For instructional purposes only • 1 st Semester SY 2020-2021
3
Lesson 3.1: Data Gathering and
Organizing Data
Lesson Summary
This lesson discusses the data collection process. It also deals with the two
major types of data, the four levels of measurement, and the different forms of
organizing and presenting data.
Learning Outcomes
At the end of the lesson, the students will be able to:
1. Differentiate between qualitative data and quantitative data.
2. Classify data according to the four levels of measurement.
3. Represent data in frequency distributions graphically, using histograms
and frequency polygons.
Motivation Question
Imagine yourself as an employee of the Philippine Statistics Authority (PSA)
and assigned to gather information about a small town in a remote province of
Southern Philippines. You are given the task of selecting the type of data that
will be useful for the community. What type of data will you collect? How are
you going to organize and interpret the data you have collected?
Discussion
Introduction
Over the years, people have been interested in determining the
occurrence of certain events at certain periods of time (i.e., birth rates,
mortality rates), crop yields, frequency of failures in school entrance exams,
etc. These activities deal with the counts or numerical measures of activities,
events, and things, which are called statistics in a limited sense.
From the research point of view, statistics is a science that deals with
the collection, presentation, analysis, and interpretation of data. Data collection
is the process of gathering and measuring information about variables being
studied in an established systematic procedure. It refers to facts or figures
from which a conclusion can be made. Data gathering involves getting
information through individual interviews, focus groups, questionnaires,
observations, experimentations, and many other methods.
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Population and Sample
A common way of expressing the fact that there are 18 565 people in a
city is to say that it has a population of 18 565. The term population, as used in
statistics, refers to a set of people, objects, measurements, or events that
belong to a defined group.
For example, the total number of trees or carabaos in a town is a
population. All the residents of Baybay City comprise a population. All the
academic staff of VSU at a specific time is a population. Quite often, we think
of populations as containing large numbers of members. Some populations do
while others do not. The distinguishing characteristic of a population is that all
members are included according to whatever defines the population.
In many research situations, it is not feasible to involve or measure all
members of a population. Hence, researchers resort to studying only a part of
the population known as the sample. A sample is defined as a subset of a
population. A sample is any subset of elements drawn by some appropriate
method from a defined population. The sample is a small but representative
cross-section of the population.
Population
-- ...
S mple
...
t
+
-
.,,.
\
IJ
Figure 1. Population and Sample
Data may be classified into two major types: qualitative and
quantitative. Quantitative data can be counted, measured, and expressed using
numbers. Contrary to quantitative data, qualitative data is descriptive and
conceptual. It can be categorized based on traits and characteristics.
Qualitative data include information on attributes such as:
•
•
•
•
•
•
Sex (male and female)
Attitude (favorable or not favorable)
Emotional condition (happy or sad)
Color (white, black, or brown)
Civil status (single or married)
Ratings (excellent, good, satisfactory, or poor)
Quantitative data, on the other hand, involve numbers and are the result of
counting or measuring. For example:
•
•
•
Vision:
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Number of students in a class
Price of a certain commodity
Age of Olympians
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•
•
•
Temperature of Coffee
Height of basketball players
Weight of an individual
Numerical data gathered about the samples are either discrete or
continuous. Discrete variables are those obtained through counting. It can only
assume a countable or finite number of values. It cannot take the form of
decimals. For instance, we can say that there are 1000 families in a certain city
and not 1000.46 families. Another example is the size of a particular family
since it can only take a specific value such as 2, 3, 4, 5, and so on. Values
between them, like 2.5 or 4.5, are not possible. We cannot have a family with
4.5 members.
Other examples of discrete variable:
•
•
•
•
Number of children in a family
Number of barangays in a city
Number of buildings in a school
Number of female employees in a company
Continuous variables are the result of a measurement. It can assume
infinitely many and continuous values. Suppose that we measure the height of
a person and we say that he is 121 centimeters in height. Does it mean that he
is exactly 121 cm. tall? Of course, he is not. In reading the scale, we merely read
the number of centimeters to which the person's height was closest.
Other examples of continuous data:
•
•
•
•
Height
Weight
Length
Temperature
Qualitath1e
Quantitative
De!l«lptililer
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m
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IJ'
11
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Four Levels of Measurement
Data such as sex distribution, age distribution, number of children in a
family, family income, and many others can be classified, measured, or labeled
in different ways. The measurement of these characteristics is classified within
a hierarchy of measurement scales that include the nominal scale, the ordinal
scale, the interval scale, and the ratio scale. It is important to know the kind of
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variable or data we are dealing with so that the data gathered can be properly
interpreted and the appropriate statistics are used.
1. Nominal Scale - are variables which can be classified into two or more
categories. The variables are grouped such that all those in a single
class are equivalent with respect to some attribute or property.
Examples:
•
•
•
•
•
•
•
Student ID Number
Sex (male/female)
Soft drinks (Coke/Pepsi/Sarsi)
Religion (Roman Catholic/Protestant/Islam)
Nationality (Filipino/American/German/Korean)
Birthplace of Respondents (Urban/Rural)
Work Station (Government/Private)
For convenience, numbers or letters are assigned for nominal
data/variables. For example, sex may be assigned as male (A) or
female (B). Another example is the smoking habits of people wherein
numbers can be assigned as follows: non-smoker (0), mild smoker (1),
and a heavy smoker (2). Arithmetic operations on these numbers have
no meaning because they are just used for identification.
2. Ordinal Scale - In this scale, there is no standard difference in
measurement. It has one additional property over those of the nominal
scale where it classifies data; however, the classification has ranks.
Examples:
•
•
•
•
•
Grading System
Military Rank
Job Position
Academic Honors
Likert Scale (Strongly Agree, Agree, Neutral, Disagree, Strong
Disagree)
3. Interval Scale - An interval scale possesses the characteristics of the
nominal and ordinal scale wherein the data are categorized and
ranked. However, this scale has the property of meaningful distance
between values. The zero point of the interval scale is just arbitrary
and does not reflect an absence of the attribute (no true zero point).
Examples:
•
•
•
Intelligence (IQ)
Test scores
Temperature in Fahrenheit and Celsius
Suppose you got a score of 60 on a Mathematics test and
your classmate got 50. It is meaningful to say that your classmate's
score is 10 points lower than yours or that your score is 10 points
higher than his/her score. And suppose a student got zero in an
English test. Does it mean that the student has absolutely no
knowledge of English? Or that he/she does not know anything in
English? No. Likewise, in temperature, 0 degree Celsius does not
mean the absence of heat.
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4. Ratio Scale - This scale takes all the properties of the interval scale
with an identifiable absolute zero point. Here, the zero point is not
arbitrary but indicates the total absence of the property measured. A
ratio variable refers to a variable where equality of ratio or proportion
has meaning.
Examples:
•
•
•
•
•
•
•
Height
Weight
Distance
Monthly Income
Number of babies in the family
Temperature in Kelvin
Money
If a person has P400 while the other has only P200, then we
can say that the former has twice as much money as the latter. And
suppose a third person has no money, then we can say that he has
zero pesos. Here, money is an example of a ratio variable. The
temperature in Kelvin is also another example. Zero Kelvin is a
meaningful concept. Zero is the absence of heat (meaning it cannot
get colder).
Forms of Data Presentation
Data gathered remain meaningless unless organized. Usually, the
information collected is translated into numerical or quantitative data. These
data can be represented by using graphs, figures, and tables.
1. Frequency Distribution Table. This is an excellent device for making
larger collections of data much more manageable. The frequency
distribution table has two parts - the frequency table and the extended
frequency table.
• A frequency table lists categories of scores along with their
corresponding frequencies. The frequency for a category or
class is the number of original scores that fall into that class.
• The extended frequency table consists of columns that can
generate various graphs or charts. It is a prerequisite for
creating graphs and charts used in statistics. It consists of the
following:
a.
b.
c.
d.
e.
f.
Class intervals (lower and upper limits)
Marks
Frequency
Cumulative frequency
Relative frequency
Cumulative relative frequency
Guidelines for frequency tables:
1. Class intervals should not overlap. Classes are mutually exclusive.
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2. Classes should continue throughout the distribution with no gaps.
Include all classes.
3. All classes should have the same width.
4. Class widths should be "convenient" numbers.
5. Use 5-20 classes.
6. Make lower or upper limits multiples of the width.
Example 1:
Given the following statistics scores of 50 students in a senior high
school class, make a frequency distribution table.
Solution:
39
33
37
37
32
30
44
45
31
26
54
35
32
40
49
38
37
22
36
32
42
20
35
48
36
32
40
30
33
36
31
47
51
44
41
32
32
38
43
38
25
26
36
38
42
37
33
35
36
39
Step 1: Arrange the raw data in descending order (highest to lowest).
54
51
49
48
47
45
44
44
43
42
37
37
37
37
36
36
36
36
36
35
42
41
40
40
39
39
38
38
38
38
35
35
33
33
33
32
32
32
32
32
32
31
31
30
30
26
26
25
22
20
Step 2: Solve for the range (R).
R
R
R
= highest score - lowest score
= 54- 20
= 34
Step 3: Determine the number of classes using Sturge's Rule. Under this rule,
the number of classes is given by:
k
k
k
k
k
= 1 + 3.3 (logn)
= 1 + 3.3 (log50)
= 1 + 3.3 (1.69897)
= 1 + 5.607
where: k = no. of classes
n = the no. of cases in the data
= 6.607
Step 4: Solve for the class size.
c
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Mission:
R
k
34
6.607
= - = -- = 5.146 or 5 (Round off to the nearest whole number)
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Step 5: Construct the frequency table.
a. Determine the lower limit (LL) of the first class.
When zero is the lowest value, then it naturally becomes the lower limit
of the lowest class. Since the lowest value in the data set is 20, then
this will serve as our starting point.
b. Enumerate the class intervals.
The succeeding lower limits (LL) can be established first by simply
adding the class size c to the preceding lower limit. The upper limit (UL)
is the step lower in the next class. Always bear in mind that classes
should not overlap. The following formula can also be used:
Upper limit
= lower limit + C - 1 unit of measure
Note: The difference between successive upper limits is also equal to c.
c. Tally the observations to determine the class frequencies.
d. For the class mark, compute the midpoint of the class limits/class
.
LL+UL
20+24
.
boundaries using the formula: m = --.
For example, m = - = 22,
2
2
then you may just add the class width c = 5 for the succeeding class
marks. That is, the class marks are 22, 27, 32, and so on.
e.
f.
For< cf, start from the lowest group frequency, then add the frequency
of each class for the succeeding classes.
For > cf, start from the highest group frequency, then add the
frequency of each class for the succeeding classes.
g. For rf, it is the f divided by n, where n is the total number of scores.
rf
Frequency
Class Interval
Class Boundaries
LL-UL
LB- UB
20-24
19.5 - 24.5
II
2
22
25-29
24.5 - 29.5
111
3
27
30-34
29.5 - 34.5
1ttk'H-1.l - 111
13
32
35-39
34.5 - 39.5
lltk-"'ti4k'ttl-k-111
18
37
40-44
39.5 - 44.5
'H-1.l- Ill
8
42
45-49
44.5 - 49.5
1111
4
47
50-54
49.5 - 54.5
II
2
52
Total
Vision:
Mission:
= [_n X 100
Tally
f
n
Class
Marks
= 50
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Class
Interval
Frequen
cy
LL- UL
f
20 -24
2
25-29
30-34
35-39
40-44
45-49
50-54
3
13
18
Less than
Cumulative
Frequency
< cf
Greater than
Cumulative
Frequency
> cf
Relative
Frequency
5
48
6%
36
32
36%
6
8%
2
18
8
44
2
50
4
48
rf (%)
50
4%
45
26%
14
2
Cumulative
Relative
Frequency
(%)
4%
10%
36%
72%
16%
88%
4%
100%
96%
2. Make Charts or Graphs
After gathering and organizing the data in a frequency
distribution, the next step is to present them in a way that is easier to
understand. One way is through graphical representation. There are a
number of graphs or charts in the presentation of the frequency
distribution. These include histogram, frequency polygon, and
cumulative frequency (ogive).
a) The histogram is a graph in which the classes are marked on the
horizontal axis (x-axis) and the class frequencies on the vertical axis
(y-axis). The height of the bars represents the class frequencies,
and the bars are drawn adjacent to each other.
Example 2:
Consider the data set of Zoe's Exam Scores. Take a look at its
frequency distribution table and create a histogram.
Solution:
Step 1: Find the class marks (midpoints) of each class.
Step 2: Draw and label the x-axis and y-axis.
Step 3: Represent the frequency on the y-axis and the midpoints on
the x-axis.
Step 4: Use the frequency to represent the height and draw the
vertical bars.
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Frequency
20
18
16
14
12
10
8
6
4
2
I
I
I
22
I
I
27
I
I
32
37
I
I
I
42
I
47
I
I
52
I
Statistics Scores
Histogram for Students' Statistics Scores
Remarks:
The graph has no gaps; it is helpful when the distribution is
interval or ratio. Histograms also illustrate central tendency, shape,
and how the data are spread out or dispersed. It may be
symmetrical, uniform, skewed, and bi-modal.
b) The frequency polygon is a graph that displays the data using points
that are connected by lines. It actually looks like a line graph. The
frequencies are represented by the heights of the points at the
midpoints of the classes. The vertical axis represents the frequency
of the distribution, while the horizontal axis represents the
midpoints of the frequency distribution.
Example 3:
Consider the frequency distribution of the previous example and
make a frequency polygon.
Solution:
Step 1: Find the class marks (midpoints) of each class.
Step 2: Draw and label the x-axis and y-axis.
Step 3: Represent the frequency on the y-axis and the midpoints on
the x-axis.
Step 4: Connect the dots. Draw a line back to the x-axis at the
beginning and end of the graph.
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Frequency
20
18
16
14
12
10
8
6
4
2
22
27
32
37
42
47
52
Statistics Scores
Frequency Polygon for Students' Statistics Scores
c) The cumulative frequency polygon or ogive (read as "oh' - jive") is
a graph that displays the cumulative frequencies for the classes in
a frequency distribution. The graph is typical "upward" in trend. It
also shows values below a certain boundary.
Example 4:
Consider the frequency distribution of the previous example
and make a cumulative frequency polygon.
Solution:
Step 1: Find the cumulative distribution of the data set.
Step 2: Draw and label the x-axis and y-axis.
Step 3: Represent the cumulative frequency on the y-axis and the
midpoints on the x-axis.
Step 4: Connect adjacent points with line segments.
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Cumulative Relative Frequency (%)
100
90
80
70
60
50
40
30
20
10
22
27
32
37
42
47
52
Statistics Scores
Cumulative Frequency Polygon for Students' Statistics Scores
Learning
Tasks/Activities
A. Determine whether the numbers obtained in the following variables are
quantitative or qualitative.
1. Address
2. Student Number
3. Weight of wrestlers
4. Happiness Rating
5. Distance of planets
B. Classify the following variables according to whether they are nominal,
ordinal, interval, or ratio data.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Vision:
Mission:
Brands of soft drinks
Birth orders of children in a family
Places in a beauty contest
Scores in an aptitude test
Attitudes toward the teaching profession
Efficiency ratings of employees
Grades in high school
Heights of students
Varieties of corn
Family Income
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C. Determine whether the numbers obtained in the following variables are
discrete or continuous.
1. Books in a library
2. Spots on a die
3. Weights of infants at birth
4. Coconut yield per hectare
5. Volume of a pail of water
6. Floor area of a classroom in square meters
7. Distance traveled by a bus in one day
8. Number of domestic animals in a barangay
9. Average temperature of a place in one year
10. Length of a residential lot
Assessment
Given below are the mathematics scores of 54 students in a high school
senior class.
71
31
72
41
51
43
77
33
74
38
46
44
68
36
66
34
42
47
64
40
63
39
46
53
55
45
61
41
51
48
50
50
60
46
58
48
45
55
56
50
59
49
40
63
50
56
52
50
35
70
46
57
47
42
Construct the following:
a.
b.
c.
d.
Frequency distribution table
Histogram
Frequency polygon
Ogive
Instructions on how to submit student output
Refer to the course policies and course content plan.
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15
Lesso n 3.2: M easu res of Ce ntra l Te n d e n cy,
Positi o n , a n d Va riation
Lesson Summary
This lesson discusses how data can be described using the measures of
averages. It deals with the three measures of central tendency, namely, the
mean, median, and mode. The measures of relative position and measures of
variation are also discussed. The measures of relative position, such as the
quartiles, percentiles, z-scores, and box- &-whisker plot, are used to describe
the location of the data value in the data set while the measures of variation,
such as the range, interquartile range, absolute deviation, variance, and
standard deviation, are used to tell how to scatter or spread out a distribution
is.
Learning Outcomes
At the end of the lesson, the students will be able to:
1.
2.
3.
4.
5.
6.
7.
8.
Find the mean, median, and mode of a given data set.
Interpret data using the measure of central tendency.
Find the range of a data set.
Find the interquartile range.
Find the variance and standard deviation.
Find the absolute deviation.
Find the first, second, and third quartiles.
Identify the position of the data value in a data set using percentiles.
Motivation Question
Measures of central tendency, position, and variation are essential topics in
statistics; why is it important to study each of them?
Discussion
Measures of Central Tendency
When given a set of observations, one of the things we would want to
know is a value that is characteristic of the group. This value must best
describe the group and be a representative of all the observations. This value
is called the measure of central tendency.
A measure of central tendency represents the center point or typical
value of a set of data. In layman's term, a measure of central tendency is an
average. In statistics, the three most common measures value of central
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Math1 1 n: Mathematics in the Modern World
tendency are the mean, median, and mode. In the following discussion, we will
look at the mean, median, and mode and learn how to calculate them and under
what conditions they are most appropriate to be used. Choosing the best
measure of central tendency depends on the type of data you have.
A. Mean
The arithmetic mean, often called the mean, is the most frequently used
measure of central tendency. The mean is the only common measure
in which all values play an equal role, meaning, to determine its value,
we need to consider all the values of any data set. It is easy to calculate:
just add up all the numbers, then divide by how many numbers there
are.
The mean is denoted by x and can be computed using the formula:
LX
-=­
x
Properties of Mean
where: x = the value of an observation
n = the total no. of observations
n
1.
2.
3.
4.
5.
A set of data has only one mean.
Mean can be applied for interval and ratio data.
All values in the data set are included in computing the mean.
The mean is very useful in comparing two or more data sets.
Mean is affected by the extremely small or large values in the data
set.
6. Mean is most appropriate in symmetrical data.
Example 1:
Solution:
Example 2:
Solve for the mean of the set 1, 5, 3, 9, 7.
_ L X 1 + 5 + 3 + 9 + 7 - 2-5 - 5
x = - = ------ 5
5
n
Suppose a basketball team has 15 players, and the heights (in
cm) are as follows:
1 80
1 93
1 88
204
1 86
1 86
1 84
1 87
Find the mean height of the players.
Solution:
_
x=
Vision:
Mission:
LX
15
=
1 89
1 82
1 90
201
1 90
1 87
200
2847
15 = 189.8 cm
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Example 3:
A teacher gave five tests in mathematics. Dina got the following
scores in the first four tests: 82, 76, 79, and 81. What must be her score
in the fifth test so that his average s 80?
Solution:
Given: x
Let x5
= 80
= Dina's score in the fifth test
- LX
x=­
n
80
82 + 76 + 79 + 8 1 + X
= ____5____5
80(5) = 318 + X5
400 = 318 + x 5
X 5 = 400 - 318
X5
= 82
Therefore, Dina's score on the fifth test is 82.
Weighted Mean
The weighted mean is particularly useful when various classes
or groups contribute differently to the total. The weighted mean is
found by multiplying each value by its corresponding weight and
dividing by the sum of the weights.
The formula for finding the weighted mean is:
_
X1 W1 + Xz W2 + X3 W3 + . . . + Xn Wn
Xw = -----------W1 + Wz + W3 + . . . + Wn
Example 4:
Joel's first quarter grade is shown in the table below. Use the
weighted mean formula to find Joel's GPA for the first quarter.
Subjects
G rade
No. of U n its
English
90
3
M ath
87
3
Filipino
88
3
Science
93
3
MAPEH
95
2
Hele
96
1
Solution:
_
Xw
Vision:
Mission:
=
90(3) + 87 (3) + 88 (3) + 9 5 (2) + 96(1)
3+ 3+ 3+2 + 1
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1, 088
xw = --
xw
12
= 90.67
Therefore, Joel's GPA for the first quarter is 90. 67.
B. Median
The median (Md ) is the middle value in a set of observations
arrange from highest to lowest or vice versa. Hence, to get the value of
the median, we arrange the observations from highest to lowest or from
lowest to highest. The observation in the middle is considered as the
median.
Properties of Median
1. The median is unique; there is only one median for a given set of
data.
2. Median is not affected by the extremely small or large values.
3. Median can be applied for ordinal, interval, and ratio data.
4. Median is most appropriate in skewed data.
Example 1:
Consider again the heights (in cm) of 15 basketball players
listed. Find the median height of the players.
180
193
Solution:
204
186
188
186
184
187
189
182
190
201
190
187
200
First, arrange the heights from the shortest to the tallest and
pick the height of the middle player.
180
200
182
201
184
204
186
186
187
1 88
189
190
190
193
Since there are 15 players in the team, the eighth observation is the
median.
Remarks:
► If n is odd, the median of the observation corresponds to the
n; i
th observation (middle-ranked).
► If n is even, the median is the average of the two middle-ranked
values.
n+ 1
Median (Rank Value)
2
=-
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Example 2:
The daily rates of the sample of eight employees at GMS Inc.
are PSS0, P42 0, P560, PS00, P700, P670, P860, P480. Find the
median daily rate of the employees.
Solution:
Step 1: Arrange the data in order.
P42 � P48� PS0� PSS� P56� P67� P70� P860
Step 2: Select the middle-rank value.
.
(
)
n+1
8+1
9
Median Rank Value = -2- = -2- = 2 = 4.5
Step 3: Identify the median in the data set.
P42 � P48� PS0� PSS� PS6� P67� P70� P860
t_ th
45
Since the middle point falls between P550 and P560, we can
determine the median of the data set by getting the average of the two
values. Then, we have:
5 5 0 + 560 1,1 1 0
Median = ---- = -2
2
= 555
Therefore, the median daily rate is P555.
C. MODE
The mode is the value that occurs the most frequently in a given
data set. Like the median and unlike the mean, extreme values in a data
set do not affect the mode. A data set that has only one value that
occurs the greatest frequency is said to be unimodal. If the data has
two values with the same greatest frequency, both values are
considered the mode, and the data set is said to be bimodal. If a data
set has more than two modes, then the data set is said to be
multimodal. There are some cases when a data set values have the
same number frequency. When this occurs, the data set is said to be no
mode.
Properties of Mode
1. The mode is found by locating the most frequently occurring value.
2. The mode is the easiest average to compute.
3. There can be more than one mode or even no mode in any given
set.
4. The mode is not affected by extreme small or large values.
5. The mode can be applied for nominal, ordinal, interval, and ratio
data.
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Example 1 :
The following data represent the total unit sales for
Smartphones from a sample of 1 0 Communication Centers for the
month of August 15, 17, 10, 12, 13, 10, 14, 10, 8, and 9. Find the mode.
Solution:
The ordered array for these data is:
8, 9, 1 0, 10, 10, 12, 13, 14, 15, 17.
Because 1 0 appears 3 times, more times than the other values,
therefore, the mode is 1 0.
Example 2:
Compute for the following data that represents the number of
LED television manufactured for the past three weeks:
20, 18, 19, 25, 20, 21, 20, 25, 30, 29, 28, 29, 25, 25, 27, 26, 22, and 20.
Find the mode of the given set.
Solution:
The ordered array for these data is:
18, 19, 2 0, 2 0, 2 0, 2 0, 21, 22, 2 5, 2 5, 2 5, 2 5, 26, 27, 28, 29, 29, 30.
There are two modes 20 and 25, since each of these values occurs four
times.
Example 3:
Find the mode of the ages of 9 middle-management employees
of a certain company. The ages are 53,45,59,48,54,46, 51 ,58, and 55.
Solution:
The ordered array for these data is:
45, 46, 48, 51, 53, 54, 55, 58, 59.
There is no mode since the data set has the same frequency.
Comparison of the Mean, Median, and Mode
When one is confronted with the question: "What is the best measure of
central tendency to use for a given data set?" No hard and fast rules can be
formulated.
When the distribution of the observations is fairly symmetric or when
there are no extreme observations, the mean is the most meaningful measure
of central tendency.
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In the case of a perfectly symmetric distribution, all three measures are
equal.
When extreme observations are found on the right end of the
distribution, these extreme values pull the value of the mean to the right;
therefore, the relationship between the three is illustrated as: x < Ma < M0
(positively skewed distribution).
If extremely low observations are present, these observations pull the
mean to the left, and the relationship between the three measures becomes:
x < Ma < M0 (negatively skewed distribution).
With the presence of extreme observations, the median is a more
meaningful measure in as much as it is not affected by these extreme values.
Measures of Relative Position
If in the measures of central tendencies, it is the relationship of the data
set around the center was described. In the measures of relative position, it is
the location of the value in the data set that is described. Measures of position,
sometimes referred to as the measure of location, are considered as the
extension of the median. It talks about the position/location of the value
relative to the other values in the data set. The common measures of position
are quartiles, percentiles, standard scores or z-scores, box-and-whisker plot.
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Calculation for Ungrouped Data:
A. Quartiles
This measure divides the observation into four equal parts. The
Q, is the middle point between the smallest value and the center value,
also called Q2. The Q2 is also called the median, Q3 is the middle value
between the median and the highest value of the data set.
Example 1:
Find the first, second, and third quartiles of the ages of 9
employees of a certain company. Their ages are: 53, 45, 59, 48, 54, 46,
51 , 58, and 55.
Solution:
Step 1: Arrange the data in ascending order.
45, 46, 48, 51, 5 3, 54, 55, 58, 59
t
t
t
7 _ 5th
5
2.S
Step 2: Select the first, second, and third quartiles value using the
formula:
where: Q k = quartile
k (n + 1)
n = no. of observations
Qk =
k = quartile location
4
th
Q1
=
➔ Qz
=
Q1
=
➔
➔
th
1(9 + 1)
4
2(9 + 1)
4
3 (9 + 1)
4
10
= 4 = 2.5
2(10)
-=5
=4
3 (10)
=- = 7.5
4
Step 3: Identify the first, second, and third quartile values in the data set.
Since the 2 . Sth falls between 46 and 48; and 7.Sth falls between
55 and 58, we can determine the first and third quartiles of the data set
by getting the average of the two values.
Therefore, Q 1
Q1
=
Q3
=
46 + 48
2
5 5 + 58
2
94
= 2 = 47
113
= 2 = 56.5
= 47, Q2 = 53, and Q 3 = 56.5.
Interpretations:
1.
2.
Vision:
Mission:
Q1
Q2
= 47 implies that one-fourth or 25% of the ages fall below 47.
= 5 3 implies that one-half or 50% of the ages fall below 53.
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3. Q 3 = 56.5 implies that three-fourths or 75% of the ages fall below
56. 5.
B. Percentiles
This divides the observation into 1 00 equal parts. It is used to
indicate how much of the observation may be found below. For
instance, if the 30 th percentile is the value, this means that 30% of the
observation may fall below it.
To find the k-th percentile:
Step 1 : Arrange the observations from lowest to highest.
Step 2: Compute the position using the formula:
L = (_!__) * n
100
Remarks:
1 . If L is whole number, k-th is midway between L and the next value.
2. If L is not a whole number, round it up to the next integer and the
value at that position is the k-th percentile.
Example 2:
Suppose the arrayed scores of 20 students in a Math 1 1 n exam
are as follows: 80, 90, 9 1 , 1 00, 1 20, 1 22, 1 23, 1 25, 201 , 9 0, 88, 98, 1 30,
1 24, 1 1 1 , 1 09, 1 40, 1 02, 85, 9 1 . Solve for the for the 40th percentile,
28th percentile.
Solution:
Step 1: Arrange the scores in ascending order.
,81[), 8 5, 8 8, 9 0 1 90, 9 1 , 9 1 , 98, 1 00 11 02, 1 0 9, 1 1 , , 1 20., 1 22, 1 23, 1 24-, 1 2 5
1 30, 1 40 20 1
,6th
8. 5t
t
f
Step 2: Compute the position.
➔
➔
* 20 = 8
L = (�)
100
(Remark 7)
* 20 = 5.6
L = (�)
100
(Remark 2)
Therefore, the 40 th percentile is the value between the 8th and 9 th
observation, and it is 99.
Interpretations:
1. Four-tenths or 40% of the scores fall below 99; or
2. Six-tenths or 60% of the scores are above 99.
Therefore, the 28 th percentile is the 6th item in the observation, which is
91 .
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Math1 1 n: Mathematics in the Modern World
Interpretations:
1. 28% of the scores fall below 99; or
2. 72% of the scores are above 99.
C. Z-Scores
A z-score measures the distance between an observation and
the mean, measured in units of standard deviation. Z-score is used to
know the position of one observation relative to others in a set of data.
A z-score indicates how many standard deviations an element is from
the mean. For instance, when the z-score is - 1, this represents that it is
one standard deviation less than the mean, z-score of 3 represents that
its 3 standard deviation above the mean.
The following formula is used to compute the z-score for a data
value x in a sample.
x-x
z = -­
s
Example 3:
The monthly expenditures of a large group of households are
normally distributed with a mean of P48,700 and a standard deviation
of P10,400. What is the z-value of monthly expenditures of P59,400 and
P38,300?
Solution:
Let x = 48,700 and s = 1 0,400. Then, using the formula of z,
determine z-values for the two x values. We have:
For x = 59,400 :
(x -­
- x)
z=s
59,400 - 48,700
= 1.00
1 0 ' 400
For x = 38,300:
(x - )
z = --x­
s
38,300 - 48,700
= - 1.00
1 0 , 400
The z of 1.00 indicates that a monthly expenditure of P59,400
for households is one standard deviation above the mean, and a z of
- 1.00 shows that a P38,300 monthly expenditure is one standard
deviation below the mean. Note that both monthly household
expenditures (P59,400 and P38,300) are the same distance (P10,400)
from the mean.
Example 4:
Anne's report card shows that his grade in Math is 98 and in
Science is 90. The mean grade in Math is 90, and a standard deviation
is 10. In Science, the mean grade is 80, and a standard deviation is 5. In
which subject does Anne perform better?
Solution:
For Math: :
For Science:
Vision:
Mission:
(x - x)
z=­
s
98 - 90
= 0.8
-1-0-
(x - x)
z = --­
s
90 - 80 2
--=
5
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The values show that Math is 0.8 higher than the mean, while Science
is 2 standard deviation higher than the mean. Thus, Anne performs
better in Science.
Example 5:
A time study report indicates that an assembly line task should
be finished at an average of 5.64 minutes, with a standard deviation of
0.97 minutes. One particular item had a z - score of 1.53. What was the
completion time of this item?
Solution:
Given: x = 5.64, s = 0.97 and z = 1.53. Substituting the given
values to determine the x value, we get:
(x - x)
Z = ---
➔
X = X
X
+ ZS
(by cross multiplication)
= x + zs
= 5.64 + (1.53) (0.97)
= 5.64 + 1.4841
= 7.1241 minutes
The item had an assembly time of 7.12 minutes.
D. Box- &-Whisker Plot
Box-&-Whisker shows the median, the quartiles, and the
extremes for a numerical set of data. The box portion contains about
50% of the data values. The two whiskers each contain about 25% of
the data values. It shows how spread out the data values are. They are
useful for comparing sets of data.
Example 6:
Suppose you have the following prices of capsule umbrella: 225,
350, 175, 450, 429, 205, 431, 250, 248.
Solution:
Step 1: Place the data in order from least to greatest:
175, 205, 225, 248, 250, 350, 429, 431, 450
Step 2: Find the lower extreme: 175
Find the upper extreme: 450
These are the endpoints of our whiskers.
Step 3: Find the median of the data: 250
Step 4: Draw a line through your median. Then, split the two halves in
half again. If there is an even number of data on each side, you need to
average the two middle numbers to find the end of the lower quartile
and the beginning of the upper quartile. See the example below:
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I I
1 75
205
225 1 248
2f0
Median
350
429
43 1 1
450
This is the halfway point - A verage
these two numbers to get the end
of the first quartile
This is the halfway point - A verage
these two numbers to get the
beginning of the last quartile
20 5 + 2 3 5
---- = 2 1 5
2
429 + 4 3 1
43
-2-- = 0
Step 5: Graph the median, the extremes, and the quartiles below a
number line. Then, draw the box and whiskers.
1 50
1 00
200
250
�.._____.---------I
300
350
400
•
450
Common Measures of Variation
Measures of central tendency describe one important aspect of a set
of data -- their middle or their average, but they tell us nothing about this other
basic characteristic. We can have data sets having the same mean, and yet
they are not identical data sets simply because of the different values the data
sets contain. Hence, we require ways of measuring the extent to which data
are dispersed or spread out. The terms variability, spread, and dispersion are
synonyms. Measures of variation or dispersion tell us how to scatter or spread
out a distribution is. One measure of dispersion is the range.
A. Range
Probably, the simplest and easiest way to determine the
measure of dispersion is the range. The range (R) is the difference
between the largest and the smallest in the given set of data.
Example 1:
Date Set
A
B
C
Values
5, 5, 5, 5, 5
3, 4, 5, 6, 7
4, 4, 5, 6, 6
Mean
5
5
5
Given the data sets A, B, and C, we find the following:
RA
RB
Re
=
=
=
5 - 5
7 - 3
6 - 4
= 0
= 4
= 2.
We note that a range of zero simply means that all the values in
the data set are the same. There is no variability in the values, or the
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variable under consideration is a constant for this data set. Also, the
larger is the difference between the two extreme values; the larger is the
range. Comparing the three data sets with respect to variability based on
the range, we can say that while data set A is perfectly homogeneous,
data set B is the most heterogeneous. Data set C ranks second to data
set B in terms of variability.
Although the range is easily computed, it is not considered to be
the best measure of dispersion because it involves only the two extreme
values. It does not tell anything about the remaining values in a set of
data.
B. Interquartile Range
Interquartile range is also called the midspread. It is the
difference between the 75th and 25th percentile or between the upper and
lower quartile. It is denoted by IQR and computed as IQR = Q 3 - Q 1 _
Example 2:
Given: {4, 5, 8, 9, 10, 11, 15}
Solution:
➔
n
=7
Q1 :
1(7
+ 1)th
➔
2nd observation
Q3 :
3(7
+ 1)th
➔
6th observation
4
4
: 6th
Q3
= 11
Therefore, !QR = 11- 5 = 6.
C. Absolute Deviation
The absolute deviation (AD) is the average of the absolute
deviation from the central point or the average of the average distance
between each data value and the mean. This is best used when the
median is the appropriate measure of central tendency (in the presence
of extreme values/skewed distributions).
Example 3:
Consider the number of blender units sold by a store for one
week: 5, 4, 2, 10, 8, 9, 6, 12.
Solution:
To calculate the absolute deviation of ungrouped data:
Step 1: Arrange the values from highest to lowest.
2, 4, 5, 6, 8, 9, 10, 12
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Step 2: Compute the mean and the absolute deviation of each value
from the mean.
l x - meanl
Unit Sold (x)
5
3
2
1
1
2
3
5
2
4
5
6
8
9
10
12
L X 56 7
:i = =-=
n
8
➔
➔
➔
12 - 7 1 = I -S I = s
14 - 7 1 = l - 3 1 = 3
I S - 7 1 = 1 -2 1 = 2
L l x - mean l = 22
Step 3: Compute the absolute deviation using the formula:
I l x - mean l
AD = ---­
n
22
AD = s
= 2.75
D. Variance and Standard Deviation
One of the most widely used measures of dispersion is the
standard deviation. The more spread apart the data, the higher the
deviation. Standard deviation is the square root of variance. The
variance of a set of numbers is the mean of the squared deviations of
these numbers from their mean.
To facilitate calculations, we have the following formula to find
the variance for ungrouped data:
I(x - x) z
sz =
n-1
Example 4:
Consider the following daily rates of a sample of eight
employees at GMS Inc.: P550, P420, P560, P500, P700, P670, P860,
P480. Find the variance and standard deviation.
Solution:
Step 1 : Compute the mean of the data set.
L x 5 50 + 420 + 5 60 + 500 + 700 + 670 + 860 + 480
x_ = =
n
Vision:
Mission:
4,740
= -- = 5 92 5
8
8
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Step 2: Calculate the mean and the deviations of each item from the
mean.
Daily Rate (x)
550
42 0
5 60
500
x-x
(x - x.) 2
670
86 0
48 0
- 42.5
- 1 72.5
- 32.5
-92.5
1 0 7 .5
7 7 .5
2 6 7 .5
- 112.5
1,8 0 6.2 5
29, 7 5 6.2 5
1, 0 5 6.2 5
8,5 5 6.2 5
11,5 5 6.2 5
6, 0 0 6.2 5
7 1,5 5 6.2 5
12,65 6.2 5
I x = 4, 7 4 0
I cx- x) = o
I ex - x.) 2 = 1 42,9 5 0
700
➔
➔
➔
(- 42.5) 2
2
(- 1 72.5)
(- 32.5) 2
Step 3: Solve for the variance and standard deviation.
}: (x - x.) 2 _ 1 42,95 0 _
s - 2 0 , 42 1. 4 3
8_1
n_ 1
2
_
Therefore, the variance is P20, 42 1. 4 3 and the standard deviation is
P142.90.
Learning Tasks/Activities
A. Measures of Central Tendency
1. In one hour of fishing, nine fishermen caught the following number
of tilapias: 7 , 4, 8, 6, 5 , 8, 1 0 , 7, 8
Find the three measures of central tendency.
2. If the smallest and largest numbers are removed from the set, which
of the three measures of central tendency would decrease? Remain
the same? Increase?
3. How many fish should a tenth fisherman catch in order to increase
the mean by 1?
B. Measures of Relative Position
Suppose the arrayed grades of 40 high school students are as follows:
89
90
80
98
75
87
88
93
79
92
87
79
77
75
85
90
92
86
79
87
Vision:
Mission:
83
85
91
81
89
82
84
91
82
93
90
87
95
77
80
84
70
76
90
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1 . Find the 1 st and 3 rd quartile and interpret the results.
2. Find the 50 th and 75th percentile and interpret the results.
3. Find the 3 8 th percentile and interpret the result.
C. Common Measures of Variation
The following scores were obtained from a 1 5 - item test in English.
Boys: 9, 10, 13, 6, 10, 7, 9, 10, 8
Girls: 9, 7, 5, 12, 3, 8, 10, 2, 8, 10, 14, 7
Find the range, absolute deviation, variance, and standard deviation.
1 . For the boys
2. For the girls
3. For the whole class
Assessment
Answer the following problems:
1 . A pizza parlor sells colas in three sizes: small, medium, and large. The
small size costs P25, the medium P3 5, and the large P50. Yesterday,
1 20 small, 250 medium, and 1 00 large colas were sold. What was the
weighted mean price per cola?
2. A class of 300 students had test scores that are adequately described
by a normal distribution with a mean of 76 and a standard deviation of
8. If a certain student had a z-score of -0.625, determine his/her test
scores.
3. The report card of Alvin shows that his grade in Math is 89 and Science
is 93. The mean grade in Math is 8 5, and the standard deviation is 5. In
Science, the mean grade is 80, and the standard deviation is 8. In which
subject does Alvin perform better?
4. Construct a boxplot for the data set 53, 45, 59, 48, 54, 46, 51 , 58, and
55. Describe the distribution of the data set.
Instructions on how to submit student output
Refer to the course policies and course content plan.
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31
Lesson 3.3: Probability, Normal
Distribution, Linear
Regression, and Correlation
Lesson Summary
In this lesson, we will study probability - its meaning and how it is computed.
This lesson also tackles the properties and characteristics of the unit normal
curve and its relation to the standard scores. Linear regression and correlation
are also covered. It discusses how to describe what type of relationship or
correlation exists between two quantitative variables.
Learning Outcomes
At the end of the lesson, the students will be able to:
1. Compare and contrast conditional probability from unconditional
probability.
2. Solve problems involving conditional and unconditional probability.
3. Discuss the characteristics of normal distribution.
4. Enumerate the steps in solving the normal distribution problem.
5. Solve problems in normal distribution.
6. Explain the concept of linear regression.
7. Discuss the concept of simple linear regression analysis.
8. Enumerate the assumptions on linear regression analysis.
9 . Solve problems involving linear regression.
10. Discuss the concept of correlation analysis.
11. Compare and contrast linear correlation analysis
12. Solve problems involving correlation analysis.
Motivation Question
What is the highest possible value of a probability?
Discussion
Probability and Normal Distribution
Probability is the chance or likelihood of an event to happen. It can be
expressed as proportions from 0 to 1 or percentages from 0% to 100%. A
probability of 0 indicates that the event doesn't have the chance to occur,
whereas a probability of 1 indicates that the event will certainly happen.
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Math1 1 n: Mathematics in the Modern World
32
Example 1: Data below show the number of gadget addictiveness among
children 5-10 years of age who are seeking medical care.
[
Boys
Girls
Totals
Age (years)
5
6
1
431
380
512
409
840
892
1
7
500
413
913
I 411
s
J 421
g
435
846
460
881
I
10
417
501
918
I Total
2560
2730
5290
Unconditional probability
P(characteristics) = number of persons with characteristics/N is called
unconditional since the denominator is the N giving each child an equal chance
to be selected.
Example 1: Using the data given above:
2 5 60
5290
? (selecting a boy) =
? (selecting a 7 years old)
= 0.484
=
Can you figure out the following?
9 13
5290
= 0.173
1. What is the probability of selecting a girl? 0. 5 1 6
2. What is the probability of selecting a 7- year old? 0. 173
3. What is the probability of selecting a boy who is 10 years of age? 0. 079
Conditional probability
A conditional probability is the probability of an event occurring, given
that another event has already occurred. The conditional probability of event
B occurring, given that event A has occurred, is denoted by and is read as
"probability of B, given A."
Example 2: What is the probability of selecting a 9-year old child given that she
is a girl?
P (9 - year old lgirl)
=
460
2730
= 0.168
This means that 16.8% of the girls are 9 years of age.
Example 3: What is the probability of selecting a boy given that he is 6 years
old?
I
P (boy 6 years old)
=
380
892
= 0.42 6
This means that 42.6% of the boys are 6 years of age.
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33
Normal Distribution
Characteristics of a Normal Distribution
Normal distributions are:
1. Symmetric: a normal distribution is perfectly symmetrical around its
center. That is, the right side of the center is a mirror image of the left
side.
2. Unimodal: There is only one mode in a normal distribution.
3. Asymptotic: The extremes come closer and closer to the horizontal line,
but it never touches.
4. Equal values of the mean, median, and mode
Steps in solving normal distribution problem:
1. Draw a picture of the normal distribution.
2. Translate the problem into one of the following: P (X < a), P (X >
b), or P (a < X < b). Shade in the area on the picture.
3. Standardize a (and/or b) to a z -score using the z -formula.
4. Look up the z -score on the z -table and find its corresponding
probability.
a. Find the row of the table corresponding to the leading digit (ones
digit) and first digit after the decimal point (the tenths digit).
b. Find the column corresponding to the second digit after the decimal
point (the hundredths digit).
c. Intersect the row and column from steps (a) and (b).
• If we need a "less-than" probability - that is, P (X < a) - then,
we're done.
• If we want "a greater-than" probability- that is, p(X > b) take one minus the result from step 4.
• If we need a "between-two-values" probability - that is,
P (X < a < b) - do steps 1 - 4 for b (the larger of the two
values) and again for a (the smaller of the two values), and
subtract the results.
Area Under the Unit Normal Curve
The area under the unit normal curve may represent several things like
the probability of an event, the percentile rank of a score, or the percentage
distribution of a whole population. For instance, the area under the curve from
z = z1 to z = zi , which is the shaded region in the figure below may represent
the probability that z assumes a value between z1 and z1 .
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Math1 1 n: Mathematics in the Modern World
The table below gives the area under the standard normal curve from
O to z. Since the curve is perfectly symmetrical, the value of z may be taken as
positive ( +) or negative (-), but the corresponding area is always taken as
positive.
Cum ulative Standard Normal Distribution
z
.DO
.01
.D2
.03
.D4
.05
.06
.07
,08
.00 3
.0003
.0003
. 0003
. 0003
.0003
.0004
.0004
. 0004
.0004
.0004
.0003
.0006
.0006
.0006
. 0006
.0005
.0005
.0005
.0007
- 3.4
.0 □ 03
.00 □ 3
.0003
- 3.3
. 0005
.0005
.0005
□
.00 □ 4
.0007
.0006
,0002
- 3. 2
.0 □ 07
- 3. 1
.0 1 0
□
.00 9
.0009
.0009
.0008
.0008
. 0008
.0008
.0007
- 3 .0
.00 1 3
.001 3
.001 3
.001 2
.001 2
.001 1
. 001 1
.001 1
.00 1 0
.001 0
- 2.9
.001 9
.0018
.001 8
.001 7
.001 6
.001 6
.001 5
.001 5
.00 1 4
.001 4
- 2.8
.0 2-6
.0025
.0024
.0023
.0023
.0022
. 0021
. 0021
.0020
.001 9
- 2.7
. 0035
.0034
.0033
.0032
.0031
.0030
. 0029
.0028
.0027
.0026
- 2.6
.0047
.0045
.0044
.0043
.0041
.0040
. 0039
. 0038
.0037
.0036
- 2. 5
.0062
.0060
.0059
.0057
.0055
.0054
. 0052
.0051
.0049
.0048
- 2.4
.0082
.0080
.0078
.0075
.0073
.0071
. 0069
.0068
.0066
.0064
- 2.3
.0107
.01 04
.01 02
.0099
.0096
.0094
.0091
. 0089
.0087
.0084
- 2. 2
.01 39
.01 36
.01 32
.01 29
.01 25
.01 22
.01 1 9
.01 1 6
.01 1 3
.01 1 0
-2.1
.01 79
.01 74
.01 70
.01 66
.0162
.01 58
.01 54
.01 50
.01 46
.0143
- 2.0
.0228
.0222
.02 1 7
.02 1 2
.0207
.0202
. 0 1 97
. 0 1 92
.01 88
,01 83
- 1 .9
.0287
.0281
.0274
.0268
.0262
.0256
.02-50
.0"1:44
.0239
.0233
- 1 .8
.0359
.0351
.0344
.0336
.0329
.0322
.031 4
.0307
.0301
,0294
- 1 .7
. 0446
.0436
.0427
.041 8
.0409
.0401
. 0392
.Ocffl4
.0375
.0367
- 1 .6
.0548
.0537
,0526
,051 6
.0505
.0495
. 0485
.0475
.0465
,0455
-1.5
.0668
.0655
.0643
.0630
.061 8
.0606
. 0594
.0582
.0571
.0559
□
□
- 1 .4
.OB08
.0793
.0778
.0764
.0749
.0735
. 0721
, 0 70B
.0694
.0681
-1.3
. 0968
.0934
.0918
.0901
.0885
.0869
.0853
,1 151
,1 1 12
. 1 093
, 1 076
. 1 056
. 1 030
. 1 0 20
.0838
. 1 00 3
.OB23
- 1 ,2
-1,1
.095"1
.1 131
. 0985
. 1 357
. 1 335
. 1 3 14
. 1 292
. 1 27 1
. 1 zs 1
. 1 Z30
1 21 0
. 1 1 90
. 1 1 70
- 1 .0
. 1 587
, 1 562
. 1 539
, 1 51 6
, 1 492
, 1 469
1 446
. 1 423
. 1 401
, 1 37 9
QJj
. 1 B4 i
. 1 81 4
. 1 788
_ ] 762
. 1 736
. 1 71 1
. 1 685
. 11:;so
. l il:?.S
. 1 61 1
- 0 .8
.21 H l
.2090
.2061
.2033
.2005
. 1 07 7
. 1 940
. i 894
.1 8il7
.2 1 77
.2 1 4 8
0.7
.2420
.2389
.2358
.2327
.2298
.2266
. 2236
. Hl22
.2206
- 0 .6
.2743
.2709
.26 76
. 2643
. 261 1
.2578
. 2546
.251 4
.24 83
. 245 1
- 0 .5
. 3085
.3050
.301 5
. 208 1
. 2946
. 29 1 2
2877
.2843
.281 0
.277 6
- 0.4
.344 6
.3409
.3372
.3336
.3300
.3264
. 3228
.3 1 92
.31 66
- 0.3
. :iis2 1
,3 783
.374S
.37 07
. 31i32
, 3:;94
.3�:i7
. 3520
- 0.2
.4 1 68
.4 1 29
.4�0
. 40$2
.40 1 3
. 3974
.39;!6
.3897
. 3859
-0. 1
.4207
.4ij lJ2
• 11 \'I
.4511 2
.452;?
.4483
. 4443
.4 4 04
4364
.4 :t25
.4286
.424 7
0.0
. SOOO
.4960
.4920
.4880
. 4840
.480 1
,.176 1
.4-72 1
.'1681
. 464 1
F� / vilf
.3 1 2 1
.34llS
I 111M - � 4�, u�" 0 000 1 ,
0
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For instructional purposes only • 1 st Semester SY 2020-2021
f Cumulative Standard Normal Distribution
z
.Oil
.01
.02
.03
.1)4
.05
.06
.07
.08
.09
0.0
.5 000
.5040
.5080
.51 20
.51 60
.51 99
. 5239
.5279
. 53 1 9
. 5359
0 .1
.5398
.5438
.5478
. 551 7
.5557
. 5596
. 5636
.5675
.571 4
. 5753
0.2
.5793
.5 832
.5871
. 591 0
. 5948
. 5987
.6026
.6064
.61 03
.61 4 1
0.3
.6179
.621 7
.6255
.6293
.633 1
.6368
.6406
.6443
.6480
.65 1 7
0.4
.6554
.6591
.6628
.6664
.6700
.6736
.6772
.6808
.6844
.6879
0.5
.691 5
.6950
.6985
.701 9
.7054
. 7088
.71 23
.71 57
.71 90
.7224
0.6
.7257
.7291
.7 324
.7357
.7389
.7422
.7454
.7486
. 75 1 7
. 7549
0.7
.7580
.761 1
.7642
.7673
.7704
.7734
. 7764
.7794
. 7823
.7852
0.8
.7881
.791 0
.7939
. 7967
.7995
.8023
. 805 1
.8078
.81 06
.81 33
0.9
.8159
.8 1 86
.821 2
. 8238
. 8264
.8289
. 831 5
.8340
. 8365
.8389
1 .0
.841 3
.8438
.8 461
.8485
. 8508
. 853 1
. 8554
.8577
. 8599
. 8621
1 .1
.8643
.8665
.8686
. 8708
.8729
. 8749
.8770
.8790
.88 1 0
.883 0
1 .2
.8849
.8869
.8888
. 8907
. 8925
. 8944
.8962
. 8980
. 8997
.90 1 5
1 .3
.9032
.9049
.9066
.9082
.9099
.91 1 5
.91 3 1
.91 47
.91 62
.91 77
1 .4
.9192
.9207
.9236
.9251
.9265
.9279
. 9382
.9394
.9429
. 93 1 9
.9370
.94 1 8
.9306
.9357
.9406
.9292
.9332
.9345
.9222
1 .5
1 .6
.9452
.9463
.9474
.9484
.9495
.9505
.951 5
.9525
.9535
.954 5
1 .7
.9554
.9564
.957 3
.9582
.9591
.9599
.9608
.96 1 6
.9625
.9633
1 .8
.9641
.9649
.9656
.9664
.967 1
.9678
.9686
.9693
. 9699
. 9706
1 .9
.971 3
.971 9
.9726
.9732
.9738
.9744
.9750
.9756
.9761
.9767
2.0
.9772
.9778
.9783
.9788
.9793
.9798
.9803
.9808
.98 1 2
. 98 1 7
.9821
.9826
.9830
.9834
.9838
.9842
.9846
.9850
. 9854
.9857
2.1
.9441
2.2
.9861
.9864
.9868
.9871
.9875
.9878
. 9881
.9884
. 9887
. 9890
2.3
.9893
.9896
.9898
.9901
.9904
.9906
.9909
.99 1 1
. 99 1 3
. 99 1 6
2.4
.991 8
.9920
.9922
.9925
. 9927
.9929
.993 1
.9932
. 9934
.993 6
2.5
.9938
.9940
.9941
.9943
.9945
.9946
.9948
.9949
.99 5 1
.9952
2 .6
.9953
.9955
.9956
.9957
. 9959
.9960
. 9961
.9962
.9963
. 9964
.9965
.9966
.9967
.9968
.9969
.9970
.997 1
.9972
.9973
.9974
2.6
.9974
.9975
.9976
.9977
.9977
.9978
.9979
.9979
.9960
.998 1
2.9
.9981
.9982
.9982
.9983
.9984
.9984
.9985
.9985
.9986
.9986
3.0
.9987
.9987
.9987
.9988
.9988
. 9989
.9989
. 9989
.9990
. 9990
3.1
.9990
.9991
.9991
.999 1
.9992
.9992
.9992
.9992
.9993
. 9993
2.7
3.2
.9993
.9993
.9994
.9994
.9994
. 9994
.9994
.9995
.9995
.9995
3.3
.9995
.9995
.9995
.9996
.9996
.9996
.9996
.9996
.9996
. 9997
3.4
.9997
.9997
.9997
.9997
. 9997
.9997
.9997
.9997
. 9997
.9998
for z v-alues greater lhan 3.49, LJSe 0.9999.
Area
0
Example 1 : Find the area between z
z
= O and z = + 1 .
Solution:
From the table, we locate z
is equal to 0.3413.
Vision:
Mission:
= 1 and get the corresponding area, which
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Math1 1 n: Mathematics in the Modern World
Example 2: Find the area between z = -1 and z =
Solution:
o.
As we can see, there is no negative value of z. Thus, we read the
positive value. Hence, the area is also 0.3413.
Example 3: Find the area below z = -1.
Solution:
Since the whole area under the curve is 1, then the whole area is divided
into two equal parts at z = 0. This means that the area to the left of z = 0 is
0.5. To get the area below z = -1 means getting the area to be the left of z
-1. The area below z = -1 is then equal to 0.5000- 0.3413 = 0.1587.
Example 4: Find the area between z = -0.70 and z = 1.25.
Solution:
The area between z = -0.70 and z = 0 is 0.2580, while that between
z = o and z = 1.25 is 0.3944. Therefore, the area between z = -0.70 and z =
1.25 is 0.2580 + 0.3944 = 0.6524. We add the two areas since the z values
are on both sides of the distribution.
Example 5: Find the area between z
Solution:
= 0 and z = 0.68 and z = 1.56.
The area between z = 0 and z = 0.68 is 0.251 8, while the area between
z = 0 and z = 1.56 is 0.4406. Since the two z values are on the same side of
the distribution, we get the difference between the two areas. Hence, the area
betweenz = 0.68 and z = 1.56 is 0.4406 - 0.2518 = 0.1888.
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For instructional purposes only • 1 st Semester SY 2020-2021
These techniques of finding areas under the curve are very useful in making
approximations, as illustrated in the next example.
Example 6:
You join in a crab catching contest. The sizes of the crabs are normally
distributed with a mean of 16 cm and a standard deviation of 4. What is the
chance of catching crabs that are less than 8 cm? What is the chance of
winning a prize if the prize is offered for any crabs over 24 cm? What is the
chance of catching crabs between 16 cm and 24 cm?
Solution:
1. Draw a picture of the normal distribution.
4
8
2 4,
ll
J.B
2. Locate the problems in the graph.
3. Translate each problem into probability notation.
•
•
•
Problem 1:
Problem 2:
Problem 3:
P (x < 8)
P (x > 24)
P (16 < x < 24)
4. Change the x values into z scores.
For x
For x
= 8:
•
•
•
Vision:
Mission:
= --- = -2
z - score
= --- = 2
= 24:
Therefore,
Problem 1:
Problem 2:
Problem 3:
8 - 16
4
z - score
24 - 1 6
4
P (x < 8) = P (x < -2)
P (x > 24) = P (x > 2)
P (16 < x < 24) = P (O
< x < 2)
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Math1 1 n: Mathematics in the Modern World
5. Identify the z values using the z table.
= -2, the table value is 0.4772.
When z = 2, the table value is 0.4772.
When z
6. Calculate probabilities
Problem 1 : P (x
> -2)
It is 0. 5, which is the area of the region lower than the mean less
0.4772 gives 0.0228. Thus, the probability of getting crabs smaller than
8 cm is 2.28%.
Problem 2: P (x < 2)
It is 0.5 less 0.4772 gives 0.0228, which implies that the
probability of catching crabs bigger than 24 cm is 2.28%
Problem 3: P (O < x < 2)
The area from o to 2 is 0.4772. It follows that the probability of
catching crabs with sizes between 16 cm and 24 cm is 47.72%, which
is also the probability of winning a prize.
Linear Regression and Correlation
Correlation is a statistical method used to determine whether a
relationship between variables exists. A variable here is a characteristic of the
population being observed or measured. For instance, the variable of interest
might be advertising expense and sales. The sample then consists of random
observations of the variable describing a given population.
Regression analysis is a statistical method used to describe the nature
of the relationship between variables, that us, either positive or negative, linear
or nonlinear. There are two types of relationships: simple and multiple. In a
simple relationship, there are two variables an independent variable (or
explanatory variable or predictor variable) and a dependent variable (or
response variable). The simple linear relationship can be positive or negative.
A positive relationship exists when either variable increases at the same time
or both decrease at the same time. On the contrary, in a negative relationship,
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as one variable increases, the other variable decreases or vice versa. The text
is limited with the discussion of a simple linear regression analysis.
A Scatter diagram is a useful tool for checking the assumptions in a
regression analysis. It can be viewed during an initial screening run of the
analysis or after the analysis. The benefit of looking at the scatter diagram
residuals in the beginning stages of analysis is that it may save a researcher's
time.
A.
Pearson - Product Moment Correlation
Pearson product-moment correlation is the most widely used in
statistics to measure the degree of the relationship between the linear
related variables. The Pearson r correlation would require both variables
to be normally distributed. Correlation refers to the departure of two
random variables from independence. For example, in the stock market, if
we want to measure how two products are related to each other, Pearson
r correlation is used to measure the degree of relationship between the
two products.
The correlation coefficient is defined as the covariance divided by
the standard deviations of the variables. The following formula is used to
calculate the Pearson r correlation:
r
(x - x) (y - y)
= ---;::::=I=======
-J[I (x - x) 2 ] [}: (y - y) 2
or
n }: xy - (L x) (}: y)
r = --;:::=============
2
-J[n(}: x
2
2
2
) - (I x) ] [n(}: y ) - (L y) ]
Pearson's product-moment correlation coefficient or simply
correlation coefficient (or Pearson's r) is a measure of the linear strength
of the association between two variables. It is founded by Karl Pearson.
The Value of the correlation coefficient varies between +1 and - 1. When
the value of the correlation coefficient lies around ±1, then it is said to be
a perfect degree of association between the two variables. As the value of
the correlation coefficient goes closer to zero, the relationship between
the two variables will be weaker. This information is summarized in the
charts below.
..
.
.
...
..
;;
•
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(r ... ....!llllll.PI
.. .
■-
-
•
;
� Ciar�llon
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The following summarizes the correlation coefficient and strength of
relationships:
0.0
±0.01 to ± 0.20
No correlation, no relationship
➔ Very low correlation, almost negligible
±0.21 to ± 0.40
➔
➔
Slight correlation, definite but small relationship
±0.41 to ± 0.70
➔
Moderate correlation, substantial relationship
±0.71 to ± 0.90
➔
High correlation, marked relationship
➔
Very
±0.91 to ± 0.99
relationship
±1.00
➔
high
correlation,
very
dependable
Perfect correlation, perfect relationship
Example 1:
The owner of a chain of fruit shake stores would like to study the
correlation between atmospheric temperature and sales during the summer
season. A random sample of 12 days is selected with the results given as
follows:
2
1
Day
Tem peratu re 79
76
Tota l
Sales
147 143
( U n its)
3
78
147
4
84
5
90
168 206
6
83
155
7
93
192
8
94
211
Step 1: Plot the data on a scatter diagram.
1J
l SO
ca
1 70
1 j!,U
1 50
uo
.�
• ••
� .... ,
Vision:
Mission:
♦
•
1 90
209
11
88
10
85
187 200
12
82
150
•
220
:rno •
9
97
•
•
�u
s
!.!Iii
a
T e11 pe ra h1:r e ,
95
t Ofl
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Step 2: Compute the coefficient of correlation r.
Day
1
2
3
4
5
6
7
8
9
10
11
12
Tota l
X
79
76
78
84
90
83
93
94
97
85
88
82
1,029
Therefore:
LX =
1,029
Ix
L Y = 2, 115
✓
x2
6,241
5,776
6,084
7,056
8, 100
6,889
8,649
8,836
9,409
7,225
7,744
6,724
88,733
y
147
143
147
168
206
155
192
211
209
187
200
150
2, 115
2
L Y2
y2
2 1,609
20,449
2 1,609
28,224
42,436
24,025
36,864
44,521
43,681
34,969
40,000
22,500
380,887
= 88,733
xy
1 1,613
10,868
1 1,466
14,112
18,540
12,865
17,856
19,834
20,273
15,895
17,600
12,300
183,222
I xy = 183,222
= 380, 887
n i xy - (I x) (I y)
r = --;:::[n(
=============
( )
(
] [n(
I x 2 ) - I x) 2
I y2)
Iy
2
✓ [12 (88,733) - (1,029) ] [ 12 (380,887) - (2, 115)
-
]
12 (183,222) - (1,029) (2, 115)
✓ [5, 95 5] [97,419]
2
2
]
22,329
r
= 0.9270572 5 54 � 0.9 3
Step 3: Interpret the result.
The coefficient of correlation, r = 0.93, between the atmospheric
temperature and total sales indicates a very high positive correlation (very
dependable relationship). That is, an increase in atmospheric temperature is
highly associated with the increase in total sales of fruit shake.
Simple Linear Regression Analysis
Regression analysis is a simple statistical tool used to model the
dependence of a variable on one (or more) explanatory variables. This
functional relationship may then be formally stated as an equation, with
associated statistical values that describe how well this equation fits the data.
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Simple linear regression is the least estimator of a linear regression
model with a single predictor (or one independent variable). The least-square
model determines a regression equation by minimizing the sum of squares of
the vertical distances between the actual y values and the prediction values of
y. Meaning, simple linear regression fits a straight line through the set of
n points in such a way that makes the sum of squared residuals of the model
as small as possible. This method gives what is generally known as the "best­
fitting" line. The difference between an observed and predicted value is called
the residual. The mean of the residuals is always zero. The points that fall
outside the overall pattern of the other points are known as outliers.
In the scatterplot, there are scores whose removal greatly changes the
regression line, which is called influential scores. In some cases, these scores
are restricted to points with extreme x-values. Some influential scores may
have a small residual but still have a greater effect on the regression line than
scores with possibly larger residuals but average x- values.
n (r xy) - Q: x) (L Y)
b 1 nQ: x 2 ) - Q: y 2 )
where: y = predicted or fitted value of y.
x
= the value of any particular observation of the independent variable.
y = the value of any particular observation of the independent variable.
b 1 = slope of the regression line.
= intercept of the regression line.
x = mean of the independent variable.
y = mean of the dependent variable.
b0
Assumptions on Linear Regression Analysis
1. The values of the independent variable X may be "fixed" that is, the
researcher may select the values of X in advance, or X could be a
random variable.
2. The values of X are measured without error.
3. The variances of the subpopulations of the dependent variable, given
different values of the independent variable, are equal. This is a
condition known as homoscedasticity.
4. The subpopulation of dependent variable Y, given different values of the
independent variable X, is normally distributed.
5. The means of the subpopulation of Y all lie on the same straight line.
This is called the assumption of linearity.
Example 2:
Using the given in Example 1, determine the regression equation, plot
the regression line, and interpret it.
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Solution:
Step 1: Obtain the sum of x, y, x 2 , y 2 and xy. Recall that we already obtain the
values in Example 1.
L X = 1,029
L X2
L Y = 2,115
L y 2 = 380, 887
= 88,733
L xy = 183,222
Step 2: Compute for the slope of the simple linear regression.
n(L xy) - ( L x) (y)
h 1 = -----2
n(L x 2 ) - ( L x)bi =
12(183,222) - (1,029) (2,115)
12(88,733) - (1,029) 2
2,198,664- 2,176,335
1,064,796 - 1,058,841
22,329
5955
= --= 3.7496
Step 3: Compute for the mean value of x and y.
x=
y=
L X 1,029
n
= � = 85.75
LY
71
2,115
= � = 176.25
Step 4: Compute for the intercept of the simple linear regression.
= 176.25- 3.7496(85.75)
= 176.25- 321.5282
= -145.2782
Step 5: Substitute the slope and intercept in the general simple linear
regression equation.
General Equation for Simple Linear Regression
The Simple Linear Regression is y = 3.749x - 145.2782.
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Step 6: G raph the least square regression line.
-
Cl)
._
ctl
._
Cl)
c..
E •
Cl)
I-
140
150
160
170
180
190
200
210
220
Sales
Thus, the regression equation is y = 3.749x - 145.2782. The b 1 of
3.7496 indicates that for each other additional temperature in Fahrenheit, sales
are expected to increase by 3.7496 units. The b 0 value of - 145.2782 indicates
that the intercept with the y-axis is below the origin. A concrete interpretation
is that if the temperature in Fahrenheit is zero, a negative 145.2782 units would
be sold.
Learning Tasks/Activities
Perform as indicated.
A. The table below shows the results of a study in which researchers
examined a child's IQ and the presence of a specific gene in the child.
H igh IQ
Normal IQ
Total
1.
2.
3.
4.
Gene present
33
39
72
Gene not present
19
11
30
Tota l
52
50
102
Find the probability that the child has a normal IQ.
Find the probability that a child has the gene.
Find the probability that a child has a normal IQ and has the gene.
Find the probability that a child has a high IQ, given that the child
has the gene.
B. If scores are normally distributed with a mean of 25 and a standard
deviation of 7.5, what percent of the scores is:
1. Greater than 40
2. Lower than 40
3. Between 28 and 42
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Assessment
A.
The average age of bank managers is 40 years. Assume that the
variable is normally distributed. If the standard deviation is 5 years,
find the probability that the age of a randomly selected bank manager
will be in the range between 35 and 46 years old.
B.
A rate analyst for LEYECO was asked to determine if there is a linear
relationship between electricity consumption and the number of
rooms in a single-family dwelling. Since electricity consumption varies
from month to month, he decided to study usage during the month of
March. He collected the following data.
No.
of
6
Rooms (x)
Ki lowatts3.5
hours (y)
10
8
7
11
5
4
3
3
6
14
7
4
12
3
2
1
1.5
6
1. Determine the regression equation.
2. Plot the regression line.
3. Interpret the result.
Instructions on how to submit student output
Refer to the course policies and course content plan.
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50
Lesson 4.1 : I ntrod uctio n to Log ic
Lesson Summary
We begin by examining closely the fundamental concepts in the study of logic.
This lesson discusses the statement forms and how logical operators affect
the truth and falsity of statements; this tal ks about the logical eq uivalences of
statements.
Learning Outcomes
1 . Determine the truth values of propositions.
2. Translate statements or propositions into symbols and vice versa.
3. Construct truth tables of proposition involving d ifferent logical
operators.
4. Establish the validity of arguments.
Motivation Question
What is the importance of studying logic in our current society?
Discussion
Propositions
Let us begin with the definition of proposition - as the building block of
our reasoning. A proposition declares that something is the case, or it says that
something is not.
Definition 1 : A proposition (or statement) is a declarative statement that is
either true or false, but not both.
The variables p, q, r, s, and t are commonly used to represent
propositions.
Example 1 :
p: 2 i s the only even prime number.
This can be read as,
p is the proposition "2 is the only even prime number. "
Definition 2: The truth or falseness of a statement is called truth value. The
truth value of a proposition is true, denoted by T if it is a true statement;
otherwise, the truth value is false, denoted by F.
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Example 2: For each of the given sentences, determine which ones are
propositions and identify its corresponding truth value.
1. Baybay city is in Region VIII.
6. Oh my God!
3 . How old are you?
8. 8 + 2 (2 + 2)
7. Humans never die.
2. 4 is a composite number.
4. The moon is a planet and caterpillars can fly.
5. Close the door when you leave.
9 . ,.Jz < 1
= 16
1 0. Zero is an integer
The sentences listed above are propositions except 3, 5, and 6.
Nondeclarative sentences like questions (3 ) , commands (5 ) , and exclamations
(6) are not propositions. It is not possible to determine its truth value. For
example, take sentence (3); we cannot determine whether "How old are you?"
is true or false.
Statements 1, 2, 8, and 10 are true, while statements 4, 7 and 9 are false.
Definition 3: A compound proposition is a proposition that is made up of
several propositions joined by logical connectors.
The following are examples of logical connectors involving
propositions p and q .
not p
p and q
p or q
If p, then q
p if and only if q
Definition 4: A proposition is simple if it cannot be broken down into more than
one component propositions.
Remark: A compound proposition is made up of more than one simple
propositions.
Example 3: From the previous example, identify the simple and compound
propositions. For each compound proposition, break it down into simple
propositions.
The propositions in the previous example are sentences 1, 2, 4, 7, 8, 9,
and 1 0. Of these seven propositions, only proposition 4 is a compound
proposition. Let p: The moon is a planet and caterpillars can fly. Proposition p
has the following simple proposition components:
q : The moon is a planet.
r: Caterpillars can fly.
Symbolically, proposition p can be expressed as "q and r".
Definition 5: An open sentence is a sentence that has a variable.
Open sentences are either true or false, depending on the value taken
by the variable. Consider the following sentences:
1. He is an honest person.
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52
2. x < 0
3. 3y
= 111
These are examples of open sentences. It is not possible to decide the
truth value of these sentences because of the presence of the variables. For
instance, sentence 1, "He is an honest person." The truth value of this sentence
depends on who the pronoun "He" refers to.
Many mathematical sentences contain variables and are therefore
neither true nor false as they are, but will become propositions when variables
take values. A common example of such a mathematical sentence is a
formula.
Quantifiers
In linguistic, a quantifier makes a sentence about something with
specific property into a sentence about the quantity (number) of the things
having that property. In mathematics, a quantifier is a word, an expression, or
a phrase that specifies the number of things a statement relates to.
There are two types of quantifiers in mathematics, namely, the universal
quantifier and the existential quantifier.
Definition 6: The universal quantifier refers to the phrase "for all," "for each," or
"for every." It is represented by the symbol "v." It states that a certain statement
holds for any value of the independent variable in its domain.
For example, x 2 � O is true for all real numbers x since the square of
any real number is nonnegative. Symbolically, 'rfx E !Rl, x 2 � O, where IRl is the
set of real numbers. This can be read as "For all x in the set of real numbers,
x 2 � O."
Other examples are listed below.
1. The statement '"r/x, 'rfy E IRl, xy = yx" (read as "For all number x and all
number y in the set of real numbers, xy = yx. ") is true since the
multiplication of real numbers is commutative.
2. The statement '"r/x E ru, 2x + 3 = 7" (read as "For each number x in the
set of natural numbers, 2x + 3 = 7. ") is false since it is only true when
x = 2. Here, ru is the set of natural numbers.
3. The statement '"r/x E l, ..Jx E IRl" (read as "For every number x in the set
of integers, the ..Jx is in the set of real numbers.") is false since -1 E l
yet R.. = i fl. IRl. Here, l is the set of integers. i is called an imaginary
number and an element of the set of complex numbers a + bi.
Definition 7: The existential quantifier refers to the phrase "for some" or "there
exists." It is represented by the symbol "3." It says that a certain statement
holds for at least one element in the domain.
Consider the third example above, if 'r/ is replaced by 3, then the
statement "3x E l such that ..Jx E IRl" (read as 'There exists a number x in the
set of integers such that ..Jx is in the set of real numbers.") becomes true since
there exists an integer 1 for which -JI. = 1 is a real number.
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Other examples are listed below.
1. The statement "3x E lffi. (read as "there exists a number x in the set of
real numbers") such that x 2 + 2x + 3 < O" is false since there does not
exist a real number for which the statement is true. For any real number,
x 2 + 2x + 3 > O. The graph of the equation x 2 + 2x + 3 = O is a
parabola that opens upward with a vertex at the point (-1,2).
-vx
is an
2. The statement "3x E (Q (read as "for some rational number x'')
irrational number" is true since 2 is a rational number and -Jz is an
irrational number. Here, (Q is the set of rational numbers.
3. The statement "3x E ru (read as "there exists a natural number x '') such
that 2x + 3 = 1" is false since the value of x for which 2x + 3 = 1 is x =
-1 but -1 is not a natural number.
Logical Operators and Equivalent Statements
Definition 8: A truth table of a proposition is a table that shows all the possible
combinations of the truth values of its component propositions.
Given a proposition p. The truth value of p is either true (T) or false (F),
but not both. Its truth table is shown below.
Table 1 Truth table of p
Given two propositions p and q . Note that both p and q could either be
true or false. The truth table of p and q has 2 2 = 4 possible combinations of
the truth values of p and q, as shown below.
T
T
F
F
T
F
T
F
both propositions are true
p is true, but q is false
p is false, but q is true
both propositions are false
Table 2 Truth Table of p and q
Given three propositions p, q, and r. This time, there will be 2 3 = 8
possible combinations of the truth values of the given propositions. That is,
p q r
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F
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Table 3 Truth Table of p, q, and r
Remark: The truth table of n different simple propositions will have z n possible
truth-value combinations.
Definition 9: A logical operator on propositions is a word or phrase that
combines propositions to make a new compound proposition.
The Negation of a Proposition
Let p be a proposition. The negation of p is the statement "not p," and
is denoted by ~ p. The symbol ~ represents the negation operation and is read
as "not" or "negation of."
Definition 1 0: If p is true, then ~ p is false; and if p is false, then ~ p is true.
p ~p
T
F
F
T
Table 4 Truth table for Negation
The negation of proposition p is the proposition "It is not the case that
p" or "It is not true that p."
Example 4: Write the negation of each of the following statements.
1. Manila is the capital city of the Philippines.
2. The product of two consecutive integers is odd.
3. Every Viscan is diligent.
For each number below, the statements written are all negations of the
given statement.
1. Given:
a.
b.
c.
Manila is the capital city of the Philippines.
Manila is not the capital city of the Philippines.
It is not the case that Manila is the capital city of the Philippines.
It is false (or it is not true) that Manila is the capital city of the
Philippines.
The given proposition "Manila is the capital city of the Philippines" is
true; hence its negation must be false.
2. Given: The product of two consecutive integers is odd.
a. The product of two consecutive integers is not odd (or even).
b. It is not true (or it is false) that the product of two consecutive
integers is odd.
c. It is not the case that the product of consecutive integers is odd.
Since the product of two consecutive integers n and n + 1 is even (if n
is odd, then n + 1 is even; or if n is odd, then n + 1 is even), then the
given proposition is false. Thus, its negation is true.
3. Given: Every Viscan is diligent.
a. It is not the case that every Viscan is diligent.
b. It is not true that every Viscan is diligent.
c. Some Viscans are not diligent.
Since there are Viscans who are not diligent, then the given proposition
must be false. Thus, its negation is true.
Remarks:
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i.
ii.
55
The negation ~ p of a proposition p is not exactly the opposite what p
declares. The negation of proposition 3 does not speak about every
Viscan. It is enough to identify even just one Viscan who is not diligent
to negate the given proposition.
The negation of the statement ~ p is the statement p, that is,
~ (~ p) = p .
Conjunction
A coniunction is a compound proposition formed by joining
propositions with the word "and." That is, if p and q are propositions, then the
conjunction of p and q is the proposition "p and q," and is denoted
p /\ q .
Definition 1 1 : If p and q are true, then p A q is true; otherwise, p A q is false.
T
q
T
p /\ q
T
F
F
F
T
F
p
F F
T
F
Table 5 Truth table for Conjunction
The propositions p and q in "p A q" are called coniuncts. A conjunction
is true only if both conjuncts are true. Other than this combination, the
conjunction is false.
Example 5: Consider the following propositions:
p: The Covid - 19 vaccine Sputnik V is from Russia.
q: H2 0 is water.
Write, as a sentence, the following conjunctions:
a. P A q
b. ~ p A q
~ p /\~ q
d. P A~ q
C.
Answers:
a. The Covid-19 vaccine Sputnik V is from Russia, and H2 O is water.
b. The Covid-19 vaccine Sputnik V is not from Russia, and H2 0 is water.
c. It is not the case that the Covid-19 vaccine Sputnik V is from Russia,
and H2 0 is not water.
d. The Covid-19 vaccine Sputnik V is from Russia, and it is not true that
H2 O is water.
Of these four conjunctions, it is only p A q that is true. Note that a
conjunction is only true when both component propositions are true.
Example 6: Determine the truth value of each proposition.
1. The moon is a star and millipedes do fly.
2. There are seven colors in a rainbow and the sun sets in the south.
3. -2 < O and (-2) 2 > o.
4. 5 is an even number and 2 x 5 is an even number.
Answers:
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1. ''The moon is a star" is false and "millipedes do fly" is also false. Thus,
the conjunction is false since both conjuncts are false.
2. ''There are indeed seven colors in a rainbow but the sun sets in the west.
Thus, the conjunction is false since one of its conjuncts is false.
3. The conjunction is true since both "- 2 < 0" and "(-2) 2 > 0" are true.
4. 5 is an odd number but 2 x 5 is an even number since 2 x 5 = 10. Thus,
the conjunction is false since it has one false conjunct.
Remark: Apart from the word "and, " the terms "but, " 'yet, " "although, " though, "
"even though, " "moreover, " "furthermore, " "however, " "whereas, " and "while"
also denote conjunctions.
Disjunction
A disiunction is a compound proposition formed by joining propositions
with the word "or." That is, if p and q are propositions, then the disjunction of p
and q is the proposition "p or q," and is denoted
p V q.
Definition 1 2: If p and q are false, then p v q is false; otherwise, p v q is true.
q
pvq
T
F
T
F
F
P
T
F
T
T
T
T
F
Table 6 Truth Table of Disjunction
The propositions p and q in p v q are called disiuncts. The disjunction
is true when at least one disjunct is true and is false only when both disjuncts
are false.
Example 7: Consider the following propositions:
p : Johnny is eating sugar.
q : Covid-19 is from abroad.
r: The Miami Heat is the 2020 NBA champion.
Write, as a sentence, the following disjunctions.
a. p V q
b. ~ p V (q v r)
C. p V~ r
Answers:
a. Johnny is eating sugar or Covid-19 is from abroad.
b. It is not true that Johnny is eating sugar, or Covid-19 is from abroad or
the Miami Heat is the 2020 NBA champion.
c. Johnny is eating sugar or it is not the case that the Miami Heat is the
2020 NBA champion.
Example 8: Determine the truth value of each proposition.
1. The moon is a star or millipedes do fly.
2. There are seven colors in a rainbow or the sun sets in the south.
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3. -2 < o or (-2) 2 > o.
4. 5 is an even number or 2 x 5 is an even number.
Answers:
1. The first proposition is false. Both disjuncts are false.
2. The second proposition is true. A rainbow has seven colors, indeed.
Regardless of whether the sun sets in the south or not, the disjunction
is true since it has at least one true disjunct.
3. The third proposition is true since both disjuncts are true.
4. The fourth proposition is true. Even if 5 is an odd number, still 2 x 5 =
1 0 is an even number. Thus, it has one disjunct that is true.
Conditional/Implication
The conditional of propositions p and q is the proposition "If p, then q"
or "p implies q," and is denoted by
p
➔
q
The proposition p is called the hypothesis. while the proposition q is called the
conclusion.
Definition 1 3: The conditional proposition is false if the hypothesis is true and
the conclusion is false: otherwise, the conditional proposition is true.
q
p --+ q
T
F
F
F
F
T
p
T
F
T
T
T
T
Table 7 Truth table for Conditional
Example 9: Determine the truth value of the given propositions.
1.
2.
3.
4.
If the moon is a star, then millipedes do fly.
If there are seven colors in the rainbow, then the sun sets in the south.
If -2 < o, then (- 2) 2 > o.
5 is an even number implies 2 x 5 is an even number.
Answers:
1. The moon is not a star, so the hypothesis is false. It means that the
conditional proposition is true whether or not millipedes do fly.
2. The hypothesis is true but the conclusion is false since the sun sets in
the west. Thus, the conditional proposition is false.
3. Both hypothesis and conclusion are true. Thus, the conditional is true.
4. 5 is an odd number; it means that the hypothesis is false. Thus, the
conditional is true.
Remarks:
In logic, the hypothesis of a conditional proposition need not cause its
conclusion. For instance, the first conditional proposition, whether or not
the moon is a star, will not cause the millipedes to fly.
ii) Aside from the "ff ... , then ... " form, conditionals can be stated in the
following forms as well:
✓ p implies q
✓ q when p
✓ If p, q
i)
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✓ p is sufficient for q ✓ q follows from p
✓ p only if q
✓ q if p
✓ q is necessary for p ✓ q whenever p
Biconditional
A biconditional is a compound statement formed by joining two
propositions with the phrase "if and only if". That is, if p and q are propositions,
then the biconditional of p and q is the proposition "p if and only if q".
Symbolically,
p H q.
Definition 1 4: The biconditional p H q of p and q is true if both p and q have
the same truth value; otherwise, it is false.
p
T
T
F
F
q
T
F
T
F
pHq
T
F
F
T
Table 8 Truth table for Biconditional
The biconditional proposition p
conditional propositions p ➔ q and q ➔ p.
H
q is the conjunction of the
Example 10: Determine the truth value of the given propositions.
1. The moon is a star if and only if millipedes do fly.
2. There are seven colors in the rainbow if and only if the sun sets in the
south.
3. -2 < O if and only if (-2) 2 > o.
4. 5 is an even number if and only if 2 x 5 is an even number.
Answers:
1. Both ''The moon is a star" and "Millipedes do fly" are false. Thus, the
biconditional proposition is true.
2. ''There are seven colors in the rainbow" is true, while ''The sun sets in
the south" is false. Thus, the biconditional proposition is false.
3. Both "-2 < O" and "(-2) 2 > O" are true. Hence, the biconditional
proposition is true.
4. "5 is an even number" is false and "2 x 5 is an even number" is true.
Therefore, the biconditional proposition is false.
Remark: The following phrases also denote the biconditional of p and q.
p iff q which is read as "p if and only if q "
"p is necessary and sufficient for q "
Constructing Truth Tables
Recall that the truth table of a proposition is a table that shows all the
possible combinations of the truth values of its component propositions.
Example 11: Construct the truth table of ~ p v q .
Answer: The given statement involves two propositions. This implies that its
truth table will have 22 rows. Begin with the columns containing p and q .
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p
q
T
F
F
F
59
T T
F
T
Then in the third column are the truth values for ~ p. Recall that ~ p is
true if p is false and ~ p is false if p is true.
p q ~p
T
T
F
T
T
T
F
F
F
F
F
T
Now, the last column will be the truth values of ~ p v q . Use the
definition of a disjunction to determine the truth value for each combination.
p q ~p ~pVq
T
T
F
T
T
T
F
F
F
F
T
F
F
T
T
T
Example 12: Construct the truth table of (p ➔ q) A (q ➔ p) .
Answer: The given statements involves two propositions, p and q . Thus,
its truth table will have 2 2 rows. The truth values of propositions p and q are
shown in the first two columns.
p
q
T
F
F
F
T T
F
T
Add as a third column the truth value of p ➔ q . Use the definition of the
conditional to determine the truth values.
p
T
T
F
F
F
F
T
F
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p➔q
T
T
q
T
T
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On the fourth column, write the truth values for q
p
q
p ➔ q
q ➔ p
T
T
T
T
T
F
T
F
F
F
F
T
F
T
Lastly, the truth values of (p
conjunction.
➔
p.
➔
T
T
q) A (q
➔
p) using the definition of
p
q
p ➔ q
q ➔ p
(p ➔ q ) A ( q ➔ p )
T
T
T
T
T
T
F
T
F
F
F
F
T
F
T
T
F
T
T
F
Example 13: Construct the truth table of (p A q) v (~ r) .
Answer: The truth table for (p A q) v (~ r) will have 2 3 rows since there
three propositions, namely, p, q, and r.
p
q
T
T
T
F
F
T
T
T
F
T
T
T
F
F
F
T
F
T
T
F
F
F
F
p
q
pAq
T
T
T
T
F
F
T
F
T
Then add the column for p A q .
T
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T
T
T
T
F
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T
F
F
F
F
T
F
F
F
F
T
F
F
F
F
Add the column for ~ r.
p
q
T
T
T
F
F
T
T
T
F
F
F
F
T
F
F
r pAq ~r
T
T
F
T
F
F
T
F
F
F
F
T
F
F
F
T
F
F
F
F
T
61
T
F
F
T
F
T
T
T
T
Finally, the column for (p A q) v ( ~ r) .
p
q
T
T
T
F
F
T
T
T
F
F
F
r p A q ~ r (p A q ) V (~ r)
T
T
F
T
T
F
F
F
T
F
F
F
F
F
F
T
F
F
F
T
F
F
F
F
T
T
F
F
F
T
T
T
T
T
T
T
T
Tautologies and Contradictions
Definition 1 5: A proposition that is always true is a tautology. while a
proposition that is always false is a contradiction.
A tautology is denoted by the small Greek letter r and the small Greek
letter cp denotes contradiction.
Example 14: Let p and q be propositions. Use the truth table to show the
following:
i)
p V~ p is a tautology
ii) p A~ p is a contradiction
iii) p ➔ (p v q) is a tautology
iv) (p A~ q) A (p A q) is a contradiction.
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Answers:
i)
p ~ p p V~ p
T
F
T
F
T
T
The truth values in the last column are all true; this shows that p v~ p
is a tautology.
ii)
p
T
F
iii)
~ p p A~ p
F
F
T
F
The truth values in the last column are all false; this shows that p A~ p
is a contradiction.
p
T
q
F
F
F
➔
T
T
T
F
pVq p
T
T
T
T
T
(p V q )
T
T
F
The truth values in the last column are all true. Thus, p
tautology.
➔
(p v q) is a
iv)
q
~q
T
F
T
F
F
p
T
F
T
T
F
p A~ q p A q
F
F
T
F
F
F
F
T
T
F
(p A~ q ) A (p A q )
F
F
F
F
Therefore, (p A~ q) A (p A q) is a contradiction.
Equivalent Statements
Definition 1 6: Two propositions p and q are said to be logically equivalent,
denoted by p q, if p and q have the same truth values for all possible truth
=
values of their components or whenever they have identical truth tables.
Example
Show that p q =~ p v q using a truth table.
1 5:
➔
Answer: To show this, we need to show that p ➔ q and ~ p v q have
identical truth tables.
p
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q
~p
p --+ q
~pVq
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T
F
T
T
T
F
F
F
F
=~
T
T
T
F
F
F
T
T
T
T
T
It can be seen that the fourth and fifth columns have identical truth values.
Hence, p ➔ q
p v q.
Example 16: Show that p H q
= (p
➔
q) A (q
➔
p) using a truth table.
Answer: We need to show that p H q and (p
identical truth tables.
p
T
1) A (q
q
T
p H q
p ➔ q
q ➔ p
(p ➔ q ) A ( q ➔ p )
T
T
T
T
F
T
F
T
T
F
F
F
F
F
T
T
F
➔
T
F
T
T
It can be observed that p H q and (p ➔ q)
Thus, the two are equivalent statements.
➔
p) have
F
A (q ➔
p) have identical truth tables.
Arguments
Definition 1 7: An argument is a compound proposition of the form
(P 1 A P2 A · · · A Pn ) ➔ q ,
where the propositions P 1 i p 2 , . . . , Pn are called premises of the argument, and q
is the conclusion.
An argument can be written in propositional form, as above, or in
column or standard form:
P1
P2
Pn
:. q
where the symbol :. is read as "therefore." The premises of an argument are
intended to act as reasons to establish the validity or acceptability of the
conclusion.
Example 17: Identify the premises and conclusion of the following arguments.
a. He's a Cancer since he was born on the first day of July.
Answer:
b.
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Premise: He was born on the first day of July.
Conclusion: He's a Cancer.
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If Covid-19 is true, then we should stay at home.
Covid-19 is true.
Therefore, we should stay at home.
Answer: The propositions "If Covid-19 is true, then we should stay at
home." and "Covid-19 is true." are the premises of this argument. While
the proposition "We should stay at home." is the conclusion.
The following phrases are used to indicate the premises and conclusion
of an argument.
Premises indicator
for
since
given that
for the reason that
it is a fact that
as shown by the fact that
granted that
Conclusion indicator
therefore
thus
hence
so
consequently
then
implies
Definition 1 8: An argument that has a conclusion that is true whenever the
premises are all true is called a valid argument.
An argument that is not valid is called an invalid argument or fallacy.
Theorem : The argument consisting of premises p1 , p2 , . . ., Pn and conclusion q
is valid if and only if the proposition (p1 A p2 A · · · A Pn ) ➔ q is a tautology.
List of Some Valid Arguments
1 . Law of Detachment (also known as Modus Ponens)
Symbolically, the argument is written as
p➔q
'[l___
p
T
q
T
p
➔
T
q
(p
:. q
➔
q) A p
T
T
F
F
F
F
F
T
F
F
T
T
[(p
➔
q) A p ]
T
➔
q
T
T
F
T
The truth table shows that the given argument is valid since
[ (p ➔ q) A p] ➔ q is a tautology.
Example 18: Verify the validity of the following arguments.
a. If Hansel studied the lesson well, then he got a good exam score.
Hansel studied the lesson well.
Therefore, Hansel got a good exam score.
Answer: If we let p to denote "Hansel studied his lesson well" and q
be "He got a good exam score." Then the given argument can be
written symbolically as
p➔q
'[l___
:. q
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By Law of Detachment, this is a valid argument.
b. If Hansel studied the lesson well, then he got a good exam score.
He got a good exam score.
Therefore, Hansel studied the lesson well.
Answer: If we let p to denote "Hansel studied his lesson well" and q
be "He got a good exam score." Then the given argument can be
written symbolically as
p➔q
fJ___
:. p
The Law of Detachment does not apply to this case. We will use a
truth table to verify the validity of this argument. That is, we need
to verify whether [(p ➔ q) A q] ➔ p is a tautology.
q
p➔q
(p ➔ q ) A q
[(p ➔ q) A p] ➔ p
T
F
F
F
T
F
F
T
F
p
T
F
T
T
T
T
T
T
T
F
T
Clearly, [(p ➔ q) A q] ➔ p is not a tautology. Thus, the given
argument is a fallacy.
2. Law of Contraposition (also known as Modus Tollens)
Symbolically, the argument can be written as
p➔q
~q
:.~ p
p q p ➔ q ~ q (p ➔ q ) A ~ q ~ p [(p ➔ q ) A ~ q ] ➔~ p
T
T
T
F
F
F
T
T
F
F
T
T
T
F
F
F
F
T
F
T
T
T
F
T
F
T
T
T
The truth table shows that the given argument is valid since
[(p ➔ q) A ~ q] ➔~ p is a tautology.
Example 19: Verify the validity of the following arguments.
a. If Hansel studied the lesson well, then he got a good exam score.
It is not the case that Hansel got a good exam score.
Therefore, Hansel did not study the lesson well.
Answer: If we let p to denote "Hansel studied his lesson well" and q
be "He got a good exam score." Then the given argument can be
written symbolically as
p➔q
~q
:. ~ p
By the Law of Contraposition, this is a valid argument.
b. If Hansel studied the lesson well, then he got a good exam score.
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Hansel did not study the lesson well.
Therefore, Hansel did not get a good exam score.
Answer: If we let p to denote "Hansel studied his lesson well" and q
be "He got a good exam score." Then the given argument can be
written symbolically as
:. ~ q
The Law of Detachment does not apply to this argument. We will
use the truth table to verify the validity of this argument. That is, to
show that [(p ➔ q) /\~ p] ➔ ~ q .
p
T
q
T
p
➔
T
T
F
F
F
F
T
F
q
F
➔
q ) /\~ p
F
T
T
T
~q
F
F
T
T
(p
~p
➔~
➔
q ) /\~ p]
T
T
T
Clearly, [(p ➔ q) /\~ p]
argument is a fallacy.
[(p
q
T
F
T
➔~
F
T
T
q is not a tautology. Thus, the given
3. Law of Syllogism
Symbolically, the argument can be written as
p
➔
p
q
T
T
T
T
T
T
F
T
F
F
T
T
T
T
T
F
F
F
T
F
F
F
T
F
F
F
F
T
p
➔
q
q
➔
T
r
(p
q
r
.: p ➔ r
q ➔
➔
q) I\ (q
T
➔
r)
p
➔
F
T
F
F
T
T
F
F
T
T
T
T
T
T
➔
T
F
T
q) A (q
T
F
T
➔
T
F
F
[(p
T
T
T
r
T
T
T
r)]
➔
(p
➔
r)
T
T
T
T
T
T
The truth table above shows that the given argument is valid, since the
compound proposition in the last column is a tautology.
Example 20: Verify the validity of the following arguments.
a. If Hansel studied the lesson well, then he got a good exam score.
If he got a good exam score, then he'll buy an ice cream.
Therefore, if Hansel studied the lesson well, then he'll buy an ice
cream.
Answer: If we let p to denote "Hansel studied his lesson well," q be
"He got a good exam score," and r be "He'll buy an ice cream." Then
the given argument can be written symbolically as
p➔q
q➔r
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:. p
➔
67
r
By the Law of Syllogism, this is a valid argument.
b. If it rains today, then I will wear my raincoat.
If I wear my raincoat, then I will stay dry.
Thus, if it rains today, then I will stay dry.
Answer: If we let p to be the proposition "It rains today.", q be "I will
wear my raincoat.", and r be "I will stay dry." Then the above
argument can be written symbolically as
p
➔
q
q➔r
:. p ➔ r
Clearly, it is a valid argument by the Law of Syllogism.
Arguments and Euler Diagrams
An Euler diagram is used to analyze or test the validity of an argument.
Example: Create Euler diagrams to test the validity of an argument.
1. All Math teachers love numbers.
Jhesi is a Math teacher.
Therefore, Jhesi loves numbers.
Answer: Begin with an Euler diagram of the first premise.
The group of Math teachers is within the group of people to love
numbers. Note that there are individuals who love numbers but are not
Math teachers.
Then, add an Euler diagram of the second premise.
people who
love numbers
Jhesi, as a Math teacher, is in the group of Math teachers.
Clearly, the Euler diagram says that Jhesi loves numbers. Therefore,
the argument is valid.
2. If it rained, then the ground is wet.
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The ground is wet.
Therefore, it rained.
Answer: The Euler diagram of this argument is shown below.
Note that there are times when the ground is wet, yet it has not
rained. Thus, rain is just one of the many reasons why the ground is
wet. This shows that the argument is invalid.
Learning Tasks/Activities
1. Determine which of the given sentences is a proposition. For each
proposition, determine its truth value.
a. O divided by O is 1.
b. Are we there yet?
c. Turn off the light when no one is using it.
d. There are three primary colors.
2. Determine the truth value of the following propositions and provide
reasons for your answer.
a. Vx in the set of real numbers, .!.X is a real number.
b. 3x in the set of integers for which 2x - 3 = -3.
3. Construct a truth table of the following propositions and determine
whether it is a tautology, a contradiction, or neither.
a. (p ➔ q) V (q ➔ p)
b. (p A ~ p) A q
4. Show that p ➔ q
q ➔ p.
5. Determine the validity of the following argument:
If two sides of a triangle are equal, then the angles opposite to these
sides are equal.
Two sides of a triangle are not equal.
Therefore, the angles opposite to these sides are not equal.
=~ ~
Assessment
1. Determine the truth values of the following propositions.
a. It is not the case that 2 is the only even prime number.
b. The number rr is an irrational number and its value is less than
3.
c. 13 is a composite number or 13 is an odd number.
d. If 2 divides all even numbers, then 22 is divisible by 2.
2. Given the following propositions,
p : Dyroth is a prince.
q : Odette is beautiful.
r:Cyclops has one eye.
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i)
Express the following propositions symbolically.
a. Cyclops has two eyes and Odette is beautiful.
b. Either Odette is beautiful or Dyroth is a prince.
c. Either Odette is beautiful or Cyclops has one eye, but
not both.
ii) Express the following propositions in words.
a. p V~ q
b. ~ q ➔~ P
3. Establish the validity of the following argument using truth table.
Roger is either a human or a werewolf.
Roger is not human.
Therefore, he is a werewolf.
Instructions on how to submit student output
Refer to the course policies and course content plan.
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Math1 1 n: Mathematics in the Modern World
Lesson 4.2: Mathematics in Finance
Lesson Summary
This lesson discusses the two primary ways of computing interest, the simple
interest, and the compound interest. This lesson includes various problem
applications that facilitate learning necessary skills in applying simple and
compound interest concepts in real-world settings.
Learning Outcomes
1 . Calculate the maturity value, principal amount, and interest rate in
simple and compound interest.
2. Calculate the time it takes for the principal to raise in simple and
compound interest.
3. Differentiate simple from compound interest.
4 . Calculate problems that involve simple and compound interest.
5. Solve real-world problem applications involving simple and compound
interest.
Motivation Question
Suppose that Ellyze won P 1 0,000. 00, and she plans to invest it for three years.
A cooperative group offers two simple interest rate per annum. A bank offers
1 . 5% compounded annually. If she wants to earn more, which will she choose,
and why?
Discussion
Everybody uses money. Sometimes we work for our money, and other
times our money works for us. For instance, unless we are attending college
on a full scholarship, it's very likely that our family have either borrowed money
or saved money, or both, to pay for our education. When we borrow money, we
usually have to pay interest. When we save money, we may lend our money to
a financial institution and expect to earn interest on our investment.
Interest is typically defined as a fee for borrowed money. We receive
interest when we let others use our money, for instance, by depositing money
in a savings account. We pay interest when we use other people's money, such
as borrowing from a bank or a friend. Interest is an expense for the person who
borrows money and income for the person who lends money.
There are two ways of calculating the amount of interest: simple
interest and compound interest. Though simple interest and compound
interest are basic financial concepts, becoming thoroughly familiar with them
may help us make wiser decisions when considering a loan or investing.
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Simple Interest
When interest is charged only on the principal amount for the entire
term, it is called simple interest. The interest under this method is not charged
on any accumulated interest. This kind of interest usually is used for loans or
investments for a short period.
Simple Interest Formula:
The simple interest amount is computed using the following formula:
I = Prt
(1)
where:
/ is the interest
P is the principal amount
r is the rate
t is time or term, in years
Before we proceed, let us be familiar with the following terms that we
will be using in the discussion:
•
•
•
•
•
•
Principal - the amount of money borrowed or invested on the origin
date
Time/term - number of unit of time for which the interest is computed
Rate of interest - usually in percent, the fractional part of the principal
that is paid on a loan or the investment
Future value or maturity value - the sum of the principal and interest,
which is accumulated at a particular time
Lender or creditor - a person (or institution) who invests the money or
makes the funds available
Borrower or debtor - a person (or institution) who owes money or avails
of the fund from the lender
Example 1 :
A bank offers a 0.22% annual simple interest rate for a particular
deposit. How much interest earned if 1 million pesos is deposited in this
savings account for one year?
Solution:
Given: P = P 1 000 000.00; r = 0.2 5% = 0.002 5; t = 1 year
Required: Interest (I)
To solve for the interest, we simply substitute the above given to the
formula I = Prt. Hence, we have
I
= (1 000 000) (0.002 5) (1)
I = 2 500
Thus, the interest earned is P 2 500. 00
Example 2:
Find the interest of a P 43 450.00 investment at a 6 . 5% interest rate for
1 ½ year.
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Solution:
Given: P =P 43 450. 00; r = 6.5 % = 0.065; t = 1.5 years
Required: Interest (I)
Solving for I, we have
I = (43 450) (0.065) (1. 5 )
I = 4 2 36.375
Thus, the interest earned in the investment is P 4 2 36.375.
Example 3:
How much interest is charged when P 60 000.00 is borrowed for ten
months at an annual simple interest rate of 10%?
Solution:
Notice that the time given is expressed in months (M). Before we can
use the formula, we need to convert it to years by t = �12
Given: P = P 60 000; r = 10% = 0. 10; t = .2..
years
12
Required: Interest (I)
Substituting the given information to the formula, we get
I
= (60 000) (0.10) ( :2 )
I = 4 500
Thus, the interest charged is P 4 500. 00.
Note:
•
In a real-life setting, we often neglect that money is rounded to the
nearest centavo. For this reason, when using the formulas for interest,
principal, and maturity value, we may round off our answers to the
nearest centavo.
Ordinary and Exact Simple Interest
When time is expressed in days, interest can be computed using the
exact number of days in a year or the approximate number of days in a year
that assumes 30 days every month. Since time (t) in the formula is expressed
in a year, the number of days must be expressed as a fractional part of the year,
which can have 365 (or 366 for leap year) for exact interest, and 360 days for
ordinary interest.
Ordinary simple interest (10 ) is a type of interest that uses 360 days as
the equivalent number of days in a year or 30 days each month. This kind of
interest is computed using the formula:
_ Pr (
Io -
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number of days
)
360
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On the other hand, exact simple interest (le ) is an interest computed based on
the exact number of days in a year, 365 days, or 366 days for leap year. Exact
simple interest is computed using the formula:
_
(number of days)
le - Pr
3 5
6
(3)
Example 4:
On May 30, 2019, a businessman loan P 45 000 in the bank to expand
his restaurant. The businessman agreed to pay the amount at a 6% rate on
August 10, 2019. How much is the ordinary simple interest to be paid?
Solution:
First, we need to compute the number of days from May 30, 2019, to
August 10, 2019. Note that May 30 is the beginning date; hence it is not
included in the counting.
May 31: 1 day
June 1-30: 30 days
July 1-31: 31 days
August 1-10: 10 days
Thus, there are 72 days from May 30, 2019, to August 10, 2019. Now, solving
for the ordinary simple interest:
10
10
= (45 000) (0.0 6) (::o )
= 540
Therefore, the ordinary simple interest to be paid is P 540.00.
Example 5:
Mr. X borrowed P 13 500.00 from his aunt last December 25, 2019. He
promised that he would pay at 8% interest on March 14, 2020. Determine the
exact simple interest to be paid by Mr. X.
Solution:
2020:
Calculating the number of days from December 25, 2019, to March 14,
December 25 - 31: 6 days
January 1-31: 31 days
February 1-29: 29 days
March 1-14: 14 days
Thus, there are 80 days from December 25, 2019, to March 14, 2020.
Now, solving for the exact simple interest, we obtain
le
= (13 5 00) (0.08)
le
=
6
2 3 .7 1
C86°5
)
Therefore, Mr. X will pay P 236.71 interest.
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Computing the Maturity Value
One may be interested to know the amount that a lender will give to the
borrower on the maturity date. For instance, we may want to know the total
amount of money in a savings account after t years at an interest rate r. This
amount is called the maturity value or future value F.
The amount of interest is just the additional amount to be paid on the
money loaned or borrowed. The maturity value or final amount is the sum of
the interest accumulated over the period and the principal amount. It is
computed using the formula:
or
where,
F
=P+I
(4)
F
= P ( 1 + rt)
(5)
F is the maturity (future) value
P is the principal amount
r is the rate
t is the time, in terms of year
I is the interest
Since the formula for solving interest is also applied in the formulas
above, time is also in years; hence, ordinary and exact simple interest can still
be used.
Example 6:
Find the maturity value if P 500 000.00 is deposited in a bank at an
annual simple interest rate of 0.22% after five years.
Solution:
Given: P =P 500 000.00; r = 0.22% = 0.0022
Required: maturity value F after five years
We will solve this problem in two methods. We can either solve the
simple interest /, and then add it to the principal amount P, that is, F = P + I,
or we can simply use formula (5). Whichever method we use, we will have the
same results.
Solving for I:
Method 1:
Method 2:
Vision:
Mission:
I
= (500 000) (0.0022) (5)
F
=P+I
= 500 000 + 5 500
I = 5 500
F
F
= 505 500
= P (1 + rt)
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= 500 000 [ 1 + (0.0022) (5) ]
F = 505 500
Thus, P 505 500.00 is the maturity value after five years.
Example 7:
A sum of P 65 500.00 is borrowed for 15 months at 8 �4 % simple
interest. Determine the future value of the loan.
Solution:
Given: P = P 65 500.00; r = 8 �4 % = 0.082 5; t = 15 months
Required: maturity value F
Substituting the given information to formula (5), we have
F
= P (1 + rt)
= (65 500) [ 1 + co.082 5)
F
= 72 2 54.6875
G�)l
Thus, P 72 254.69 is the future value of the loan.
Computing the Principal Amount
Computation for principal amount is applied for planning purposes.
This approach is more of a progressive viewpoint. If someone expects to earn
a certain amount from investment or pay a loan or debt, determining what
amount to borrow or invest will give the borrower or investor the idea to start
to achieve the amount expected to pay or earn.
Recall that the interest is obtained by multiplying the principal, the
interest rate, and time; that is I = Prt. Therefore, dividing the interest by the
product of the interest rate and time gives us the formula for the principal
amount (P).
p
I
=rt
(6)
=F-1
(7)
F
1 + rt
(8)
Moreover, the future amount or the maturity value is the sum of the
principal amount and the interest. Thus, the principal is the difference between
the future amount and the interest; that is,
P
If only the future amount, time, and interest rate are given, we can
manipulate formula (2) above to calculate the principal.
p
Example 8:
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What is the principal invested at 4% for eight months if the interest is P
200.00?
Solution:
Given: r = 4% = 0.04; I = P 200.00; t = 8 months
Required: principal (P)
Since interest, rate, and time are given, we use formula (6) to compute
the principal amount. Hence, we have
I
P=­
rt
200
8
(0.04) ( 12 )
P = 7,500
Therefore, P 7 500.00 is the principal amount to be invested.
Example 9:
Emyat paid her friend P 42 400.00 on the money she borrowed. If there
is an interest of P 2 400, how much did she borrow?
Solution:
Given: F = P 42 400.00; I = P 2 400.00
Required: Principal (P)
Since the future value and the amount of interest are given, we use
formula (7) to solve for P. Thus, we have
=F-1
= 42 400 - 2, 400
P = 40 000
P
Therefore, Emee Li borrowed P 40 000.00 from his friend.
Example 10:
Leoma Rich wants to earn P 1 million in 10 years. How much should he
invest in the bank if the interest rate is 2. 5%?
Solution:
Given: F = P 1 000 000.00; t = 10; r = 2.5% = 0.02 5
Required: Principal amount (P)
Applying formula (5), we have
F
P = -1 + rt
(1 000 000)
= 1 + (0.02 5) (10)
P
= 800 000
Therefore, Leoma Rich must have an initial investment of P 800 000.00.
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Computing the Time
Formulas being used in solving interest has years as the unit of time, so
it must be observed to have consistent units of time to avoid miscalculations.
It should also be noted that the rate should be in decimal form before using it
in the formula.
Formula:
Example 11:
I
t=­
Pr
(9)
The interest earned on an P 80 000.00 deposit is P 12 250.00. How long
was the term if the interest rate applied is 5%?
Solution:
Given: P = P 80 000.00; / = P 12 500.00; r = 5% = 0.05
Required: time (t)
Substituting the given to formula (9), we have
I
t=­
Pr
t
=
(12 500)
(80 000) (0.05)
3.125
Thus, the amount was deposited for 3.125 years or 3 years, 1 month, and 15
days.
Example 12:
A particular organization loaned P 14 000.00 from a bank at 9% interest.
They will pay a total of P 20 300.00 within the agreed date of payment. How
long will the organization need to pay the amount?
Solution:
Given: P = P 14 000.00; F = P 20 300.00; r = 9% = 0.09
Required: time (t)
To use formula (9), we need to solve first the interest I.
Solving for the interest:
I = F-P
= 20 300 - 14, 000
Now, solving for t:
I = 6 300
I
t=­
Pr
(6 300)
(14 000) (0.09)
t=5
= ------
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Hence, the organization must pay the amount after five years.
Example 13:
Gia Ellyzz withdraws P 3 220.00 from a bank after she invested her
money at 7.5% interest. If she receives an interest of P 420.00, for how long
was the investment made?
Solution:
Given: F = P 3 220.00; / = P 420.00; r = 7.5% = 0.075
Required: time (t)
First, we need to solve for the principal amount P for us to use the
formula (9) in solving for t.
P
Hence, solving for t, we have
=F-l
= 3 2 2 0 - 420
P = 2 800
I
t=­
Pr
(420)
(2 800) (0.075)
t=2
Thus, the amount was invested for two years.
Computing the Simple Interest Rate
Simple interest rate is computed using the formula:
I
Example 14:
r=Pt
(1 0)
A loan of P 170 500.00 was charged P 10 500.00 for 2. 5 years. What
was the simple interest rate applied to the loan?
Solution:
Given: P = P 170 500.00; I = P 10 500.00; t = 2.5 years
Required: interest rate (r)
Applying the formula, we have
I
r=Pt
1 0 500
(170 500) (2.5)
= ------
r = 0.0246334
Thus, approximately 2.46% was applied to the loan.
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Example 15:
Suppose that P 60 000.00 was deposited in the bank. At the end of three
months, the amount became P 61 625.00. Find the interest rate.
Solution:
Given: P = P 60 000. 00; F
Required: interest rate (r)
= P 61
625.00; t
= 3 months
To solve for the interest rate using formula (1 0), we need the interest
/. Since I = F - P, we have
I
r=Pt
F-P
= Pt
(61, 625 - 60, 000)
= ------3
(60, 000) (1 2)
1, 625
1 5, 000
r = 0.108 3 3 3 3
Thus, the interest rate is approximately 1 0.83%.
Compound I nterest
As discussed earlier, simple interest is generally used for investments
or loans for a short period. For more extended periods, compound interest is
usually being used. With compound interest, interest is paid or charged on
interest as well as on the principal amount. The interest earned in the period
will be added to the principal. Both will earn an interest in the next period. This
process is called compounding. The compound amount is the accumulated
value of the principal amount and all interest amounts of prior periods. In other
words, the compound amount is the sum of the principal and all compound
interests.
For instance, if P 100 000.00 is deposited at 5% interest for one year, the
interest at the end of the year is P 100 000.00(0.05) (1) = P 5, 000.00. the
balance in the account is P 100 000.00 + P 5 000.00 = P 105 000.00. Suppose
that this amount is left at 5% interest for another year. The interest is computed
on P 105 000.00 instead of the original P 100 000.00. This implies that the
account at the end of the second year has the amount of P 1 0 5 000.00 +
P 105 000.00(0.05) (1) = P 1 1 0 2 50.00. Note that simple interest would yield a
total amount of only
P 100 000 [ 1 + (0.05) (2) ]
=
P 1 1 0 000.00.
The additional P 250.00 is the interest on P 5 000.00 at 5% for one year.
Interest can be compounded more than once per year. The following
are standard compounding periods:
•
•
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Semi-annually - two periods per year
Quarterly - four periods per year
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•
Math1 1 n: Mathematics in the Modern World
Monthly - twelve periods per year
The interest rate per period, i, is obtained by dividing the annual interest, r, by
the number of compounding periods, m, per year.
Compound Amount Formula
The formula below computes compound interest:
where i
= !:..
and n = mt,
m
C = P(1
+ On
(1 1)
C is the compound amount or the future value
P is the present value or the principal amount
i is the interest rate per period or the periodic interest
n is the total number of compounding periods
m is the number of compounding periods in a year
r is the nominal rate or the annual interest rate
t is the number of years
Example 16:
An amount of P 2 0 0 0 0 . 0 0 is invested at a rate of 2% compounded
yearly. After two years, how much will the amount be?
Solution:
Given: P = P 20 000.00; r = 2% compounded yearly; t = 2 years
Required: Compound amount or future value C
Since the nominal rate is 2% compounded yearly, so m
0.02
i = -- = 0.02
1
= 1. Hence,
and
Thus, solving for c
n = 1 (2) = 2.
C = P (1 + O n
= 2 0 000(1 + 0.02) 2
C = 2 0 808
Therefore, the amount becomes P 20 808.00 after two years.
Example 1 7:
Ms. Lea M . Arich loaned P 50 000.00 from a bank and agreed to pay the
amount at 6% compounded quarterly for three years. How much will she pay
after three years?
Solution:
Given: P = P 50 000.00; r = 6% compounded quarterly; t = 3 years
Required: Compound amount or future value C
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Note that the nominal rate is 6% compounded quarterly; hence m
Thus, solving for the future value, we have
C
= P (l + O n
C
= 59 780.90857
= 4.
0.06 4 (3 )
= 50 000 (1 + 4)
Therefore, Ms. Lea will pay P 59 780.91 after three years.
Computing the Principal Amount
Similar to simple interest, computing the principal amount under
compound interest anticipates the planned or expected future amount would
be paying or receiving.
With the formula for the compound amount, the principal amount is
computed as follows:
C
P = --(1 + on
Example 18:
(12)
A father wants to give his son P 100 000.00 when he turns 21 years
old. How much will he invest in the bank at 4% interest compounded semi­
annually if his son 13 years old now?
Solution:
Given: C = P 100 000.00; r = 4% compounded semi-annually; t
Required: Principal amount (P)
Compounded semi-annually implies m
C
= 8 years
= 2. Now solving for P,
=
p (1 + on
100 000
P
=
0 04 Z ( S)
( 1 + -·2-)
72 844.58137
Therefore, the father must invest P 72 844. 58 in the bank.
Example 19:
XYZ company paid a total of P 150 million for the 6-hectare lot loan
paid for ten years at 6% compounded monthly. How much was the price of
the lot?
Solution:
Given: C = P 150 million; r = 6% compounded monthly; m
Required: Principal amount (P)
= 12; t = 10 years;
Applying the formula for P, we obtain
C
P = --(1 + on
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150 000, 000
0.06 1 2 ( 1 0 )
( 1 + """I2)
= 82 444 910
P
Therefore, the price of the lot was P 82 444 910.00.
Computing the Compound Interest Rate
Notice that in computing the compound amount, there are two interest
rates to be considered:
1. Nominal interest (r)
2. Periodic interest (i)
Also, recall that
nominal rate (r)
periodic rate (i) = --------------number of compounding period in a year (m)
(1 3)
From the formula in finding the compound amount, the formula in
finding the periodic interest rate is expressed as
i
= (�)11 - 1
(14)
Example 20:
At what periodic rate, compounded quarterly, will P 175 000.00
become P 322 500.00 at the end of 12 years?
Solution:
Given: C = P 322 500.00; P
Required: Periodic rate (i)
= P 175, 000.00; t = 12 years;
m
=4
Solving for i:
1
i
= (�)11 - 1
1
322 500 4 1 2
= (175 000) ( ) - l
i = 0.012817219
Thus, the periodic rate is approximately 1.28%.
Example 2 1:
Emly borrowed P 10 000.00 from a friend, and compounded interest is
done quarterly. If she paid a total of P 11 948.30 after two years, how much is
the interest rate?
Solution:
Given: C = P 11 948.30; P
Required: nominal rate (r)
Vision:
Mission:
= P 10 000.00; t = 2 years;
m
=4
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To solve the nominal rate, we will use the formula r
Hence, we need i.
Solving fo r i:
i
=
=
i
Solving for r:
(�t
= im, from (13).
1
-1
1
11 948.30 4(2)
_l
(
)
10 000
= 0.022499877
r = im
r
= 0.022499877 (4)
= 0.08999511
Therefore, her friend charges approximately 9% interest compounded
quarterly.
Learning Tasks/Activities
1. Complete the following tables below. Show all necessary solutions
neatly and systematically. Unorganized solutions will not be
considered.
Table 1
Rate (r) Time (t)
Principal
Maturity
Interest (I)
(P)
Value (F)
P 70,000
4%
8
a) _
b) _
c) _
e) _
1 2%
5
P 1 5,000
d) _
1 0. 5%
f) _
P 1 57,500
P 457,500
Table 2
p
r
l
m
i
p
2,000
a)_
Semiannually
b)_
c)_
p
5,000
f)_
quarterly
g)_
h)_
n
le
C
2
d)_
e)_
p
2,800
5 yrs
and 6
months
i)_
P 500
j)_
t
Legend:
P - principal amount; r - nominal rate; l - interest compounded;
m - frequency of conversions in a year; i - interest rate per period;
t - time in years; n - total number of conversions;
le - compound interest; C - the compound amount
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Assessment
A. Read each problem carefully, then solve for what is being asked. Show
all necessary solutions neatly and systematically. Unorganized
solutions will not be considered.
1. Cassy Nillo invested an amount of P 10 000.00 for two years at 8%
simple interest. Jay makes the same investment on the same terms,
but interest is compounded annually. What is the difference in
earnings between the two investments made?
2. Ernie Lee invested P 48 000.00 in the money market at a 10% interest
rate last March 18, 2018. She plans to get back the money by
January 28, 2021. Suppose the borrower insists on paying exact
interest instead of ordinary interest. How much does he lose or gain
in interest?
3. Suppose that Weng has P 85,000.00. He decided to deposit it in a
bank and will not withdraw from it for five years. A bank offers two
types of compound interest accounts. The first account offers 5%
interest compounded semi-annually and the second account offers
4.5% interest compounded monthly. Which account will he choose if
he wants his money to earn more?
4. At what simple interest rate will an amount of money double itself in
10 years?
B. In your own words, discuss the difference between simple interest and
compound interest.
Instructions on how to submit student output
Please refer to the course policies and course content plan.
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85
Lesson 4.3: Mod u lar Arithmetic
Lesson Summary
This lesson covers the principle of modular arithmetic such as operations on
cong ruence, finding the least residue of a number, and some mod u l a r
arithmetic appl ications t o include-finding a day in a week, check digit schemes,
and basic concept of cryptology. Also, students a re provided with ample
exam ples to appreciate the use of mathematics in their l ife.
Learning Outcomes
1.
2.
3.
4.
Perform operations on mathematical expressions correctly.
Find the least residue of a n u m ber.
U s e modular arithmetic t o solve problems related t o real-life scenarios.
Apply mathematical principles in cryptog raphy.
Motivation Question
Without referri ng to a calendar, answer the fol l owing questions.
1 . If it is now Sunday, what day will it be 1 000 days from now?
The answer is Saturday, but the interesting fact is that we didn't arrive at
the answer by starting with Sunday and counting off 1 000 days (it is not
reasonable to do so). Instead, we simply observe that 1000 = 142 (7) + 6,
and we count six days from Sunday.
2. If it is now Novem ber, what month will it be 1 60 months from now?
Similarly, if it is now November, it is easy to see that 1 50 months from now
will be March. This time we get the answer by noting that 160 = 13(12) +
4, so we add four months to November instead of counting 1 60 months.
This simple idea has various essential mathematics and computer science
appl ications that will be discussed i n this l esson.
Discussion
Modular Arithmetic
Modular Arithmetic is a form of arith metic deal ing with the remainders
after integers are d ivided by a fixed positive i nteger m, called the mod u lus.
I n this concept, we only deal with i ntegers and rema inders. The
operations used are addition, su btraction, multiplication, and d ivision. In
modular a rithmetic a l l operations are performed rega rding the mod u l u s or a
d ivisor.
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As follows, we will learn the notion of divisibility. If we divide an integer
by a positive integer, we can get a quotient and a remainder. These remainders
lead us to modular arithmetic, which plays a vital role in mathematics.
*
If a and b are integers with a O, we say that a divides b if there is an
integer c such that b = ac, or equivalently, if � is an integer.
When a divides b, we say that a is a factor or divisor of b, and that b is a
multiple of a. We write a I b to mean that a divides b . We write a t b
means that a does not divide b .
Example 1: Determine whether 3 1 14 and if 3 1 18.
Solution: It is easy to see that 3 t 14 since 3 is not an integer. On the other
14
hand, 3 I 18 since
18
3
= 6.
When an integer is divided by a positive integer, there is a quotient and
a remainder. This is shown in the following result.
The Division Algorithm
Let a be an integer and d be a positive integer. Then there are unique
integers q and r, with O � r < d, such that a = d q + r.
Example 2: Find the quotient q and the remainder r when 101 is divided by 11
according to the Division Algorithm.
Solution: By the Division Algorithm, 101 = ._____,___,
11(9) + 2 since 101 + 11 = 9 r. 2.
Hence q = 9 and r = 2.
99
Example 3. Find q and r when -38 is divided by 7 according to the Division
Algorithm.
Solution: The Division Algorithm requires that the remainder r must be
nonnegative but less than d (divisor). Since d = 7, then a = -38 = 7 (-6) + 4,
._____,___,
-42
so that q = -6 and r = 4.
In some situations, we only care about the remainder of an integer when
it is divided by a specific positive integer.
The notation a (mod m) represents the remainder when an integer a is
divided by the positive integer m. In what follows, we will learn the concept of
congruence, indicating that two integers have the same remainder when they
are divided by the positive integer m.
Congruence
Definition. Let a and b E Z, and m E N. We say "a is congruent to b modulo
m", and write "a = b (mod m)", if m l (a - b) ( reads as m divides (a - b) ).
In other words, if (a - b) is divisible by m. The integer m is called the
modulus of the congruence. If a is not congruent to b modulo m, we write
a '$. b (mod m).
Z-set of integers
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N-set of natural/counting numbers
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Example 4: Verify if the following are true:
a. 2
b. 1
c. 2
Solution:
= 5 (mod 3)
= 10 (mod 3)
= 7 (mod 3)
=
a. Since 3 I (2 - 5), then 2 5 (mod 3) .
b. Since 3 I (1 - 10), then 1 10(mod 3) .
c. Since 3 � (2 - 7), then 2 $. 7 (mod 3) .
=
Also, if m > 0 and r is the remainder when the Division Algorithm is used to
divide b by m, then r is called the least residue of b (mod m) .
Example 5: Find the least residue of the following using the division algorithm:
a. 15 (mod 11)
b. - 13(mod 5)
c. 101(mod 16)
d . 928 (mod 27)
e. 136(mod 200)
f. 157 (mod 8 1)
Solution: Use the division algorithm to find the remainder or the least residue.
a. The least residue of 15 (mod 11) is 4 since 15 = 11(1) + 4. Hence,
15 (mod 11) 4.
b. The least residue of - 13(mod 5) is 2 since - 13 = 5 (-3) + 2 . Hence,
- 13(mod 5) 2 .
c. The least residue of 101(mod 16) is 5 since 101 = 16(6) + 5 . Hence,
101(mod 16) 5.
d. The least residue of 928 (mod 27) is 10 since 928 = 27(34) + 10.
Hence, 928 (mod 27) 10.
e. The least residue of 136 (mod 200) is 0 since 136 = 200(0) + 136.
Hence, 136 (mod 200) 0.
f. The least residue of 157 (mod 8 1) is 76 since 157 = 8 1(1) + 76.
=
=
=
=
=
=
For letter b, - 13 (mod 5) 2 since - 13 = 5 (- 3) + 2 . Here, we have to
multiply the modulus by a quotient that will result to a number greater than 13
(with the negative sign). That is 5 times - 3, which is - 15. In this way, we can
get a positive remainder, which is 2. Also, when we add - 15 and 2 the result
should be equal to - 13.
In letter e, notice that 136 is less than the modulus 200. In this case,
136 is just the remainder.
Take note that a remainder is a value greater than or equal to zero but
less than the divisor. We cannot have a negative number as a remainder.
Properties of Congruences
Congruence is useful because it can be manipulated like ordinary
equations. Congruences to the same modulus can be added, multiplied, and
taken to a fixed positive integral power, i.e., for any a, b, c, d E Z and m, n E N .
=
1. If a = b (mod m) and c d ( (mod m), then
a. a + c b + d (mod m)
b. a - c b - d (mod m)
c. ac bd (mod m)
2. If a = b (mod m), then an = b n (mod m) .
3. If a = b (mod m), then na nb (mod m).
==
=
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4. If a = b (mod m), then f(a) = f(b) (mod m) for any polynomial f (n)
with integer coefficients.
5. If a = b (mod m), then a = b (mod n) for any positive divisor d of m.
6. Reflexive: a a(mod m) .
7. Symmetric: If a = b (mod m), then b a(mod m).
8. Transitive: If a b (mod m) and b c (mod m), then a c (mod m) .
=
Arithmetic Operation
=
=
=
=
In arithmetic operation, we perform the addition, subtraction, division or
multiplication and then divide by the modulus. The answer is the remainder.
Thus the result of an arithmetic operation mod m is always a whole number
less than m. When we wish to compute ab (mod m) or a ± b (mod m), and a or
b is greater than m, it is easier to "mod first", then perform the operations on
the results. However, we may also use the other way around.
1. a + b (mod m)
(Addition)
2. a - b (mod m)
(Subtraction)
4. a + b (mod m)
(Division)
3. a x b (mod m)
(Multiplication)
Example 6: Evaluate each of the following:
a. 54 + 19 (mod 1 1)
Solution: Add 54 and 19 to get 73. Then divide the sum by the modulus 11.
The answer is the remainder.
54 + 19 (mod 1 1)
73 (mod 1 1)
73 + 1 1 = 6 r. 7
73 = 1 1 (6) + 7 Using Division Algorithm
Hence, 54 + 19 (mod 1 1) 7
=
b. 1 1 1 - 40 (mod 6)
Solution: Get the difference between 1 1 1 and 40, which is 7 1 . Then divide by
the modulus, 6. We may also use the Division Algorithm to show the least
residue.
7 1 (mod 6)
7 1 = 6(1 1) + 5
Hence, 1 1 1 - 40(mod 6)
=5
c. 39 x 48 (mod 5)
Solution 1 : Multiply 39 and 48 to produce 1872, then divide by the modulus,
5. We may also use the Division Algorithm to get the least residue.
1872 (mod 5)
1872 + 5 = 3 74 r. 2 or
1872 = 5 (3 74) + 2 Using Division Algorithm
Hence, 39 x 48 (mod 5)
Vision:
Mission:
=2
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Solution 2: Instead of performing multiplication, we work with the "mod" first.
Observe that
39 (mod 5)
48 (mod 5)
=
=4
=3
since 39
since 48
=
= 5 (7) + 4
= 5 (9) + 3
and
Thus, 39 x 48 (mod 5) 4 x 3 (mod 5) 12 (mod 5) . Observe that 12 is
greater than 5, so we still need to divide 12 by 5 to get the remainder.
12 + 5 = 2 r. 2 or using the Division Algorithm
Consequently, we can write 12 (mod 5) 2 .
Hence, 39 x 48 (mod 5)
=
= 12 (mod 5) = 2
12 = 5 (2)
+ 2.
Note: This process is also applicable for addition or subtraction in modulo m.
However, we can also get the same answer by performing the operation
directly.
For division modulo m, we need the notion of the multiplicative inverse. But
first, we look at the concept of the additive inverse.
Additive Inverse
If the sum of two numbers is O, then the numbers are additive inverses of
each other.
For instance, 4 + (-4)
the additive inverse of 4.
= O, so 4 is the additive inverse of -4, and -4 is
=
The same concept applies to modular arithmetic. For example, 4 + 3
O (mod 7). 0 here is the remainder when 4 + 3 is divided by 7. Hence, in mod 7,
3 is the additive inverse of 4, and 4 is the additive inverse of 3. Here, we consider
only those whole numbers smaller than the modulus.
Multiplicative Inverse
If the product of two numbers is 1, then the numbers are multiplicative
inverses of each other.
For instance, 6 · ¼ = 1, so 6 is the multiplicative inverse of ¼ and ¼ is the
multiplicative inverse of 6.
The same concept applies to modular arithmetic (in this case, the
multiplicative inverses will always be natural numbers). For example, in mod 7
arithmetic, 5 is the multiplicative inverse of 3 (and 3 is the multiplicative inverse
of 5) because 5 · 3 l (mod 7) . 1 here is the remainder when 1 5 is divided by
7. (Again, we will concern ourselves only with natural numbers less than the
modulus). The multiplicative inverse of a number b is usually denoted by b - 1 .
=
Example 7:
1. Find the additive inverse of 9 in mod 15 arithmetic.
2. What is the multiplicative inverse of 9 in mod 13 arithmetic?
3. Does 4 have a multiplicative inverse in mod 6 arithmetic?
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Solution:
1. 6 is the additive inverse of 9 in mod 15 because 9 + 6
= 0 (mod 15).
2. 3 is the multiplicative inverse of 9 mod 1 3 since 9 · 3
l (mod 13) .
=
= 2 7 and 2 7
3. 4 does not have a multiplicative inverse in mod 6 arithmetic. If 4 has a
multiplicative inverse, it could be one of the numbers 0, 1, 2, 3, 4, and 5.
However, none of them will give a remainder of 1 when multiplied to 4
modulo 6.
In general, if a is a whole number, then a has a multiplicative inverse
modulo m whenever a and m are relatively prime. That is, a and m
have no common factor except 1 .
Now, how to perform division in modular arithmetic?
If a, b, and m are whole numbers, m > 1, then
a + b (mod m) = a · b - 1 (mod m)
provided b- 1 exists.
Example 8:
a. Evaluate (100 + 81) mod 1 3 .
Solution: To get the answer, we need to find the multiplicative inverse of
=
=
81. Observe that 8 1 - 1 (mod m) 9 since 8 1 x 9 = 729 l (mod 13) .
(Multiply 81 by 9 to get 729, then divide 729 by 13. The remainder is 1). So
the multiplicative inverse of 81 (in mod 13) is 9. Hence,
100 + 8 1 (mod 1 3) (100 · 8 i - 1 ) mod 13
=
= (100 · 9) mod 13
= 900 (mod 1 3)
since 900 + 1 3 = 69 r. 3
100 + 8 1 (mod 1 3) = 3
b. Evaluate (3 5 + 24) mod 1 1 .
Solution: Find the multiplicative inverse of 24. Try all possible numbers
from O to 10. Observe that 24- 1 (mod 1 1) 6 since 24 x 6 = 144
l (mod 1 1) . 144 + 1 1
Hence,
=
=
= 1 3 r. 1 . Then, the multiplicative inverse of 24 is 6.
= (3 5 · 24- 1 ) mod 1 1
= (3 5 · 6) mod 1 1
= 2 1 0 (mod 1 1)
(35 + 24) mod 1 1 = 1
since 2 1 0 + 1 1 = 19 r. 1
(3 5 + 24) mod 1 1
c. Evaluate ( 49 + 12) mod 8.
Solution: 12 and 8 are not relatively prime, so 12 has no multiplicative
inverse. Thus, the division ( 49 + 12) mod 8 is not possible.
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Example 9: a.) What is the last digit when the number 6
b.) What about the last digit of 3
Solution:
66
2 6 5 5 3 908 ?
66
66
6
is divided by 1 O ?
6
a). When we input 6
, it seems impossible to get the answer directly in the
calculator and divide it by 10 since it is of big value. But using the concept of
modular arithmetic and applying patterns, we can have
61 = 6
6 2 = 26
3
6 = 216
66
:=:
6 = 46656
6
Notice that any power of 6 results in a number with last digit of 6, which makes
the last digit 6
66
66
6
when divided by 10 is of 6. Hence, the last digit is 6.
b). Now for 32 6 5 5 3 908 , notice that
31 = 3
32 = 9
3 3 = 27
34 = 81
3 5 = 243
36 = 729
.-.
We can observe that the last digit of 3 as it raises to any number, has a pattern
of 3, 9, 7, 1, 3, 9, 7, 1, and so on as the remainders. From this, the four numbers
are our choices for the unit digit. To solve which of these is the last digit, we
can use the concept of modular arithmetic. Since there are four numbers, then
m = 4, that is (mod 4). Now, we get the exponents and find the remainder when
we divide it by four.
=== 132 (mod
(mod 4)
4)
== 01 (mod
4)
(mod 4)
=2
26553908 = 0 (mod 4)
f)_ 3
32
33
34
35
36
►1
=9
= 27
= 81
= 243
= 729
:
=
2
3
4
5
6
:
(mo d 4)
(mo d 4)
Notice that the number 26553908 ends with 8 which is divisible by 4 (giving as
a remainder of 0). Note, that 34 = 81 has the last digit of 1 (by applying patterns
as well), we can say that 32 6 5 5 3 908 has a unit digit of 1.
We can also write,
==
=
x y (mod m) � m divides x - y
3 y (mod 4)
3 -1 (mod 4)
Since 3 - y = 4, then y = -1. Now, since 26553908 is even then
32 6 5 5 3 908 (mod 4) ( - l ) 26 5 5 3 908 (m od 4)
1 (mod 4)
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Applications of modular arithmetic
We will start with some questions, as follow:
1. If today is Thursday, what day will it be 1000 days from now?
2. If it is now November, what month will it be 500 months from now?
Without referring to a calendar, we can get the answer to each question.
Modular arithmetic has something to do with these problems. This is just one
of the applications of modular arithmetic to real-life situations.
Example 1 :
a. If today is Thursday, what day will it be 1000 days from now?
Solution: Take note that there are seven days in a week. If we add seven
days to Thursday, the result will still be Thursday. So we will divide 1000
days by 7 to get the remainder and use this to identify the day of the week.
1000 + 7 = 142 r. 6
The remainder is 6. Start counting 6 days from Thursday, the result is
Wednesday. Therefore, if today is Thursday, 1000 days from now will be
Wednesday.
b. If it is now November, what month will it be 500 months from now?
Solution: If we add 12 months to November, the result will still be
November. So, we will divide 500 by 12 to get the remainder.
500 + 12
= 41 r. 8
Start counting 8 months from November, and that is July. Therefore, if it is
now November, the month it will be 500 months from now is July.
Example 2: Disregarding a.m. or p.m., if it is now 2 o'clock,
a. What time will it be 45 hours from now?
Solution: Observe that a clock has 12 hours. If we add 12 hours after 2
o'clock, it will still be 2 o'clock. Thus, we will use arithmetic modulo 12.
Once we get the remainder, we will start counting off from 2 o'clock. So,
45 + 12
= 3 r. 9
Count off 9 hours starting 2 o'clock; the result is 11 o'clock. Therefore, if it
is now 2 o'clock, 45 hours from now will be 11 o'clock.
b. What time was it 304 hours ago?
Solution: We will also use the modulus 12. Since we are task to get the time
304 hours ago, we will count backward using the remainder. Divide 304 by
12 to get the remainder.
304 + 12 = 25 r. 4
Starting at 2 o'clock, count 4 hours backward, and it is 10 o'clock.
Therefore, it is now 2 o'clock; 304 hours ago, it was 10 o'clock.
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Tip: There is a faster way to calculate the remainder. For example, find
the remainder when 304 is divided by 12. Using a calculator, input the
following:
304 + 12 = 2 5.33333333
Now, subtract the whole number from the quotient to let the decimal
remains.
2 5.33333333 - 2 5 = 0.33333333
The decimal is actually the remainder when we multiply it with the
divisor 12.
0.3 3 3 3 3 3 3 3 X 12
=4
The remainder when 304 is divided by 12 is 4. You may try another
examples.
The next application is about finding a day in a week. We need the following
ideas below before we solve some examples.
Finding a day in a week:
Table 9 Number represents the day in a week
Each corresponding number represents the day of the week.
Sunday
O
Monday
1
Tuesday
2
3
.... Wednesday
I Thursday
------<
4
Friday
5
Saturday
6
Table 10 Number represents the month in a year
Each corresponding number represents the month of the year.
January
February
March
April
May
June
+
+
11
12
1
2
3
July
August
September
October
November
December
5
6
7
8
9
10
Observe that there's a twist: we start counting with March = 1 so that
January and February are the 11th and 12th months of the previous year,
respectively. It ensures that days will not be affected because of a leap year (a
year when February has 29 days; it happens every 4 years) . Of course, the
arithmetic modulo to use is 7 because there are seven days in a week.
There is a formula that will give us the day of the week for any date in
history and even in the future. It is called Zeller's formula- an algorithm
devised by Christian Zeller to calculate the day of the week for any Julian or
Gregorian calendar date.
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To find the day of the week, here is the Zeller's formula:
W
=
D
+ [ 2 .6m - 0. 2 ] - 2 C + Y + [�] + [�] (mod 7)
Where:
W-a day of the week
Y-the particular year of the century
(last two digits of the year)
D -a day of the month
C-century (obtained as the first two
digits of the year)
m -a month of the year
Let us have some examples. We can pick any date.
Example 1 : We know that December 25, 2019, was Wednesday (you may verify
by looking at your calendar for this first example). Throughout the solution, the
result should be Wednesday.
Solution:
The date is December 25, 2019. Refer to Table 10 for the corresponding
number of December. So,
m=10
C
December ......,
2 5 , 20 1 9
......,
D
y
m = 10
D = 25
C = 20
Y
= 19
Simply substitute the values to the formula, then simplify.
W
W
W
W
=
=
==
D
+ [ 2 .6m - 0. 2 ] - 2 C + Y + [�] + [�] (mod 7)
1
2
°]
+ [ 2 .6(10) - 0. 2 ] - 2 ( 2 0) + 19 + [ : ] + [
4
2 5 + [ 2 6 - 0. 2 ] - 40 + 19 + 4.75 + 5 (mod 7)
2 5 + 2 5.8 - 40 + 19 + 4.75 + 5 (mod 7)
25
(mod 7)
Before simplifying further, each term is calculated as an integer result. It means
any remainder is discarded (simply disregard the decimal point). In the last
equation, 2 5.8 will be 2 8, and 4.75 will be 4. Simplifying further,
W
W
== 385 (mod
+ 5 - 40 + 19 + 4 +
7)
2
2
5 (mod 7)
Add the numbers
Now, evaluate 38 (mod 7) by finding the remainder using Division Algorithm:
38 = 7 (5) + 3 since 38 ...,.. 7 = 5 r. 3 . The remainder is 3. Hence, 38 (mod 7)
3. Consequently,
W
= 38 (mod 7)
=
W :::: 3
Now, as shown in Table 9, 3 is Wednesday. It means that December 25, 2019
was Wednesday, and we got the correct day using the Zeller's formula.
Example 2: Find the day of the week of January 1, 1899.
Solution: Observe that the given month is January. Based on the condition in
Table 10, January is the 11th month of the previous year, meaning that the year
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to use is the year before 1899, which is 1898 (the result is still the day of the
week of January 1, 1899). The condition is also the same as that of February.
January 1 , 18 98
'-,,J
D=l
.._,..,
D
C
Y
m = ll
C
m= 1 1
y
= 18
= 98
Using the formula; we have
= D + [2.6m - 0.2] - 2 C + Y + [;] + [�] (mod 7)
8
8
W = + [2.6( ) - 0.2] - 2 ( 8) + 8 + [ ] + [ ] (mod 7)
4
4
W = + [28.6 - 0.2] - 36 + 8 + 24.5 + 4.5 (mod 7)
W = + 28.4 - 36 + 8 + 24.5 + 4.5 (mod 7)
Disregard the decimal point:
W = + 28 - 36 + 8 + 24 + 4 (mod 7)
Add the numbers
W=
(mod 7)
W
1
11
1
1
9
1
9
9
1
9
1
9
119
Evaluate 1 1 9 (mod 7) . That is, 1 1 9 = 7 ( 1 7) + 0 since
remainder is 0. Hence, 1 1 9 (mod 7) 0. Consequently,
W
=
1 1 9 (mod
W=0
7)
=
1 19
+ 7 = 1 7 r. 0. The
Thus, January 1, 1899, was Sunday. You can check the result by looking at a
calendar.
=
If ever you got a negative sum, for example W - 1 5 (mod 7), simply use
the division algorithm to get the remainder. That is, - 1 5 = 7 (- 3) + 6.
Then the dav of the week is Saturdav.
The next application we have is about error-detecting codes.
Check Codes
Code
It is a system of words, letters, figures, or symbols used to represent others,
especially for the purposes of secrecy.
Check Digit Schemes and Error-Detecting Codes
Many methods are being used to produce unique identification
numbers. Some of these are Universal Product Code (UPC) for consumer
products, the European Article Number (EAN), the Credit Card, and the
International Standard Book Number (ISBN). The UPC, EAN-13, Credit Card
number and ISBN use check digits as their last digit to ensure that the number
is valid. (Sirug, 20 1 8)
What is a check digit?
Check digits are an integral part of error-detecting codes. Check digit schemes
are numbers appended to an identification number that allows the accuracy of
the information stored to be checked by an algorithm. It is usually the last digit
of an identification number. Error-detecting schemes and check digits are
found in the following:
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1 . UPC
The Universal Product Code (UPC) is used by many of the products sold
in supermarkets and shopping malls, containing identification numbers coded
with bars. These barcodes can be read by an optimal scanner.
The U PC is a 12-digit code consisting of two blocks of five digits
preceded and a single digit. The first six identify the country and the
manufacturer of the product, and the next five identify the product itself. The
final digit (last digit) is the check digit.
o
1 2000
Check digit
�
1
Figure 2 Sample UPC barcodes
Retrieved: https://www.leadtools.com/help/sdk/v20/dh/to/upc-ean-barcodes-in-leadtools.html
2. European Article Number (EAN)
The European Article Number (EAN), also known as International Article
Number, is a standard numbering system used globally to identify specific
retail product types, packaging configurations, and manufacturers. The most
commonly used EAN standard is the EAN-13.
1 11 11 1
/1 Prod
1u ct �
N umber Mfg
Check
7
Syste m
50 1054
Cod e
53 0 107
Code
0
1 11 1
07 567 8
1 1
l6 4l2 5
Dig it
Figure 3 Sample EAN-1 3 barcodes
Retrieved: http://www. barcodeisland.com/ean1 3.phtml
3. Credit Card Numbers
Companies that issue credit cards also use modular arithmetic to
determine whether a credit card number is valid. This is especially important in
online bank transactions, where credit card information is frequently sent over
the Internet.
The primary coding method is based on the Luhn algorithm, which uses
mod 10 arithmetic. Credit card numbers are normally 13 to 16 digits long. The
first one to four digits are used to identify the card issuer.
Figure 4 Sample credit card number
Retrieved: https://grit.ph/credit-cards/
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4. ISBN
The International Standard Book Number (ISBN) is a 13-digit number
created to help ensure that orders for books are filled accurately and that books
are catalogued correctly. It is usually found on the last cover page.
ISBN is a string of ten digits or thirteen digits. The first ten digits of an
ISBN-1 O identifies the book, and the last digit refers to the check digit.
I
I I
I S B N 9 78 - 1 - 9 1 1 2 2 3 - 1 3 - 9
9 78 1 9 1 1 223 1 39
Figure 5 Sample ISBN 1 3 and ISBN 1 0
Retrieved: https://selfpublishi ngadvice.org/new-createspace-barcode-policy/
https://www. isbn-1 3 . i nfo/example
Formulas used in the different applications of modular arithmetic:
1. Universal Product Code (U PC) - Check Digit Formula
d1 2
= 10 - (3d+ +3dd
1
2
+ 3d 3 + d4 + 3d 5 + d 6 + 3d 7 + d 8 + 3d 9 + d 1 0
11)
mod 10
2. European Article Number (EAN-13) - Check Digit Formula
d1 3
= 10 - (d ++ d3d ++3dd +) (mod
3d + d + 3d
10)
2
1
11
4
3
5
12
6
+ d 7 + 3d 8 + d 9 + 3d 1 0
3. Credit Card Number - Check Digit Formula using Luhn Algorithm
Note: Add all digits, treating two-digit numbers as two single digits.
d1 6
= 10 - (2d
1 + d 2 + 2d 3 + d 4 + 2d 5 + d 6 + 2d 7 + d 8 + 2d 9 + d 1 0
+ 2d 1 1 + d 1 2 + 2d 1 3 + d 1 4 + 2d 1 5 ) (mod 10)
4. International Standard Book Number (ISBN) with 10 digits - Check Digit
Formula
d1 0
= 11 - (10d
1
+ 9d 2 + 8d 3 + 7 d4 + 6d 5 + S d 6 + 4d 7 + 3d 8 +
2d 9 ) (mod 11)
5. International Standard Book Number (ISBN) with 13 digits - Check Digit
Formula
d1 3
Vision:
Mission:
= 10 - (d
1
+ 3d 2 + d 3 + 3d4 + d 5 + 3d 6 + d 7 + 3d 8 + d 9 + 3d 1 0 +
d 1 1 + 3d 1 2 ) (mod 10)
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Example 1 : Determ ine the check d i g it for the new prod uct release of ABCD I nc.
The first 1 1 digits are 7 - 87634 - 73 - 2 14-?.
Solution: To solve for the check d i g it 7 - 87634 - 73 - 2 14-?, we need to use
the formula:
d1 2
= 10 - (3d++3dd
1
+ 3d 3 + d4 + 3d 5 + d 6 + 3d 7 + d 8 + 3d 9 + d 1 0
1 1 ) mod 10
2
Let d 1 = 7 , d 2 = 8, d 3 = 7 , d4 = 6, d 5 = 3, d 6 = 4, d 7 = 7 , d 8 = 3, d 9 = 2, d 1 0 =
1, d 1 1 = 4.
Substitute the values to the form ula, then use the d ivision algorithm to get the
check d i g it.
= 10 - (3d+ +3dd +mo3dd +10d + 3d + d + 3d + d + 3d + d
d = 10 - [3(7) + 8 + 3(7) + 6 + 3(3) + 4 + 3(7) + 3 + 3(2) + 1
+ 3 (4)] (mod 10)
d = 10 - (2 1 + 8 + 2 1 + 6 + 9 + 4 + 2 1 + 3 + 6 + 1 + 12) (mod 10)
d = 10 - 112 (mod 10)
Now, evaluate 112 (mod 10) . Using the d ivision algorithm, 112 = 10(11) + 2
since 112 ...,.. 10 = 11 r. 2 . Thus, 112 (mod 10) = 2 . Simpl ifying fu rther,
d = 10 - 112 (mod 10)
d = 10 - 2
d =8
d1 2
1
2
3
4
5
11)
7
6
9
8
10
12
12
12
12
12
12
Hence, the check d i g it is 8, and the U PC of the new product is 7 - 87634 - 73 2 14 - 8.
Example 2: Determ ine the check digit of EAN-1 3 n u m ber 9 - 310779 30000-?.
Solution: We let, d 1 = 9, d 2 = 3, d 3 = 1, d4 = 0, d 5 = 7, d 6 = 7, d 7 = 9, d 8 =
3, d 9 = 0, d 1 0 = 0, d 1 1 = 0, d 1 2 = 0.
Applyi ng the form u l a for EAN-1 3, we get
= 10 - (d ++ 3d3d )+(mod
d + 3d + d + 3d + d + 3d + d + 3d + d
10)
d = 10 - [ 9 + 3 (3) + 1 + 3 (0) + 7 + 3 (7) + 9 + 3(3) + 0 + 3 (0) + 0
+ 3(0) (mod 10)
d = 10 - (9 + 9 + 1 + 0 + 7 + 2 1 + 9 + 9 + 0 + 0 + 0 + 0) (mod 10)
d = 10 - 65 (mod 10)
Evaluate 6 5 (mod 10) . Using the d ivision algorithm, 65 = 10(6) + 5 since 65 ...,..
10 = 6 r. 5 . So, 65 (mod 10) = 5 .
d = 10 - 65 (mod 10)
d = 10 - 5
=5
d1 3
1
2
3
4
5
6
7
8
9
10
11
12
13
13
13
13
13
d1 3
Hence, the check d i g it is 5, and the EAN-1 3 is 9 - 310779 - 30000 - 5 .
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Mission:
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Example 3: Evaluate whether 4404 - 4507 - 7611 - 3863 is a valid cred it card
n u m ber.
Solution: Since the Cred it card is composed of 1 6 digits, we will apply the check
d i g it
d1 6
= 10 - (2d
+ d 2 + 2d 3 + d4 + 2d 5 + d 6 + 2d 7 + d 8 + 2d 9 + d 1 0 + 2d 1 1
+ d 1 2 + 2d 1 3 + d 1 4 + 2d 1 5 ) (mod 10)
1
Where
d 1 = 4, d 2 = 4, d 3 = 0, d4
7 , d 1 0 = 6, d 1 1 = 1, d 1 2 = 1, d 1 3 = 3, d 1 4
d1 6
d1 6
d1 6
= 4, d 5 = 4, d 6 = 5 , d 7 = 0, d 8 = 7, d 9 =
= 8, d 1 5 = 6.
= 10 - (2d+ +d d ++2d2d ++ dd ++2d2d +)d(mod+ 2d10)+ d + 2d + d + 2d
= 10 - [2 (4)+ 1+ +4 2+(3)2 (0)+ (8)
+ 4 + 2 (4) + 5 + 2 (0) + 7 + 2 (7) + 6 + 2 (1)
+ 2 (6)] (mod 10)
= 10 - (8 ++412)+ 0(mod
+ 4 + 8 + 5 + 0 + 7 + 14 + 6 + 2 + 1 + 6 + 8
10)
1
2
3
12
4
5
14
13
7
6
8
9
10
11
15
Note that we will use the Luhn Algorithm, add a l l dig its, and treat two-d igit
num bers as two single digits. From the last equation, 1 4 will be (1 + 4) and
1 2 will be (1 + 2)
= 10 - (8 ++41+2)0(mod
+ 4 + 8 + 5 + 0 + 7 + 14 + 6 + 2 + 1 + 6 + 8
10)
d = 10 - (8 + 4 + 0 + 4 + 8 + 5 + 0 + 7 + (1 + 4) + 6 + 2 + 1 + 6 + 8
+ (1 + 2)) (mod 10)
d = 10 - (8 + 4 + 0 + 4 + 8 + 5 + 0 + 7 + 5 + 6 + 2 + 1 + 6 + 8
+ 3) (mod 10)
d = 10 - 67(mod 10)
Evaluate 67(mod 10) . Using d ivision algorithm 67 = 10(6) + 7. So,
67(mod 10) = 7.
d = 10 - 67(mod 10)
== 10 - 7
3
d
d1 6
16
16
16
16
d1 6
16
Thus, the check d i g it is 3, and the credit card n u m ber 4404 - 4507 - 7611 3863 is val i d .
Example 4 : Check whether ISBN 978 - 62 1 - 406 - 064 - 1 o f t h e book
"General M athematics for Senior High School" by Winston Si rug is val id.
Solution: Since the ISBN is com posed of 1 3 digits, we will apply the check digit
d1 3
= 10 - (d ++ 3d3d )+(mod
d + 3d + d
10)
1
2
3
12
Where d 1 = 9, d 2 = 7 , d 3 = 8, d4
6, d 1 0 = 0, d 1 1 = 6, d 1 2 = 4.
d1 3
d1 3
d1 3
d1 3
Vision:
Mission:
4
5
+ 3d 6 + d 7 + 3d 8 + d 9 + 3d 1 0 + d 1 1
= 6, d 5 = 2, d 6 = 1, d 7 = 4, d 8 = 0, d 9 =
= 10 - (d ++ 3d3d )+(mod
d + 3d + d + 3d + d + 3d + d + 3d + d
10)
= 10 - [9 ++33(7)(4)+(mod
8 + 3(6) + 2 + 3(1) + 4 + 3 (0) + 6 + 3 (0) + 6
10)
== 1010 -- 89(9 (mod
+ 2 1 + 8 + 18 + 2 + 3 + 4 + 0 + 6 + 0 + 6 + 12) (mod 10)
10)
1
2
3
4
5
6
7
8
9
10
11
12
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1 00
Evaluate 89(mod 10). 89
d1 3
d1 3
d1 3
= 10(8) + 9. So, 89(mod 10)
== 1010- 89(mod 10)
=1 9
= 9.
Thus, the check digit is 1, and the ISBN 978 - 621 - 406 - 064- 1 is a valid
number.
Example 5: Determine the ISBN check digit for the book "Discrete
Mathematics" by Kenneth A. Ross and Charles R.B. Wright. The first 9 digits of
the ISBN are 0-13 - 215286-?.
Solution: Since the ISBN comprises 10 digits, we will apply the formula below
to find the check digit.
d1 0
= 11- (10d+ 2d+ 9d(mod
+ 8d + 7d
11)
1
2
9)
3
+ 6d 5 + 5d 6 + 4d 7 + 3d 8
4
Where d 1 = O, d2 = 1, d 3 = 3, d4 = 2, d 5 = 1, d 6 = 5, d 7 = 2, d 8 = 8, d 9 = 6
= 11- (10d + 9d(mod
+ 8d + 7d + 6d + 5d + 4d + 3d
11)
+ 2d
d 0 = 11 - [ 10(0) + 9(1) + 8(3) + 7(2) + 6(1) + 5(5) + 4(2) + 3(8)
+ 2(6) (mod 11)
=
11(O
+
9 + 24 + 14 + 6 + 25 + 8 + 24 + 12) (mod 11)
d 0
d 0 = 11 - 122(mod 11)
Evaluate 122(mod 11). 122 = 11(11) + 1. So, 122(mod 11) = 1.
d 0 = 11- 122(mod 11)
d 0 = 11- 1
d = 10
d1 0
1
2
4
3
5
9)
7
6
8
1
1
1
1
1
10
We cannot write 10 in the check digit since it is a two-digit number. Instead, we
write its Roman number equivalent, X. Thus, the check digit is X, and the ISBN
of the book is 0-13 - 215286 - X.
With all of these computations, we have learned how to find or verify the
check digit of some barcodes or identification numbers. There is actually a
shortcut in finding the check digit. However, we still use the check digit
formula. It is just presented in an easier way.
We will use the given examples above, then compare the results:
Example 1 : Determine the check digit for the new product release of ABCD Inc.
The first 11 digits are 7 - 87634- 73 - 214-?.
Solution: Note that UPC contains 12 digits. Using the formula in easier way, we
will multiply the odd positioned digits by 3 and then add all the terms.
3
,....,,
3
,.....,
3
,....,,
3
,....,,
3
,.....,
3
,....,,
7876347 3 214
21 + 8 + 21 + 6 + 9 + 4 + 21 + 3 + 6 + 1 + 12
=
= 112
Now, divide 112 by 10 or evaluate 112(mod 10) to get the remainder. So, 112 +
10 = 11 r. 2. The remainder is 2. Thus, 112(mod 10) 2
d 12
Vision:
Mission:
= 10- 112(mod 10)
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1 01
= 10- 2
d1 2 = 8
d1 2
Hence, the check digit is 8, and the UPC of the new product is 7 - 87634- 73 214- 8.
Example 2: Determine the check digit of EAN-13 number 9 - 310779 30000-?.
Solution: Note that EAN-13 contains 13 digits. Using the formula in easier way
again. This time, multiply the even positioned digits by 3, then add all the terms.
3
,-,,
,.,.,3 ,.,.,3 ,.,.,3
3
,-,,
,.,.,3
931077930000
9 + 9 + 1 + 0 + 7 + 21 + 9 + 9 + 0 + 0 + 0 + 0 = 65
65(mod 10)
d1 3
= 5 since 65 = 10(6) + 5
= 10- 65 (mod 10)
d = 10- 5
d =5
13
13
Hence, the check digit is 5, and the EAN-13 is 9 - 310779 - 30000 - 5.
Example 3: Evaluate whether 4404- 4507 - 7611 - 3863 is a valid credit card
number.
Solution: Since the Credit card is composed of 16 digits, we will multiply the
odd positioned digits by 2, then add all the terms using Luhn algorithm.
,...,2 ,...,2 ,...,2 ,...,2 ,...,2 ,...,2 ,...,2 ,...,2
440445 07 7 6 1 1 3 8 6
8 + 4 + 0 + 4 + 8 + 5 + 0 + 7 + 14 + 6 + 2 + 1 + 6 + 8 + 12
8 + 4 + 0 + 4 + 8 + 5 + 0 + 7 + 5 + 6 + 2 + 1 + 6 + 8 + 3 = 67
6 7 (mod 10)
d1 6
= 7 since 67 = 10(6) + 7
= 1010)
d = 10 - 7
d =3
6 7 (mod
16
16
Thus, the check digit is 3 and the credit card number 4404- 4507 - 7611 3863 is a valid number.
Example 4: Check whether ISBN 978 - 621 - 406 - 064- 1 of the book of
"General Mathematics for Senior High School" by Winston Sirug is valid.
Solution: Since the ISBN is composed of 13 digits, multiply the even positioned
digits by 3, then add all the terms.
,.,.,3 ,.,.,3 ,.,.,3 ,....,3 ,.,.,3
3
,-,,
97 8 6 2 1 40 60 64
9 + 21 + 8 + 18 + 2 + 3 + 4 + 0 + 6 + 0 + 6 + 12 = 89
89(mod 10)
d1 3
= 9 since 89 = 10(8) + 9
= 10- 89(mod 10)
d = 10 - 9
13
Vision:
Mission:
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Development of a highly competitive human resource, cutting-edge scientific knowledge
and innovative technologies for sustainable communities and environment.
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1 02
d1 3
=1
Thus, the check digit is 1 and the ISBN 978 - 62 1 - 406 - 064 - 1 is a valid
number.
Example 5: Determine the ISBN check digit for the book "Discrete
Mathematics" by Kenneth A. Ross and Charles R.B. Wright. The first 9 digits of
the ISBN are 0 - 13 - 2 15286-?.
=
Solution: Since the ISBN is composed of 10 digits, notice the formula d 10
11 - (10d 1 + 9d 2 + 8d 3 + 7 d4 + 6d 5 + 5 d 6 + 4d 7 + 3d 8 + 2d 9 ) (mod 11) . We
will multiply the 1st digit by 10, 2 nd digit by 9, and so on, then add all the terms.
10 9 8 7 6 5 4 3 2
,..., ,..., ,..., ,..., ,...,
0+9
,-.
,..., ,..., ,...,
0 1 3 2 1 5 2 8 6
+ 24 + 14 + 6 + 2 5 + 8 + 24 + 12 = 122
122 (mod 11) 1 since 122 = 11(11) + 1
d 10
=
= 11 - 122 (mod 11)
= 11 - 1
d = 10
=X
d 10
10
d 10
Thus, the check digit is X, and the ISBN of the book is 0- 13 - 2 15286 - X.
Comparing the results, we got the same answers with that of the long process
presented using the formulas.
Now, let us have another application of modular arithmetic, which is cryptology.
Cryptology
If an individual wanted to secretly store or communicate messages,
they make use of cryptology. It involves a technique to obscure a message so
outsiders cannot read the message. The use of codes is ancient. Julius Caesar
used this simple encryption scheme by jumbling the alphabet letters according
to some rule. (Sirug, 20 1 8)
Why would anybody use a cryptosystem ? There are several possibilities: (Tiborg,
1 999)
1. Confidentiality. When transmitting data, one does not want an
eavesdropper to understand the contents of the sent messages. The
same is true for stored data that should be protected against
unauthorized access, for instance, by hackers.
2. Authentication. This property is the equivalent of a signature. The
receiver of a message wants proof that a message comes from a
particular party and not from somebody else (even if the original party
later wants to deny it).
3. Integrity. This means that the receiver of specific data has evidence
that no changes have been made by a third party.
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Mission:
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1 03
First, we define some terminology.
1. Encryption-is the process of using an algorithm to transform
information into a format that cannot be read.
2. Decryption-is the process of using an algorithm to transform
encrypted information back into a readable format.
3. P/aintext-is the message that you wish to put into a secret form.
4. Cipher-the method for altering the plaintext.
5. Encipher-changing from plaintext to cipher text.
6. Cipher text-is the secret version of the plain text.
7. Decipher-changing from cipher text to plain text.
8. Key-information that will allow someone to encipher the plaintext and
also decipher the cipher text.
One of the examples of cryptology where modular arithmetic is used is the
simple substitution cipher by Julius Caesar.
Simple Substitution Cipher
A substitution cipher replaces each
letter in the message with a different letter,
following some established mapping or key.
One good example of a substitution cipher is
the Caesar Cipher (or shift cipher). In this cipher,
each letter is replaced with a letter of some
fixed number of positions later in the alphabet.
(Sirug, 20 1 8)
For example, if we use a shift of 6, then
the letter A would be replaced with letter g, the
Figure 6 Caesar Cipher Wheel
letter 6 positions later in the alphabet. In fact, a
simple substitution cipher may be viewed as a https://www.walmart.com/ip/Cla
{A, B, C, . . . , X, Y, Z} ➔
rule
or
function
ssic-Caesar-Cipher-Medallion­
{a, b, c, ... , x, y, z} assigning each plain text in Silver Decoder-Ring/204507760
the domain a different cipher letter in the
range. A function with this property is said to be one-to-one. (Sirug, 20 1 8)
Figure 7 shows the plain text (original text) to ciphertext. The plain text
is the big letters (outer circle) while the ciphertext is the small circle (the inner
circle).
Figure 7 A Cipher Wheel with shift 6
Retrieved from: https://caesar-cipher-disk.en.aptoide.com/app
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Example 1 : Use the Caesar Cipher with shift 6 to encrypt the message: SEND
THE SECURITY.
Solution: Using the mapping below to encrypt the message, SEND THE
SECURITY.
Plain text Letter (Original)
A B C D E F G H I J K L M N O P Q R s T u V
g h
j k I m n o p q r s t U V W X y z a b
Cipher text Letter
wx
C
y z
d e f
S gets replaced by y, E by k, N by t, D by j, and so on.
t:!�t�t�l t:t�l�I t:t�!�1� f=t�!!l!1
Giving the encrypted message: yktj znk ykiaxoze
Example 2: Decrypt the message amm gwc vmfb emms if it was encrypted
using a shift cipher with shift of 8.
Solution: We start by writing out the character mapping by shifting the
alphabet, with A mapping i, eight characters later in the alphabet.
A
Plain text Letter (Original)
B C D E F G H I J K L M N O P Q R S T U V W X Y Z
j
k l
m n o p q r s t u v w x y z a b c d e f g h
Cipher text Letter
We now work backward to decrypt the message. The first letter a is
mapped to S, so S is the first character of the original message, m is mapped
to E. If we continue, the decrypted message is SEE YOU NEXT WEEK.
[
a
S
f
E E
Im m
I
Iw J c
Y O U
g
1
1
m I f fb
J
N E X T J
v
I
m [m f5
l
W E E K
e
1
There is another way (a shortcut) to encrypt or decrypt a message
without using a table. Instead, we will use a Caesar cipher wheel to encrypt or
decrypt a message with respect to a given key or shift. Take note, we will
actually use modular arithmetic using the Caesar cipher wheel more efficiently.
Example 3: Encrypt the message using Caesar cipher disk.
a. Message: CANCEL THE CONTRACT, with shift 3
b. Message: ATTACK AT NIGHT, with shift 13
c. Message: TAKE CONTROL, with shift 4
How to Encrypt with the Cipher Wheel?
Spin the inner wheel around until the key to use will line up the letter A of
the outer circle (with a dot below it). Then, match up the outer letters with the
inner letters (as shown in Figure 8 on the next page). To encrypt using the
cipher wheel, take note, start from the OUTER circle to the INNER circle.
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a. Message: CANCEL THE CONTRACT, with shift 3
Align letter A to D (encircled), since
D has the number 3. Then match
up the outer letters to the inner
Hence, the encrypted message is:
fdqfho wkh frqwudfw
Figure 8 Caesar cipher wheel with shift 3
b. Message: ATTACK AT NIG HT, with shift 13
Align letter A to N, since N has the
number 13. Then match up the
outer letters to the inner letters.
l�l�l�l�l�l�I
[Kill
�
l�l�l�l�l�I
Hence, the encrypted message is:
nggnpx ng avtug
Figure 9 Caesar cipher wheel with shift 13
c. Message: TAKE CONTROL, with shift 4
Align letter A to E, since E has the
number 4. Then match up the outer
letters to the inner letters.
Hence, the encrypted message is:
xeoi gsrxvsp
Figure 10 Caesar cipher wheel with shift 4
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Example 4: Decrypt the message using Caesar cipher disk.
a. Message: svclsf, with shift 7
b. Message: toxgzxkl, with shift 19
c. Message: lwpi xh ndjg upkdgxit bdkxt? with shift of 15
How to decrypt with the Cipher Wheel?
Line up the letter A on the outer circle (the one with the dot below it) over
the letter on the inner circle with the key (as shown in Figure 11). To decrypt,
go from the INNER circle to the OUTER circle.
a. Message: svclsf, with shift 7
Align letter A to H, since H has the
number 7. Then match up the inners
letters to the outer letters.
Hence, the decrypted message is:
LOVELY
b. Message: toxgzxkl, with shift 19
Figure 1 1 Caesar cipher wheel with shift 7
Align letter A to T, since T has the
number 19. Then match up the
inners letters to the outer letters.
Hence, the decrypted message is:
AVENGERS
Figure 12 Caesar cipher wheel with shift 1 9
c. Message: lwpi xh ndjg upkdgxit bdkxt? with shift of 15
Align letter A to P, since P has the
number 15. Then match up the
inners letters to the outer letters.
X t
E
Hence, the decrypted message is:
WHAT IS YOUR FAVORITE MOVIE?
Vision:
Mission:
Figure 13 Caesar cipher wheel with shift 1 5
A globally competitive university for science, technology, and environmental conservation.
Development of a highly competitive human resource, cutting-edge scientific knowledge
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1 07
-
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Figure 15 The inner circle of the cipher wheel cutout
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Mission:
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Development of a highly competitive human resource, cutting-edge scientific knowledge
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Vision:
Mission:
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Development of a highly competitive human resource, cutting-edge scientific knowledge
and innovative technologies for sustainable communities and environment.
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For instructional purposes only • 1 st Semester SY 2020-2021
Learning Tasks/Activities
Please refer to the Course Information (on Course Policies) on how to present
your answers.
A. Verify whether each congruence is true or false. Write TRUE if true;
otherwise, FALSE. Show your solution.
____1. 2 5
____2. 54
____3 . 3 3
== 11(mod
8)
4(mod 10)
= 15 (mod 9)
____4. 45
____ 5. 67
____ 6. 68
= 5 6 (mod 2 3)
98 (mod 31)
34)
== 170(mod
B. Find the least residue of the following using division algorithm. Show
your solution.
= ____
47 (mod 5) = ---69 (mod 19) = ___
= ___
-24(mod 3 3) = ___
42 3 (mod 62) = ___
1 . 88 (mod 8)
4. -2 (mod 2 1)
3.
6.
2.
5.
C. Use modular arithmetic to determine each of the following.
1. Give three memorable dates (birthdays, etc.) that happened in your life
(indicate the event). Then, use the Zeller's formula to find the day of the
week of these dates. Show your solutions.
Zeller's formula: W
= D + [2.6m - 0.2] - 2 C + Y + [�] + [�] (mod 7) .
2. Look for items at your home with barcodes for UPC and EAN-13.
Indicate the items you will use. Verify if the barcodes are valid. (You
may use the shortcut in presenting the solutions.)
a. Two items for UPC (e.g., powdered milk (Birch Tree), canned goods
(Lucky 7 Carne Norte), etc.)
b. Two items for EAN-13 (e.g., alcohol, perfume, etc.)
3. Give two books with ISBN-13 or ISBN-10. Identify if the ISBN for each
book is valid. Indicate the name of the book. (You may use the shortcut
in presenting the solutions.)
If books are not available at home, you may use the following examples
instead. Find the check digit for each.
a. 978 - 843 - 627 - 836-?
b. 0 - 3 990 - 2827-?
4. Determine if each of the following credit cards is valid.
a.
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Mission:
b.
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110
D. Decipher the following messages using Caesar Cipher.
1. Message: pscepxc (shift of 4)
Plaintext: ____________
2. Message: i lwzijtm sqbbg (shift of 8)
Plaintext: ____________
3. Message: ed u iytut bycyj (shift of 1 6)
Plaintext: ___________________
4. Message: cpgoqpg (shift of 2)
Plaintext: ___________________
5. Message: ldgrthitg hwxgt (shift of 1 5)
Plaintext: ___________________
6. Message: bg beuvabynel atbybtvfg (shift of 1 3)
Plaintext: ___________________
7. Message: omdtl n m ntksqzl h bq n rbnohbrhkh b n u n kbzm n b n m h n rh r
(shift o f 25)
Plaintext: ____________________
Assessment
TEST I. MULTIPLE CHOICE. Write only the CAPITAL LETTERS of your answer
and SHOW YOUR SOLUTIONS (for items that need solution) on a separate
sheet of paper(s). NO SOLUTION MEANS WRONG. If the answer is not on the
choices, just write your final answer. (30 points)
1. Evaluate (134 + 65) (mod 15).
ABCD. 3
ACBD. 4
ABDC. 5
2. What is the multiplicative inverse of 7 in mod 11?
ABCD. 6
ACBD. 7
ABDC. 8
3. Evaluate 156 + 17 (mod 20).
ABCD. 0
ACBD. 1
ABDC. 5
4. Find the least residue of (35) (30) 2 (mod 33).
ABCD. 31
ACBD. 27
ABDC. 18
ACDB. 6
ACDB. 9
ACDB. 8
ACDB. 15
5. If it is now May 2018, what will be the month and year 500 months from
now?
ABDC. December 2019
ABCD. January 2019
ACDB. December 2059
ACBD. January 2060
6. If yesterday was Saturday, what will be the day of the week 999 days
from now?
ABCD. Friday
ABDC. Monday
ACBD. Wednesday
ACDB. Thursday
7. What is the least residue of -49(mod 11)?
ABDC. 4
ACBD. 5
ABCD. 6
ACDB. 3
8. Determine the check digit for ISBN 0-06-045471-?
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For instructional purposes only • 1 st Semester SY 2020-2021
ABCD. 4
ACBD. 7
AB DC. 3
ACDB. 5
9. What is the least residue of - 2 2 5 (mod 1 3)?
ABCD. 6
ACBD. 7
AB DC. 8
ACDB. 9
10. If it is now Sunday, what day of the week was it 100 days ago?
ABDC. Friday
ABCD. Sunday
ACDB. Thursday
ACBD. Monday
11. Disregarding a.m. or p.m., if it is now 4:00 o'clock, what time will it be
76 hours from now?
ABDC. 6: 00 o'clock
ABCD. 2 : 00 o'clock
ACBD. 4: 00 o'clock
ACDB. 8: 00 o'clock
66
66
is divided by 7?
12. What is the remainder when the number 6 6
ABCD. 1
ACBD. 3
ACDB. 7
AB DC. 6
13. Decipher the choices to answer this math riddle:
"When I take five and add six, I get eleven, but when I take six and
add seven, I get one. What am I?" Use shift of 25
ABCD. z mtladq
ABDC. z bknbj
ACBD. z rnmf
ACDB. z khd
14. What are the last two digits of the number 99 1 8 9 given that 99 1 =
99, 99 2 = 9,801, 99 3 = 9 70, 299 and so on?
ABCD. 0 1
ACBD. 8 9
ACDB. 0 9
AB DC. 9 9
15. Decipher the following math riddle and answer it (key = 10):
"grkd ny wkdrowkdsmc dokmrobc vsuo dy okd?"
ABCD. Pie
ACBD. Cake
AB DC. Fruit
ACDB. Ice cream
Test II. Use modular arithmetic to answer the following items.
1. Determine the check digit of the following:
a. U PC: 8-94085-74904-?
b. ISBN: 9 - 5 0 1 1 0 1 - 5 3 000-?
C. Credit Card: 8030-3892-2992-827?
d. ISBN: 978-006-447-373-?
2. What day of the week on which January 1, 2030 will fall?
3. Valentine's Day (February 14) fell on a Tuesday in 2006. On what day of
the week will valentine's Day fall in 2025?
Test Ill. Decrypt the following jumbled words to form a question then answer it
in not less than 40 words. (Use SHIFT OF 1 4)
mci igs kwzz vck wb hvwg aohvsaohwqg hwas ct qfwgwg
Instructions on how to submit student output
Refer to the course policies and course content plan.
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Mission:
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Development of a highly competitive human resource, cutting-edge scientific knowledge
and innovative technologies for sustainable communities and environment.
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