduced to two by subtracting a con-
stant from each series. This operation will have no effect on the
value of the correlation coefficient or the standard deviation, but
will affect the mean. If a constant is subtracted from each score in
the original distribution to give a new distribution, then, to find
the mean of the original distribution, the same constant must be
added to the mean of the new distribution.
Computational formulas for the means, standard deviations,
regression coefficients, and correlation coefficient are as follows:‡
(1) means:
x = /\/\
N
ΣΧ;
Y
1/4
N
ΣΥ
* See reference 16, p. 49-51; reference 18; reference 20.
† See reference 16, p. 49.
‡ See reference 15, p. 672; reference 16, p. 51; reference 17, p. 292; reference 19,
p. 115, 226, 239–40.
161
162
Machine Computation of Elementary Statistics
(2) standard deviations:
byx
where
Sx
(3) regression coefficients:
Sy
=
1
*
N √ N ZX ²
N EX² — (EX)²;
*
(4) correlation coefficient:
Txy
*
See p.
formula
1
NVN ZY²
Ν ΣΧΥ (EX)(EY)
Σχ2
S VN − 1
Ν ΣΧ2 – (ΣΧ)2 ; bxx
-
(ΣΥ)2
EX + ΣY = 2(X + Y);
(5) check formulas:
ΣΧ ΣΥ
2
2(X - Y)2}, or
Σ(X + Y)² + 2(XY)² = 2 EX² + 2 ZY2
ΣΥ
EXY = {2(X + Y)² −
ΣΧΥ 1
ΣΧΥ } {Σ(X + Y)² -
(2X) (EY) } {[E(X + Y)]²
=
EX²
2 EX2 - 2 ZY2}
ΣΥ}
— [Σ(X — Y)]²} †
-
Ν ΣΧΥ – (ΣΧ)(ΣΥ)
Ν ΣΥ – (ΣΥ)2
Ν ΣΧΥ - (ΣΧ)(ΣΥ)
√N EX² — (EX)² √N EY²
ΣΥ – (ΣΥ)2
-
2X - Y = 2(X — Y)
Sy
r²xy = byxbxy; byx = TxY S1; bxx = rxx
Sx
Sx
X = mean of distribution of X scores
Y = mean of distribution of Y scores
X
= a raw score
Y = a raw score
N =
number of cases in the distribution
sx = standard deviation of distribution of X scores
standard deviation of distribution of Y scores
Sy
Sy =
byx regression coefficient used in the equation to predict
Y from X
bxy = regression coefficient used in the equation to predict
X from Y
4 of computing sheet if N-1 is used as denominator of definitional
†This check was suggested by Professor Helen M. Walker of Teachers
College, Columbia University.
Product-Moment Correlation
163
Txy correlation coefficient indicating the correlation between
X and Y
=
On the automatic calculating machine, it is possible to compute
in one set of operations the following quantities: ZX, 2X², ZY,
ΣY², and ΣXY. It is advisable to use a list type adding machine
to compute the (X + Y) and (X - Y) check values; then the
2(X + Y), 2(X + Y)², 2(X − Y), and Σ(X - Y)2 values can be
found quickly on the automatic calculator.
EXAMPLE: Compute the coefficient of correlation between test
scores of 40 male college freshmen on a mathematics achievement
test and quantitative aptitude measure. The following data are
needed. The data are divided into two groups, 1 and 2, for facility
in handling and checking.
Group 1
Case
IQBHBOTPOINSER CND2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Mathe-
matics
Achieve-
ment
X
JERORANKINBOJKAONOJ:
60
70
42
52
55
63
28
39
60
57
60
61
39
Scores
Quanti-
tative
Apti-
tude
Y
2007 KROFONO ***2.
74
19
19
31
74
16
87
70
9
74
70
97
89
16
74
48
74
85
79
31
Check Steps
(X – Y)
(X + Y)
135
70
78
91
144
71
144
128
51
126
125
160
153
44
113
108
131
145
140
70
G
- 13
32
40
- 4
29282282
39
-30
-12
33
- 22
| |
G
- 15
-34
- 25
12
-35
12
-17
-25
-18
8
Table continued on following page
164
Machine Computation of Elementary Statistics
Group 2
Case
72****7*2* -≈≈385389
21
23
24
25
26
28
29
30
31
32
34
36
37
40
Mathe-
matics
Achieve-
ment
X
❤ERIHUO******AN JHANO
43
47
75
63
54
61
49
63
57
49
58
53
58
50
57
61
42
39
52
60
Scores
Quanti-
tative
Apti-
tude
Y
HELEN NON≈ 8*?****
11
11
96
74
27
94
82
60
82
23
96
43
19
27
31
98
48
6
87
97
Check Steps
* For EXAMPLE, see p. 163.
† Patterned after Dunlap, reference 16, p. 51.
(X + Y)
54
58
171
137
81
155
131
123
139
72
48FF 28
77
77
88
159
90
45
139
157
(XY)
32
36
-21
- 11
27
-33
-33.
3
— 25.
26.
-38.
10
39,
23
26.
-37
- 6
33.
-35
-37
K
Friden*
1. Set up computing sheets† (Appendix B) and enter values
1
1
. Then compute quantities as follows
for N, N'
and √N(N − 1)
and enter on sheets. For facility in checking, work with groups of
20 variates. The sheets are set up for working with larger numbers,
so use only the necessary spaces.
2. Clear the machine and put add key down.
3. All scores in this problem are positive and of one or two
digits. Set up data sheets so that the total number of pairs of
Product-Moment Correlation
165
variates is divided into groups of 20. For a larger number of pairs,
20 to 25 may be a preferred group size. It is helpful to put a red
line across the page at the end of each group.
4. Place decimal markers as follows:
upper dials: between 12 and 13, 6 and 7
lower dials: between 6 and 7
keyboard: between 6 and 7, 2 and 3
multiplier proof register: over 6, over 2
5. Enter the X₁ value, 61, in the multiplier. There are four
columns between the markers on the keyboard, so enter four O's
in the multiplier, then the Y₁ value, 74. The multiplier proof
register then reads 61000074, with the two decimal pointers be-
tween 1 and the first 0 and between the fourth 0 and the 7; this
aids in the checking of the values entered in the multiplier.
[CORE]
1
6. Enter the X₁ value 61 in keyboard columns eight and seven,
so that the units digit is immediately to the left of the decimal
marker. Enter the Y₁ value 74 in keyboard columns two and one,
so that the tens digit is directly to the right of the decimal marker.
7. Multiply. In the dials read these values:
1
X32
1
9,028 = 2 X₁Y₁
5,476 = Y
61 X1
Y₁
74
upper dials 16 to 13: 3,721
10 to 7:
4 to 1:
lower dials 8 to 7:
2 to 1:
*
=
8. Enter the X2 (51) and the Y2 (19) values in the multiplier
and keyboard in exactly the same manner, with four O's between.
Check the multiplier proof register to be sure the X2 and Y₂ values
are correctly placed.
2
9. Touch the ACCUM MULT key. In the dials read these values:
1
6,322 ΣΧ X² + X2
10,966 = 2EXY = 2(X₁Y₁ + X2Y2)
5,837 = ZY² = Y} + Y²
112= EX = X₁ + X2
ΣΧ
93 ZY Y₁ + Y₂
=
= 1
=
upper dials 16 to 13:
11 to 7:
4 to 1:
lower dials 9 to 7:
2 to 1:
10. Repeat steps 5 to 7 for X3, X4, ・・・, X20 and Yз, Y4, ・・・,
Y20. Be sure to touch the ACCUM MULT key each time. Note that
•,
166
Machine Computation of Elementary Statistics
for case number 9, the Y value is a single digit, 9. This must be
entered in the multiplier as 09 so that both the X and Y values
are correctly placed in the multiplier.
11. Read these values in the dials:
upper dials 17 to 13: 61,495 ΣX² = X} + X²
=
+ ··· + X 320
12 to 7: 129,972 = 2 EXY = 2(X₁Y₁
+ X2Y2 + · · + X20Y20)
1
5 to 1: 81,086 = ΣY² = Y² + y²
ΣΥ
lower dials 10 to 7: 1,091
1,091 = EX =
ΣΧ
4 to 1: 1,136
1,136 = ΣY = Y₁ + Y₂
ΣΥ
1
2
+
+ ··· + Y 20
X₁ + X₂
+ ··· +X20
+Y 20
Enter these as group 1 values on the computing sheet in the
proper columns. Blanks are arranged for convenience in checking.
Be sure to enter the right values in the right columns. Compute
ΣΧΥ from 2 ΣΧΥ.
•
12. Set X21 and Y21 in the multiplier and keyboard as before,
and touch the MULT key. This clears the group 1 values out of the
dials and enters the first set of group 2 values in the machine.
13. Set the X22 and Y22 values in multiplier and keyboard, touch
the ACCUM MULT key.
14. Continue as before through X40 and Y40, the second group.
Read these values in the dials:
upper dials 17 to 13: 60,905 = 2X² = X21 + X322
+・
lower dials 10 to 7: 1,091
12 to 7: 128,562 = 2ZXY = 2(X21Y21
+ X 22Y 22 + ··· + X40Y40)
5 to 1: 84,114
84,114 =
2Y2 = Y₁₁ + Y 22
ΣΥ2
Y21+
+
1,091 = 2X = X21 + X22
ΣΧ
+
1,112 = 2Y = Y21+ Y22
ΣΥ
+
4 to 1: 1,112
+ X 240
· + Y 2420
Yo
+X40
+ Y 40
Enter these as group 2 values on the computing sheet. Compute
ΣΧΥ from 2 ΣΧΥ.
Product-Moment Correlation
167
15. Check steps:
(X + Y): (process of getting sums and sums of squares)
(a) Enter (X + Y), in the multiplier and in keyboard
columns eight and seven. Multiply.
1
(b) Enter (X + Y)2 in the multiplier and keyboard, and
touch ACCUM MULT.
(c) Continue to accumulate multiply through (X + Y)20.
In the dials read these values:
upper dials 12 to 7: 272,553
lower dials 4 to 1: 2,227
upper dials 12 to 7: 273,581
lower dials 4 to 1: 2,203
=
Record as group 1 check values on the computing sheet.
(d) Repeat for (X + Y)21, (X + Y) 22,···, (X + Y)40.
In the dials, read these values:
Σ(X + Y)
= 2,227
=
=
=
Σ(X + Y)²
2
Σ(X + Y)
Record these values as group 2 check values.
(X – Y): (sums and sums of squares of positive and nega-
tive numbers)
(a) Procedure is the same as for (X + Y) values, except
where (X — Y) is negative. When this is so, pull down
counter control key before touching ACCUM MULT key.
Be sure it is in the normal (up) position for positive
values of (X Y). If desired, do all positive values;
then reverse counter control key and do negative values.
(b) Read and enter these values for group 1:
Σ(X — Y)²
upper dials 11 to 7: 12,609
lower dials 4 to 1: (9,955)
(c) Do the same for group 2, and
upper dials 11 to 7: 16,457
lower dials 4 to 1: (9,979)
Σ(X + Y)²
Σ(X + Y)
= -45 =
Σ(X — Y)
enter these values:
Σ(X - Y)²
= -21
2
Σ(X — Y)
=
Substitution in check formulas
(a) Work on each group separately. In this way, if there is
an error, it is more easily found. For group 1:
(1) ΣΧ + ΣΥ
1,091 + 1,136
=
(check)
168
Machine Computation of Elementary Statistics
(2) ΣΧ – ΣΥ
-
-
Σ(Χ Y) = 1,091 -
1,091 - 1,136
-45
(4) ΣΧΥ
ΣΥ
(3) Σ(X + Y)² + 2(X − Y)² = 2 EX² + 2 EY²
272,553 +12,609
II
=
=
=
2(61,495) + 2(81,086)
285,162
(check)
2
1 {E(X + Y)² - Σ(XY)2}
(272,553 - 12,609)
1 (259,944)
64,986
Txx: 0.580
X: 54.6
Y: 56.2
(check)
=
(check)
(b) Repeat for group 2. When all values check, sum values
for groups 1 and 2 and enter total values on computing
sheet as final values. Note this provides checks for the
total values.
byx: 1.97
bxy: 0.171
Marchant*
16. In computation of final values, substitute quantities as out-
lined on computing sheet, and enter values found in spaces pro-
vided. Since there are no automatic checks provided for these last
steps, all computations should be performed twice.
17. Final values should be:
Sx: 9.2
Sy: 31.2
1. Set up computing sheets (Appendix B) and enter values for
1
1
N, Ñ'
and
√N(N − 1)
Then compute quantities as follows and
enter on sheets. For facility in checking, work with groups of 20
variates. The sheets are set up for working with larger numbers,
so use only the necessary spaces.
2. Clear the machine. Depress upper green carriage shift key
for multiplication from the left.
3. All scores in this problem are positive and of one or two
digits. Set up data sheets so that the total number of pairs of vari-
ates is divided into groups of 20. For a larger number of pairs,
* For EXAMPLE, see p. 163.
Product-Moment Correlation
169
20 to 25 may be a preferred group size. It is helpful to put a red
line across the page at the end of each group.
4. Place decimal markers as follows:
upper dials: between 7 and 8, 4 and 5
middle dials: between 14 and 15, 7 and 8
keyboard: between 7 and 8, 2 and 3
5. Depress selective tab key eight. Note that the keyboard
decimal marker between columns 7 and 8 lines up with the lower
dial decimal marker between dials 14 and 15.
1
6. Set the X₁ value 61 in columns nine and eight, using the
decimal marker as guide to denote the units position, and the Yı
value 74 in columns two and one, using the decimal marker as
guide to denote the tens position.
7. Since there are two digits in the X₁ value, depress selective
tab key nine to position the carriage for multiplication by the first
two-digit number. Note that the tab key depressed is under the
first digit of the X₁ value in the keyboard. Multiply by the X1
value, 61.
1
8. Since there are two digits in the Y₁ value, depress selective
tab key two to position the carriage for multiplication by a two
digit number. Note that the tab key depressed is under the first
digit of the Y₁ value in the keyboard. Multiply by the Y₁ value, 74.
9. In the dials read these values:
upper dials
9 to 8:
2 to 1:
middle dials 18 to 15:
11 to 8:
4 to 1:
61 = X₁
74
upper dials 10 to 8:
2 to 1:
middle dials 18 to 15:
12 to 8:
4 to 1:
=
=
Y₁
=
X2
3,721
9,028 2 X₁Y1
5,476 = Y}
10. Depress keyboard dial key and selective tab key nine. Set the
X2 value 51 in columns nine and eight, and the Y₂ value 19 in
columns two and one.
11. Multiply by the X2 value, 51; tab to position two and multi-
ply by the Y₂ value, 19.
2
12. In the dials read these values:
1
112 = EX = X₁ + X2
ΣΧ
93 =
6,322
ZY = Y₁+ Y2
ΣΥ
10,966 2 EXY = 2(X₁Y₁ + X₂Y2)
=
1
5,837
5,837 = ΣY² = Y} + Y₁₂
ΣΥ
ΣX² = X² + X²
ΣΧ2
170
Machine Computation of Elementary Statistics
13. Repeat steps 6 to 9 for X3, X4,
Y 20.
Be sure to clear the keyboard ONLY between each set of accumu-
lative multiplications. Note that for case number 9, the Y value is
a single digit, 9. Therefore depress selective tab key one for the Y
multiplication.
14. In the dials read these values:
upper dials 11 to 8:
4 to 1:
, X20 and Y3, Y 4,···,
1,091 = EX = X₁ + X₂
ΣΧ
1,136 = ΣY =
ΣΥ
middle dials 19 to 15: 61,495 = EX² =
ΣΧ2
5 to 1: 81,086
13 to 8: 129,972 = 2 2XY = 2(X₁Y₁
1
+X2Y2 + ··· + X20Y20)
.
4 to 1:
+ ··· + X20
Y₁ + Y₂
+ ··· +Y 20
X + X3
+···· + X320
81,086 = ZY² = Y} + Y²
ΣΥ
+ Y². ·· + Yo
15. Enter these as group 1 values on the computing sheet, in the
proper columns. Note that blanks are arranged for convenience in
checking; be sure to enter the correct values in the correct columns.
Compute ΣΧΥ from 2 ΣΧΥ.
16. Clear the keyboard and dials and repeat the process for
X21, X22,・・, X40 and Y21, Y22,, Y40, the second group. In
the dials, read these values:
upper dials 11 to 8:
ΣΧ
21
1,091 = EX = X21 + X22
+...+Xeo
Y21+ Y22
++Y40
middle dials 19 to 15: 60,905 = ΣX² = X2₁₁ + X22
ΣΧ2
+ ··· + X 20
13 to 8: 128,562 = 2 ΣXY =
ΣΧΥ 2(X21 Y 21
+ X 22Y 22 +・・・ X40X40)
+
5 to 1: 84,114
84,114 = 2Y2 = Y₁₁ + Y² 2
ΣΥ
+ ··· + Y 2420
21
22
1,112 = 2Y =
ΣΥ
40
17. Enter these as group 2 values on the computing sheet.
Compute ΣΧΥ from 2 ΣΧΥ.
Product-Moment Correlation
171
18. Check steps:
(X + Y): (process of getting sums and sums of squares)
(a) Enter the (X + Y), value, 135, in keyboard columns
three through one; depress selective tab key three;
square the value.
(b) Continue by accumulate multiplying (X + Y)2,
(X + Y)3, · · ·, (X + Y)20.
(c) In the dials read these values:
upper dials 4 to 1: 2,227
middle dials 6 to 1: 272,553
= E(X + Y)
Σ(X + Y)²
Record as group 1 check values on the computing sheet.
(d) Repeat for (X + Y)21, (X + Y)22,···, (X + Y)40.
In the dials read these values:
lower dials 4 to 1: 2,203
middle dials 6 to 1: 273,581
Substitution in check formulas
=
=
Σ(X + Y)
= 2,227
=
Record as group 2 check values on the computing sheet.
(XY): (sums and sums of squares of positive and negative
numbers)
*******
(a) Procedure is the same as for (X + Y) values, except
where (XY) is negative. When this is so, pull the
counter control down. Be sure it is in the normal (up)
position for positive values of (XY). If desired, do
all positive values; then reverse counter control lever
and do the negative values.
(b) Read and enter these values for group 1:
=
upper dials 4 to 1: (9,955) -45 = Σ(X — Y)
middle dials 5 to 1: 12,609 = Σ(X Y) 2
(c) Read and enter these values for group 2:
upper dials 4 to 1: (9,979) -21 =
middle dials 5 to 1: 16,457
Σ(X — Y)
=
= 2(X — Y)²
Σ(X + Y)
Σ(X + Y)²
(a) Work on each group separately. In this way, if there is
an error, it is more easily found. For group 1:
(1) ΣΧ + ΣΥ
1,091 +1,136
=
―
(check)
172
Machine Computation of Elementary Statistics
(2) EX - ΣY = 2(X − Y)
ΣΧ ΣΥ Σ(X — Y) = 1,091
= 1,091 1,136
-45
(3) E(X + Y)² + 2(X − Y)² = 2 EX² + 2 ZY2
(4) ΣΧΥ
2
=
} {Σ(X + Y)² - E(XY)²}
=
Txy: 0.580
X: 54.6
Y: 56.2
272,553 + 12,609
=2(61,495) + 2(81,086)
= 285,162
(check)
(272,553 – 12,609)
(259,944)
64,986
(check)
(b) Repeat for group 2. When all values check, sum values
for groups 1 and 2 and enter total values on computing
sheet as final values. Note that this provides a check for
the total values.
(check)
19. In computation of final values, substitute quantities as out-
lined on computing sheet, and enter values found in the spaces
provided. Since there are no automatic checks provided for these
last steps, all computations should be performed twice. Here it is
advisable to ask: Can a coefficient of correlation (or standard
deviation or mean) of this size be reasonably expected for these
data? In this case, 0.580 is a reasonable coefficient of correlation
between scores on a mathematics achievement test and a quantita-
tive aptitude test.
20. Final values should be:
Sx: 9.2
Sv: 31.2
byx: 1.97
bxy: 0.171
Monroe: Automatic*
1. Set up computing sheets (Appendix B) and enter values for
1
1
N'
N, √, and
√N(N − 1)
Then compute quantities as follows and
enter on sheets. For facility in checking, work with groups of 20
variates. The sheets are set up for working with larger numbers,
so use only the necessary spaces. .
2. Put repeat key down and carriage in position one.
* For EXAMPLE, refer to p. 163.
Product-Moment Correlation
173
3. All scores in this problem are positive and of one or two
digits. Set up data sheets so that the total number of pairs of
variates is divided into groups of 20. For a larger number of pairs,
20 to 25 may be a preferred group size. It is helpful to put a red
line across the page at the end of each group.
4. Place decimal markers as follows:
right upper dials: between 6 and 7
lower dials:
keyboard:
5. Enter the X₁ value 61 in keyboard columns eight and seven,
using the keyboard marker to mark the units place; and the Yı
value, 74, in columns two and one, using the keyboard marker to
mark the tens place. Set up. On CAA machines, use the setup key
to lock the numbers in the keyboard as well as in the machine as
the multipliers.
between 12 and 13, 6 and 7
between 6 and 7, 2 and 3.
6. On machines other than the CAA model, reenter the X₁ and
Y₁ values in the same keyboard columns as in step 5.
7. Multiply. In the dials read these values:
right upper dials
lower dials
lower dials
8 to 7:
2 to 1:
16 to 13: 3,721
10 to 7: 9,028 = 2 X₁Y₁
4 to 1: 5,476
1
Y?
61 = X1
Y₁
X2
9 to 7:
2 to 1:
16 to 13:
11 to 7:
74 =
8. Clear the keyboard. Enter the X2 (51) and Y2 (19) values in
keyboard columns eight and seven, and two and one respectively.
Setup. Reenter the same values in the same columns on machines
other than CAA models.
9. Multiply. In the dials read these values:
right upper dials
=
-
-
112
ΣΧ
X₁+ X₂
93 ZY
= =
ΣΥ
=
1
Y₁+ Y₂
6,322 ΣΧ2 x² + X2
10,966 2EXY = 2 (X₁Y₁
=
=
=
4 to 1: 5,837 = ZY²
+ X2Y2)
ΣY² = Y + Y²
174
Machine Computation of Elementary Statistics
10. Repeat steps 5 to 7 for X3, X4,・・・, X20 and Y3, Y4,
Y20. Be sure to accumulate multiply.
11. In the dials read these values:
right upper dials 10 to 7:
lower dials
4 to 1:
1,091
'lower dials
: ΣΧ
ΣX = X₁ + X2
+ ·
-
1,136 = 2Y = Y₁ + Y₂
ΣΥ
17 to 13: 61,495 = EX² = X + X32
+ ··· + X 320
•
1,091
12 to 7: 129,972 = 2EXY = 2(X₁Y₁
+ X2Y2 + ..
5 to 1: 81,086 = ZY²
+ X20Y20)
Enter these as group 1 values on the computing sheet in the
proper columns. Note that the blanks are arranged for convenience
in checking. Be sure to enter the right values in the right columns.
Compute ΣXY from 2 ZXY.
12. Clear the machine.
13. Repeat for X21, X22,..., X40 and Y21, Y 22,
obtain group 2 values.
14. In the dials read these values:
right upper dials 10 to 7:
+ X20
+ ··· +Y 20
+.
=
=
Y₁ + Y2
+ ··· + Y 20
4 to 1: 1,112 = ΣY
=
.
Y 40,
ΣΧ
ΣX = X21 + X22
to
= ZY = Y21 + Y 22
21
•
+・・・+Y 40
17 to 13: 60,905 = ΣX² = X2₁ + X 22
ΣΧ2
+ ··· +X20
12 to 7: 128,562 = 2 EXY = 2(X21Y21
+ X 22Y 22 + ··· +X40X40)
5 to 1: 84,114
Y21+ Y22
+ ··· + Y 40
ΣΥ
+ ··· + X 40
•
Enter these as group 2 values on the computing sheet. Compute
ΣΧΥ from 2 ΣΧΥ.
15. Check steps:
(X + Y): (process of getting sums and sums of squares)
(a) Enter the (X + Y)₁ value, 135, in columns three through
one. Square it.
1
Product-Moment Correlation
175
(b) Clear keyboard. Enter the (X + Y)2 value, 70, in col-
umns two and one. Square it.
(c) Continue to accumulate multiply through (X + Y)20.
In the dials read these values:
upper dials 4 to 1: 2,227 =
lower dials 6 to 1: 272,553
Record as group 1 check values on the computing sheet.
(d) Repeat for (X + Y)21, (X + Y)22,
the dials read these values:
=
upper dials 4 to 1: 2,203 Σ(X + Y)
lower dials 6 to 1: 273,581 = Σ(X + Y)² 2
Substitution in check formulas
=
Record these values as group 2 check values.
(X – Y): (sums and sums of squares of positive and nega-
tive numbers)
(a) Procedure is the same as for (X + Y) values, except
where (XY) is negative. When this is so, put change
lever to divide before accumulate-multiplying. Be sure
it is in the multiply position for positive values of
(X – Y). If desired, do all positive values; then put
change lever at divide, and do negative values.
(b) Read and enter these values for group 1:
(1) ΣΧ + ΣΥ = Σ(X + Y)
= 2,227
Σ(X — Y)
-45
Σ(X + Y)
Σ(X + Y)²
upper dials 4 to 1: (9,955) ·45 = E(X
lower dials 11 to 7: 12,609 = Σ(X — Y)²
(c) Do the same for group 2, and enter these values:
upper dials 4 to 1: (9,979) = -21 Σ(Χ
lower dials 11 to 7: 16,457
2(XY)
Σ(X — Y)²
(2) ΣΧ – ΣΥ
A
·, (X + Y)40. In
=
(a) Work on each group separately. In this way, if there is
an error, it is more easily found. For group 1:
1,091 + 1,136
=
-
E(XY)
1,091 - 1,136
(check)
(check)
176
Machine Computation of Elementary Statistics
(3) Σ(X + Y)² + Σ(X — Y)² = 2 EX² + 2 ΣY2
= 272,553 + 12,609
ΣΥ
2(61,495) + 2(81,086)
285,162
(check)
(4) ΣΧΥ
11
Txy: 0.580
X: 54.6
Y: 56.2
=
} {Σ(X + Y)² - Σ(XY)²}
1 (272,553 – 12,609)
(259,944)
(check)
(b) Repeat for group 2. When all values check, sum values
for groups 1 and 2 and enter total values on computing
sheet. Note this provides a check for total values.
16. In computation of final values, substitute quantities as out-
lined on computing sheet and enter values found in spaces provided.
Since there are no automatic checks provided for these last steps,
computations should be performed twice.
17. Final values should be:
**********
= 64,986
Sx: 9.2
Sy: 31.2
byx: 1.97
bxy: 0.171
Monroe: Semiautomatic and K Models*
1. Set up computing sheets (Appendix B) and enter values for
1
1
N,
and
'N' √N(N − 1)
. Then compute quantities as follows and
enter on sheets. For facility in checking, work with groups of 20
variates. The sheets are set up for working with larger numbers,
so use only the necessary spaces.
2. Clear the machine, depress repeat key, lock right upper
dials, and put counter control lever at multiply position.
3. All scores in this problem are positive and of one or two
digits. Set up data sheets so that the total number of pairs of vari-
ates is divided into groups of 20. For a larger number of pairs,
20 to 25 may be a preferred group size. It is helpful to put a red line
across the page at the end of each group.
* For EXAMPLE, see p. 163.
Product-Moment Correlation
177
4. Place decimal markers as follows:
lower dials: between 12 and 13, 6 and 7
right upper dials: between 6 and 7
left upper dials: between 6 and 7
keyboard: between 6 and 7, 2 and 3
5. Note that upper dial, lower dial, and keyboard decimal
markers are lined up.
1
6. Set the X₁ value, 61, in columns eight and seven, using the
decimal marker as guide to denote the units position and the Yı
value, 74, in columns two and one, using the decimal marker as
guide to denote the tens position.
7. Move the carriage so that the lower dial decimal marker
between dials twelve and thirteen lines up with the keyboard
decimal marker between columns six and seven (position seven).
Multiply from the right by the X₁ value 61 noting that the left
upper dial decimal marker between dials six and seven is immedi-
ately to the right of the units digit of the multiplier.
1
8. Move the carriage to position one and multiply from the
right by the Y₁ value, 74. Note that the multiplier 74 appears in
left upper dials two and one.
1
9. In the dials read these values:
right upper dials
lower dials
lower dials
8 to 7:
2 to 1:
9 to 7:
2 to 1:
16 to 13:
10 to 7:
4 to 1:
16 to 13:
11 to 7:
4 to 1:
61 = X₁
74 = Y₁
X2
3,721
9,028 = 2 X₁Y₁
5,476
Y?
10. Clear keyboard and left upper dials and return carriage to
position seven. Set the X2 value, 51, in columns nine and eight,
and the Y2 value, 19, in columns two and one.
=
11. Multiply from the right by the X2 value, 51; move carriage
to position one, and multiply from the right by the Y₂ value 19.
12. In the dials read these values:
2
right upper dials
=
=
X₁ + X2
112 ΣΧ
93 = ΣY Y₁ + Y₂
ΣΥ
6,322 = EX² = X} + X32
1
2
10,966 = 2 ZXY = 2(X₁Y₁ + X2Y2)
5,837 = ΣY² = Y² + Y²
ΣΥ
=
=
178
Machine Computation of Elementary Statistics
13. Repeat steps 6 to 9 for X3, X4, ..., X20 and Y3, Y4,
Y20. Be sure to clear ONLY the keyboard and left upper dials after
each set of accumulative multiplications.
14. In the dials read these values:
right upper dials 10 to 7:
lower dials
4 to 1:
lower dials
1,091 = 2X = X₁ + X₂
ΣΧ
+ + X 20
1,136 = ZY = Y₁ + Y₂
ΣΥ
++Y 20
2
17 to 13: 61,495 EX2 X3+ X323
2
+ X20
=
+
12 to 7: 129,972 2 ΣΧΥ
+ X2Y2 + ···
1
2(X.Y.
+ X20Y20)
5 to 1: 81,086 = ZY²
81,086 Y² = Y + Y2
+
+ Yo
right upper dials 10 to 7: 1,091
4 to 1:
15. Enter these as group 1 values on the computing sheet, in
the proper columns. Note that blanks are arranged for convenience
in checking; be sure to enter the correct values in the correct col-
umns. Compute ZXY from 2 ΣXY.
16. Clear the keyboard and dials (unlock right upper dials,
clear, then lock again), and repeat the process for X21, X22, ·,
X40 and Y21, Y22, ., Y40, the second group. In the dials, read
these values:
1,091 = EX
ΣΧ
X21 + X22
+ ··· + X 40
1,112 = 2Y = Y21+ Y22
ΣΥ
+・・・+Y40
17 to 13: 60,905 EX2 X21 + X22
ΣΧ2
+ + X240
12 to 7: 128,562 2EXY = 2(X21Y21
+ X 22Y 22+ ··· + X40Y 40)
5 to 1: 84,114
Y21 + Y22
+ ··· +Y 20
•
ΣΥ
=
=
==
=
=
•
=
=
=
17. Enter these as group 2 values on the computing sheet.
Compute ΣΧΥ from 2 ΣΧΥ.
18. Check steps:
(X + Y): (process of getting sums and sums of squares)
Product-Moment Correlation
179
(a) Be sure right upper dials are cleared and locked. Enter
the (X + Y), value, 135, in keyboard columns three
through one, and square. Clear upper dials.
(b) Continue by accumulate multiplying (X + Y)2,
(X + Y)3, (X + Y)20.
(c) In the dials read these values:
right upper dials 4 to 1: 2,227
lower dials 6 to 1: 272,553
=
Record as group 1 check values on the computing sheet.
(d) Repeat for (X + Y)21, (X + Y)22, ·
In the dials, read these values:
(X + Y)40.
right upper dials 12 to 7: 273,581
lower dials
4 to 1: 2,203
=
Substitution in check formulas
Σ(X + Y)
2
Σ(X + Y)²
=
=
Record as group 2 check values on the computing sheet.
(XY): (sums and sums of squares of positive and negative
numbers)
=
=
(a) Procedure is the same as for (X + Y) values, except
where (XY) is negative. When this is so, put the
counter control key to divide. Be sure it is in the normal
(multiply) position for positive values of (X — Y). If
desired, do all positive values; then reverse counter con-
trol lever and do the negative values.
(b) Read and enter these values for group 1:
Σ(X + Y)²
Σ(X + Y)
right upper dials 4 to 1: (9,955)
lower dials 5 to 1: 12,609
(c) Read and enter these values for group 2:
right upper dials 4 to 1: (9,979) = −21 = 2(X — Y)
lower dials 5 to 1: 16,457 = 2(X — Y)²
-45
Σ(X — Y)²
=
Σ(X — Y)
(a) Work on each group separately. In this way, if there is
an error, it is more easily found. For group 1:
(1) EX + ΣY = 2(X + Y)
Σ(X + Y) = 1,091 + 1,136
2,227
(check)
180
Machine Computation of Elementary Statistics
(2) EX – ΣY = 2(X — Y) = 1,091 — 1,136
ΣΥ
(check)
-45
(3) Σ(X + Y)² + 2(X — Y)² = 2 EX2 + 2 ΣY2
272,553 +12,609
−
ΣΥ
= 2(61,495) + 2(81,086)
285,162
(check)
(4) ΣXY = { {Σ(X + Y)² − 2(X — Y)²}
ΣΧΥ
1(272,553 – 12,609)
(259,944)
64,986
|| ||
S
(check)
(b) Repeat for group 2. When all values check, sum values
for groups 1 and 2 and enter total values on computing
sheet as final values. Note that this provides a check for
the total values.
Txy: 0.580
X: 54.6
Y: 56.2
=
19. In computation of final values, substitute quantities as out-
lined on computing sheet, and enter values found in the spaces
provided. Since there are no automatic checks provided for these
last steps, all computations should be performed twice. Here it is
advisable to ask: Can a coefficient of correlation (or standard devi-
ation or mean) of this size be reasonably expected for these data?
In this case, 0.580 is a reasonable coefficient of correlation between
scores on a mathematics achievement test and a quantitative
aptitude test.
20. Final values should be:
Sx: 9.2
Sy: 31.2
byx: 1.97
bxy: 0.171
A. PERCENTILES
A percentile is a score in a distribution below which a specified
percentage of scores lie. For example, forty-three per cent of scores
in a distribution lie below the 43rd percentile; fifty per cent lie
below the 50th percentile (median).
The following formula and procedure for determining a percentile
are modified from Dunlap:*
where
P;
L
=
CHAPTER 9
OTHER COMMON STATISTICS
=
percentile sought
Lower limit of step interval in which the score lies
i = width of step interval
P₁ = L +
P;
=
i(pN — fc)
fi
p = per cent, corresponding to desired percentile
N
total number of scores in the distribution
f. = cumulative frequency for all intervals below that in which
the score lies
fi frequency of the interval in which the score lies
=
EXAMPLE: Find P10, P25, P50, P75, P90 for the following distribution
of scores of 78 introductory psychology students on a speed of
addition test.
* See reference 16, p. 33.
181
182
Machine Computation of Elementary Statistics
Score
18
BOOK HR2E CO∞
IRIQ ++RE ON∞ ∞∞∞
L43
17
16
15
14
13
12
11
10
9
8
7
6
5
Total = N
Frequency
―――
1
3
N
1
2
4
4
9
11
10
12
4
8
3
3
1
2
78
Cumulative
Frequency
78
CRN2 753 3827
= 78
77
74
73
71
67
63
54
43
33
21
17
9
6
3
1. Set up a work sheet providing spaces for recording N and i
(see 5 below), and for columns in which to enter values for substi-
tution in the formula, and the percentile values (see 5 below).
2. Tabulate frequencies as suggested above, making provision
for a cumulative frequency column. Enter N and i values in spaces
provided. N = 78, i= 1.
ON
3. Starting at the low end of the distribution, accumulate fre-
quencies, entering each cumulative frequency in the proper space.
4. Multiply N by each p separately, and enter in pN column.
5. Working from the pN column, determine by inspection of the
frequency table the values of L, fe, fi for each P, value. For example,
for P10, the pN value is 7.8. This means that P10 is the 7.8th score
in the distribution. Read the cumulative frequency column to find
that the 7.8th score falls in the interval 5.5 to 6.5; therefore
L = 5.5, fc 6, and f; 3. Do the same for the other P; values.
i = 1
2
Percentiles; Standard Scores
183
Percentile
Sought
P10
P25
P50
P75
P90
L
5.5
7.5
9.5
11.5
13.5
and for P50:
P10 = 5.5 +
219ROR:
10
P50 9.5 +
25
50
75
90
PN
7.8
19.5
39.0
58.5
70.2
1(7.8 — 6)
3
Z
6. Substitute the L, i, pN, fe, and f: values in the formula for
each P₁, and solve; e.g., for P10:
fo
6
17
33
54
67
5.5 + 1 (1.8)
1(39 — 33)
10
7. Verify the other P values in the score column.
= 9.5 +
10x
S
+50
fi
(1.8) = 6.1
1
10
(6)
3
4
10
9
4
w
10.1
Score
B. STANDARD SCORES
EXAMPLE: Convert the raw scores of 78 introductory psychology
students on a speed of addition test (p. 182) into scaled scores with
a mean of 50 and a standard deviation of 10.
6.1
8.1
10.1
12.0
14.3
Two methods for computing standard scores are described. The
first method is based upon the definition of a standard score as
follows:*
where
Z = a standard score in a distribution with a mean of 50 and a
standard deviation of 10
x = deviation of a raw score X from the mean of its distribution
s = the standard deviation of the distribution
* See reference 19, p. 137.
184
Machine Computation of Elementary Statistics
The second method is based upon this formula, and uses in addi-
tion the following definition:*
X = a raw score
Sx =
IK
where
z = a standard score in a distribution with a mean of 0 and a
standard deviation of 1
X
=
N
11
1
Sx
3. Find
X
the standard deviation of the distribution of X scores
the mean of the distribution of X scores
1 X
Sx
(a) First method
1. Make a frequency distribution of the scores.
2. Compute X and s (p. 153). Here X = 10.1 and s = 3.2.
-
1
1- V
Here ² = 0.3125.
·
S
S
4. Set up a table, listing the scores in sequence from high to
low in the first column; their deviations from the mean (x) in the
10x
second column; values in the third column; and 50 + val-
10x
S
S
S
ues in the fourth (see 9 below).
1
5. Lock in the machine. Determine the number of decimal
places to be expected in the products, and in order to multiply
each product by 10, set the decimal marker for the product dials.
for one less than the number of places expected in the products.
Here five (4 + 1 = 5) are expected. Therefore, set product dials
for 4.
6. Multiply each a value by the constant, enter each prod-
X
10x
uct in the column, rounding each to one decimal place. Note
S
that the values for scores below the mean are negative.
7. Add 50 to each value in the
10x
S
column, remembering that
some of the values are negative, rounding the final results to whole
numbers.
* See reference 16, p. 82.
Percentiles; Standard Scores
185
8. With practice, the
Raw
Score
•
BOOK HR2 GOOD
added mentally to each
directly.
9. Conversion of raw scores of 78 introductory psychology
students on a speed of addition test into standard scores with a
mean of 50 and a standard deviation of 10:
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
(b) Second method *
10x
S
10x
S
2
X
2 =
7.9
6.9
5.9
4.9
3.9
2.9
1.9
0.9
-0.1
— 1.1
-2.1
-3.1
-4.1
-5.1
-6.1
-7.1
* For EXAMPLE, refer to p. 182.
† See reference 16, pp. 82-83.
column may be omitted, and 50
value in the machine and recorded
X-X
Sx
X - M₂
σε
;
10x
S
2
24.7
21.6
18.4
15.3
1
Sx
12.2
9.1
5.9
2.8
-0.3
-3.4
-6.6
-9.7
- 12.8
- 15.9
- 19.1
— 22.2
X
1
=X
1 x
Sx
σχ
10x
S
The following procedure is modified from Dunlap,† and is based
on his formula No. 25‡
1
στ
+ 50
∞ ∞ ∞ ∞ # # ** **** RØNN
Mx
72
68
65
62
59
50
47
+ Here symbols used by Dunlap are changed to conform with the others used
in this manual. He writes the formulas
43
40
37
34.
31
28
186
Machine Computation of Elementary Statistics
1. Divide 1 by sx rounding quotient to six decimal places.
1
Here
0.312500.
3.2
2. Set decimal markers to point off as many decimals as there
are in both!
are in both - and X. Here 6 + 1 = 7:
S
1
3. Lock ¹ (here 0.312500) in the keyboard.
S
4. Negative multiply by X (here 10.1). The result is a negative
number, the compliment of the number appearing in the ma-
chine, ··996.8437500.
1
5. Multiply by the largest X value. Here X = 18.0. The
S
corresponding z value 2.4687500 appears in the dials (Friden
upper; Marchant middle; Monroe lower). Remember to allow for
the decimal place in the multiplier. Round to two decimal places
and record in the table in the z column (see 9).
6. Line up the product dial and keyboard decimal points.
1
Subtract successively, and record each difference in sequence
S
until the next subtraction will result in a negative number (Friden
and Monroe use minus bar; Marchant subtract by using REVERSE
and 1 of multiplier keyboard, with NONSHIFT key depressed).
7. Clear dials. On Marchant, depress upper green carriage
shift key.
8. Multiply (still in keyboard) by X. The result 3.4375+
1
Friden: upper dials
Marchant: middle dials
Monroe: lower dials
—
S
appears in the product dials.
9. Negative multiply by the raw score immediately below the
mean. Round the result to two decimal places, 0.03, prefix a
negative sign, and record in the z column as -0.03. Line up the
1
product dial and keyboard decimal points. Add successively,
S
rounding each sum to two decimal places. Prefix a negative sign,
and record in sequence (Friden and Monroe use plus bar, Mar-
chant use 1 of multiplier keyboard with NONSHIFT depressed).
Percentiles; Standard Scores
187
10. For a distribution of scaled scores with X = 50 and s = 10,
multiply each z value by 10, round to whole number, and add 50.
Remember that the numbers below the mean are negative and
must be subtracted from 50.
11. When the procedure has been learned, set the decimal
10x
places for one less than desired to multiply by 10, so that ap-
pears with each subtraction and multiplication; 50 may be added
10x
S
S
to each value and recorded directly. In this case, for the addi-
tion and subtraction of 1, the keyboard decimal will be one place
to the LEFT of one product dials decimal.
12. Conversion of raw scores of 78 introductory psychology
students on a speed of addition test into standard scores:
Raw Score
BEUR HELE GOOD OBO H∞
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
x
7.9
6.9
5.9
4.9
3.9
2.9
1.9
0.9
-0.1
-1.1
-2.1
-3.1
-4.1
-5.1
-6.1
-7.1
N
2.47
2.16
1.84
1.53
1.22
0.91
0.59
0.28
-0.03
-0.34
-0.66
-0.97
- 1.28
- 1.59
-1.91
-2.22
10x
S
24.7
21.6
18.4
15.3
12.2
9.1
5.9
2.8
-0.3
-3.4
-6.6
-9.7
-
12.8
- 15.9
-19.1
— 22.2
10x
S
+ 50
No co co co ✡ + + **** RZNJ
72
68
62
59
56
53
50
47
43
40
37
34
31
28
REFERENCES
COMPUTATION
Manuals and Methods Published by Calculating
Machine Companies
Friden Calculating Machine Company
1. Friden Automatic Calculator: Classroom Methods. San Leandro, Cali-
fornia: Friden Calculating Machine Company, 1942.
2. Friden Automatic Calculator: Operating Instructions: Friden Full Auto-
matic Tabulating Calculator Model "ST." San Leandro, California:
Friden Calculating Machine Company.
3. Friden Automatic Calculators: Effort Savers. San Leandro, California:
Friden Calculating Machine Company.
Marchant Calculating Machine Company
4. Marchant Calculators: Operating Instructions. Oakland, California:
Marchant Calculating Machine Company.
5. Marchant Methods and Tables Issued to 1946 Relating to Basic and Statis-
tical Mathematics. Oakland, California: Marchant Calculating Ma-
chine Company, various dates to 1947.
This is a series of methods and tables published in separate short pamphlets.
Monroe Calculating Machine Company
6. Instruction Book: Monroe Adding-Calculator: M, Series 3 Models. Orange,
New Jersey: Monroe Calculating Machine Company, 1936.
7. Instruction Book: Monroe Adding-Calculator: Series 3 KA Models.
Orange, New Jersey: Monroe Calculating Machine Company, 1943.
8. Monroe: Application to Simple Correlation. Orange, New Jersey: Ac-
counting Service Department, Monroe Calculating Machine Com-
pany, 1937.
9. Monroe: Machine Methods for the Extraction of Square Root. Orange,
New Jersey: Monroe Calculating Machine Company, 1933.
10. Monroe: School Manual of Instruction. Orange, New Jersey: Monroe
Calculating Machine Company, 1930.
11. Monro-Matic: Monroe Adding-Calculator CAA Model: Operating Instruc-
tions. Orange, New Jersey: Monroe Calculating Machine Company,
1948.
188
References
189
12. Operating Instructions for Monroe Adding-Calculator: Model A-1.
Orange, New Jersey: Monroe Calculating Machine Company, 1940.
13. Operating Instructions: Monroe Adding-Calculator: Model AA-1.
Orange, New Jersey: Monroe Calculating Machine Company, 1941.
14. Teachers Guide for Educator Monroe Adding-Calculator. Orange, New
Jersey: Monroe Calculating Machine Company, 1947.
Other Sources
15. Croxton, Frederick E. and Dudley J. Cowden, Applied General Statistics.
New York: Prentice-Hall, 1946.
16. Dunlap, Jack W., Computation of Descriptive Statistics. New York:
Ralph C. Coxhead Corporation, 1937.
17. Garrett, Henry E., Statistics in Psychology and Education. New York:
Longmans, Green and Co., 3rd edition, 1947.
18. Tremmel, E. E. and C. C. Weidemann, A Machine Method of Calculating
the Pearson Correlation Coefficient (University of Nebraska Publication
Number 72). Lincoln, Nebraska: University of Nebraska Extension
Division, June 1930.
19. Walker, Helen M., Elementary Statistical Methods. New York: Henry
Holt, 1943.
20. Wallace, H. A. and George W. Snedecor, Correlation and Machine Cal-
culation (Iowa State College Official Publication 30, No. 4). Ames,
Iowa: Iowa State College of Agriculture and Mechanic Arts, 1931.
TABLES
21. Comrie, L. J. ed., Barlow's Tables of Squares, Cubes, Square Roots, Cube
Roots and Reciprocals of All Integer Numbers up to 12,500. New York
City: Chemical Publishing Company, 4th edition, 1944.
22. Dr. A. L. Crelle's Calculating Tables Giving the Products of Every Two
Numbers from One to One Thousand. (Revision by Dr. C. Bremiker.)
London: David Nutt, 1st English Edition, 1902.
23. Dunlap, J. W. and A. K. Kurtz, Handbook of Statistical Nomographs,
Tables, and Formulas. Yonkers-on-Hudson, New York: World Book
Company, 1932.
24. Fisher, Ronald A. and Frank Yates, Statistical Tables for Biological,
Agricultural and Medical Research. New York: Hafner Publishing
Company, 3rd edition, 1948.
25. Holzinger, Karl J., Statistical Tables for Students in Education and
Psychology. Chicago, Illinois: University of Chicago Press, 3rd edition,
1931.
26. Kelley, Truman L., The Kelley Statistical Tables, Revision of 1948.
Cambridge, Massachusetts: Harvard University Press, 1948.
APPENDIX A
GLOSSARY OF SYMBOLS
↑
:
A Arbitrary origin of a frequency distribution; here taken as
the midpoint of the lowest step interval.
bxy Regression coefficient used in the equation to predict X
from Y.
*
GLOSSARY OF SYMBOLS *
byx Regression coefficient used in the equation to predict Y
from X.
f A frequency.
i
Width of a step interval.
N Number of cases under consideration.
p Per cent.
Correlation coefficient indicating the correlation between
X and Y.
Standard deviation of an observed sample.
Σ Sign of summation.
TXY
s
X, Y Gross scores.
x Deviation of a score X from the mean of its distribution.
x' Deviation of a score X from a reference point other than
the mean of its distribution.
X Mean of the distribution of X scores.
Ỹ Mean of the distribution of Y scores.
* See reference 19, pp. 350-359, except per cent.
193
Į
A
.
APPENDIX B
COMPUTING SHEETS
1
f
MEAN AND STANDARD DEVIATION FROM GROUPED DATA
COMPUTING SHEET
Study:
Variate:
Formulas:
(1) mean:
(3) check formulas:
Basic data:
N
A₂(for x' + 1)
Σf(x' + 1)²
NΣf(x+1)²
$
1
IX
Check steps: Substitution
Σf(x+1)=Σfx' + N =
Σf(x' + 1)²
11
x = A + — (Zƒx')
N
1.
8
I
Computation of final values:
Result
Substitution
NΣf(x')² - (Σƒx')²
√NΣƒ(x')² — (Σƒx')²
N
—
N
Σf(x' + 1) = Efx' + N; Σf(x' + 1)² = Σƒ(x')² + 2 Zƒx' + N
-
i.
i
N
(Σfx')
(Zfx')2
Date:
Computer:
Σfx'
4A4
√NZƒ(x')² — (Σƒx')² =
=
A₁ + √√ (Efx') =
N
Σf(x')² + 2Σƒx' + N =
[Σf(x′ + 1)]
i
212
X A₂+ √ √ (2ƒ(x² + 1)] =
N
N
NΣf(x + 1)² - [Zƒ(x' + 1)]² =
√NΣf(x' + 1)² − [Eƒ(x′ + 1)]²
√NΣf(x' + 1)² — [Zƒ(x' + 1)]²
(2) standard deviation:
i
Σf(x)2
ΝΣf(x)2
Final values: X
8
S
—√NΣf(x')² — (Zfx')²
A₁(for x')
Σf(x' + 1)
[Σf(x' + 1)]²
Value
Value
Value
Check
Check
197
MEAN AND STANDARD DEVIATION FROM GROUPED DATA
COMPUTING SHEET
Study:
Variate:
Formulas:
(1) mean:
(3) check formulas:
Σf(x' + 1)
Basic data:
N
A2(for x' + 1)
Σƒ(x' + 1)²
NΣf(x + 1)²
1
S
|
X = A + ² (2ƒx')
| | is
1
Computation of final values:
Result
Substitution
IK
X
Check steps: Substitution
Σƒ(x' + 1) = Σƒx' + N
Σƒ(x' + 1)² = Σƒ(x')² + 2Σƒx' + N =
N
X A₁ +
N
i
i
N
=
(fx')
Date:
Computer:
―
(Σƒx')²
Nef(x') (2ƒx')² =
√NΣƒ(x')² — (Eƒx′)²
Σfx'
√NΣf(x')³ — (Σƒx')² =
(Zfx')
Σƒx' + N; Σf(x' + 1)² = Σƒ(x')² + 2 Σƒx' + N
zie.
N
i
N
✩ [Zƒ(x' + 1)] =
A2 + ~ [Zƒ(x' + 1)] =
N
(2) standard deviation:
i
N
Σf(x)2
NΣf(x')²
NΣƒ(x' + 1)² — [Eƒ(x' + 1)]²
√NΣƒ(x' + 1)² − [Eƒ(x' + 1)]²
√NΣƒ(x' + 1)² — [Σƒ(x' + 1)]²
Final values: X
198
S VNZf( ) - (fx)?
S
A₁(for x')
Σf(x' + 1)
[Σƒ(x² + 1)]²
Value
Value
Value
Check
Check
PRODUCT MOMENT CORRELATION COEFFICIENT FROM
Study:
Variates: X
Date:
Y =
Formulas:
(1) means:
Final values: rxy
X
Ỹ____
Computer:
x = = = ΣΧ;
Directions: Divide data into subgroups of 20 to 25. Compute quantities for
basic data and checks and enter in proper spaces on page 2. Sum, and enter
TOTAL values in proper spaces at bottom of page 2.
(2) standard deviations:
1/4
N
Sx =
If using N in the denominator of the definitional formula for the standard
deviation, compute final values as directed on page 3; if N − 1, use page 4.
√N2X²
TXY =
(3) regression coefficients:
byx =
RAW SCORES
COMPUTING SHEET
-
(4) correlation coefficient:
ΝΣΧΥ - (ΣΧ)(ΣΥ)
NEX² (EX)²
(ZX)²;
Sx
2
TXY
r²xy = byxbxy;
SY
(5) check formulas:
ΣΧ ΣΥ
2X + 2Y = Σ(X.+ Y);
ΝΣΧΥ (ΣΧ)(ΣΥ)
√NEX² — (EX)² √NZY² — (≥Y)²
(ΣΧ) (ΣΥ)2
Y
8y
bxy
byx
bxy
ΣΧ – ΣΥ
Σ(X + Y)² + Σ(X — Y)² = 2ΣX² + 2ΣY2
2XY = {{2(X,+ Y)² — 2(X - Y)2}
ΣΧΥ
EXY = }{2(X + Y)² − ΣX² — ΣY²}
ΣΧΥ
ΣΧ
(EX)(2Y) = }{[E(X + Y)]² − [E(X — Y)]²}
-
SY
bxy
byx = rxy
88'
or
ΣΥ
N
√NZY2
Σ(X — Y)
ΝΣΧΥ (2X)(EY)
ΝΣΥ - (ΣΥ)
TXY
(ZY)²
8x
8Y
199
Basic Data and Checks: (set up for N up to 125)
Check column
ΣΧ + ΣΥ 2(X + Y)
Group ΣΧ
1
2 3 4 5
Total
Group ΣΧ
1 2 3
4
5
Total
Group 2 ΣΧΥ
N
1 2 3 4 5
Total
ΣΥ
ΣΥ
ΣΧΥ
2 ΣΧ2
+2ΣY2
|| ||
Check column
|| || ||
Σ(X+Y)²
+2(X-Y)2
TOTAL Values for Computation of Final Values:
1
N
Check column
ΣΧ – ΣΥ
-
Σ(X+Y)²
Check column
Check column
Σ(X+Y)² − 2(X−Y)² | 2(X + Y)² — ΣX² ΣΥ2
4
2
1
√N(N − 1)
Σ(X — Y)
2(X-Y)2
Enter values from TOTAL rows and compute new values where necessary:
ΣΧ
ΣΥ
2 ΣΧΥ
ΣΧ2
ΣΥ
NEX2
NZY2
(ZX)2
(ZY)²
ΣΧΥ
ΝΣΧΥ
(ΣΧ)(ΣΥ)
200
Computation of Final Values: N in denominator of definitional formula for
the standard deviation:
Result
(ΣΧ)(ΣΥ)
Numerator
(Nsx)²
Nsx
(N8y)2
Nsy
TXY
8X
SY
IK
X
Y
ΝΣΥ2
Denominator √NEX²
byx
bxy
Substitution
}{[2(X + Y)]²
Checks:
— [E(X — Y)]}
-
ΝΣΧΥ
NEX2
(EX)²
√NEX² - (EX)2
(ΣΥ)2
NEY2
S
N
(ZY) 2
(EX)²
X √NZY² - (EY)²
Numerator
Denominator
11/12x =
NEX
(ΣΧ)(ΣΥ)
<
√NΣX² .
G
ΣΥ Ξ
G
— √N₂Y² — (±Y)²
-
go2 Xxx
XY
Numerator
(N8x)2
Numerator
(Nsy)2
= byxbxy
byx = rxy
SY
bxy = rxy
8X
'8X
(EX)²
8Y
•
Value
Check
201
Computation of Final Values: N - 1 in denominator of definitional formula
for the standard deviation:
Result
(ΣΧ)(ΣΥ
Numerator
(Nsx)2
Nsx
(Nsy)²
Nsy
TXY
SX
SY
Denominator √NZX²
||
X
Y
byx
Oxy
Substitution
} {[2(X + Y)]²
Checks:
—[≥(X — Y)]²}
NEXY — (EX)(EY)
NEX2 (EX)²
VNEX2
ΝΣΥ - (ΣΥ)
VΝΣΥ - (ΣΥ) =
G
(EX)2
X VNEY 2
N
Numerator
Denominator
1
1
√N(N − 1)
ΣΧ
-
ΣΥ
r2
-
× √ѲX² – (EX)²
2
1
√Ñ(N − 1)
X VNEY² - (ΣY)²
Numerator
(Nsx)²
Numerator
(Nsy)²
XY
byx
************
-
(ZX)²
C
byxbxx
SY
TXY
SX
bxy = TXY
-
SX
*****
SY
(ΣΥ)2
Value
Check
202
APPENDIX C
ANSWER KEY
1
”
!
ANSWER KEY
Addition and Subtraction
Addition. Pages 28 to 40.
(1) 915
(2) 1,081
(3) 1,158
(4) 2,117
1,312
(5)
(6) 1,978
(7) 13,987
Subtraction. Pages 40 to 44.
(1) 78
(2) 505
(3) 839
(4) 1,090
(1) (…..99877)
(2) (…..99679)
(9) 1.0
Complements. Pages 46 to 51.
- 123
-321
(5) (...997747.693)
118,404
(1)
(2) 1,367,400
(3) 262,224
(4)
184,437
(5) 4,395,801
24,082
(8)
(9) 123,478
(10) 1,098,426
(11)
(12)
(13)
(14)
(3)
(4)
(5) 22,936,320
686.1369
4,520.44237
7,461.4348
14,601.424
(5) 6,270
(6) 3,471
(7) 534
(8) 9,129.47
=
(3) (...99634)
(4) (...99768.09)=-231.91
Multiplication
(a) and (b). Multiplication of whole numbers. Pages 53 to 59.
-2252.307
(6) 30,261,654
(7) 32,285,124
1,090,050
(c) Multipliers with digits 6, 7, 8, 9. Page 60.
(1) 20,592
(2)
458,271
498,112
867,573
(8)
(9) 259,257,680
(10) 496,950,300
-366
205
206
Machine Computation of Elementary Statistics
(d) Multiplication of numbers with decimals. Pages 60 to 70.
(1) 396,772.48
0.000100004
(2)
(3)
(4)
1,352.25055
3,902.5686
18,727.0
(5)
59.8026
(6) 624.49396
(7)
0.03115380
(1) 2,575
(2) 27,342
(3) 166,581.95
(4) 1,088
(5) 62,376
(6) -0.6372
(7) ΣΧ
278
(e) Accumulative and negative multiplication. Pages 72 to 89.
(10) EX = -247
ΣΧ
129.8091
ΣΧ2
(11) ΣΧ -0.42
ΣΧ2 136.6012
EX² =
ΣΧ2
(8) EX =
ΣΧ
ΣΧ
(9) ΣΧ
ΣΧ2
—
=
(1) 1,568
(2) 2,914
(1) 213
(2) 132
(3) 163
(4) 95
(5) 121
(3) 4,802
(4) 2,303
(5) 5,727
(6) 4,598
(7) 13,824
18,342
1,448
982,750
1.89
1.7587
(8)
(9) 448.1568
(10) 0.180632846
(11)
0.0625
(12) 0.1600
0.0016
0.0000156816
(13)
(14)
(15) 9,801
(12)
Use of Tables of Products. Page 93.
(8)
1,404
(9) 105,625
(10) 19,321
(11) 30.25
(12)
=
1089 2804
1385 1366
(13)
(14)
10417 -888'
4736 1833
(13)
(14) 3,337
(15) 4,827
3.7636
9,801
9,216
Division
Division of whole numbers. Pages 95 to 99.
(6) 99
(7) 78 and 367/865
(8) 75 and 7/795
(9) 301 and 65/173
(10) 138 and 31/269
Answer Key
207
Division of numbers with decimals. Pages 100 to 111.
(a) Numbers in extreme left of keyboard.
(1) 119.3196187
(3) 0.424891234
(2) 268.788969
(4) 0.995403381
(5) 0.0160552503
(b) Constant or preset decimal setup.
(1) 338.0249135 rounded to 338.024914
(2) 176.5840220 rounded to 176.584022
(3) 14.3470041 rounded to 14.347004
(4) 6.0989983 rounded to 6.098998
(5) 0.5551004 rounded to 0.555100
(6) 805.5409081 rounded to 805.540908
(7) 0.0060173 rounded to
(8) 0.1356304 rounded to
(9) 0.0086206 rounded to
0.006017
0.135630
0.008621
Reciprocals. Pages 113 to 122.
(1) 0.00272479564
(2) 0.000683994528
(3) 0.000158428390
(4) 0.000159948816
(5) 0.0156813549
(6) 0.0202591551
(13) Given two numbers having identical integers
in identical order, but with the decimal points
in different positions, the reciprocals of the
two numbers differ only in magnitude.
(14) Cumulative percentages: 100.0, 98.4, 95.9,
90.2, 83.7, 78.0 74.0, 69.1, 57.0, 41.5, 27.6,
21.1, 16.3, 9.8, 6.6, 1.6
(1) 1.621
(2) 0.5495
(7) 11.1607143
(8) 0.0100100100
(9) 0.00917431193
(10) 91.7431193
(11) 0.290782204
(12) 0.00290782204
Use of Tables of Quotients. Page 122.
(5) 0.0241
(3) 0.7848
(4) 0.9877
208
Machine Computation of Elementary Statistics
Square Root
(a) Subtraction of successive odd numbers. Pages 126 to 141.
(1) 94
(2) 52
(3) 82
(4) 99
(5) 89
(11) 9.21023343
(12) (a) 100N
(b) a²N² =
(13) 1.861209714
(14) 2.632148020
(17) If b
=
=
(1)
(2)
(3)
(4)
(5)
(6) 366.+
(7) 31,000.+
(8) 0.238+
29.48+
(6) 15.1
(7)
(8) 665.206734
2a, then √ō = √2a
0.704+
31.6+
0.284
0.619806421
(9) 92.1023343
(10) 29.1253154
√100 √Ñ = 10√Ñ
(aN)²
(b) Use of tables of square roots. Page 150.
Estimation
732
(15) 2.514866854
(16) 3.556558810
1.414 Va
=
Computation
732.
29.4867309
0.704228656
31.6918285
0.284
366.563773
31,026.02778
0.238520439
*
•
|
•
*
}
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UNIVERSITY OF MICHIGAN
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