Measures of Central
Tendencies
What Is This Module About?
Have you encountered a situation wherein you have to get the average of a given set of
numbers or a given set of data?
For instance, you have a sari sari store and you want to know how much is your average
profit per month. How then will you compute for the average profit? The average or the mean for
a set of given numbers or data is just one of the three measures of central tendency that you will
learn about in this module.
While the measures of central tendency give information about the average or the center of a
given distribution of data, the measures of dispersion are used to analyze the variations of the data
in the given set of distribution. In this module, you will learn further about their differences and
how they are applied in problem solving.
This module is divided into two lessons. These are:
Lesson 1 – Measures of Central Tendency
Lesson 2 – Measures of Dispersion
Wait!
Before studying this module, be sure that you have completed the module on Graphs and
Frequency Distribution.
What Will You Learn From This Module?
After studying this module, you should be able to:
♦
describe the differences between the mean, median and mode;
♦
use the mean, median and mode to analyze and interpret data to solve problems in
daily life;
♦
describe the differences between the range, mean deviation and the standard
deviation; and
♦
use the range, mean deviation and standard deviation to solve problems in daily life.
1
Let’s See What You Already Know
A.
B.
The following is the frequency distribution of 50 scores in a Science test.
Interval of scores
Frequency
95 – 99
2
90 – 94
6
85 – 89
6
80 – 84
8
75 – 79
7
70 – 74
6
65 – 69
4
60 – 64
5
55 – 59
4
50 – 54
2
1.
Find the mean.
2.
Find the median.
3.
Find the mode.
The following are the test scores of Jeremy and Theresa in four of their subjects.
Jeremy
Theresa
Physics
Chemistry
Biology
78
79
61
65
80
91
Geology
89
85
Complete the table below.
R an g e
V arian ce
Jerem y
(1)
(2)
(3)
T heresa
(4)
(5)
(6)
S tan d a
D eviati
Well, how was it? Do you think you fared well? Compare your answers with those in the
Answer Key on page 37 to find out.
If all your answers are correct, very good! This shows that you already know much about
the topics in this module. You may still study the module to review what you already know. Who
knows, you might learn a few more new things as well.
If you got a low score, don’t feel bad. This means that this module is for you. It will help you
understand some important concepts that you can be apply in your daily life. If you study this
module carefully, you would learn the answers to all the items in the test and a lot more! Are you
ready?
You may now go to the next page to begin Lesson 1.
2
LESSON 1
Measures of Central Tendency
Measures of central tendency provide information about the averages or centers of a
distribution of data. It is the number that best represent a given set of numbers or what we call a
group of scores.
This lesson will introduce you to the three measures of central tendencies called the mean,
median and the mode; how they are computed and how they are used in everyday life.
After studying this lesson, you should be able to:
♦
describe the differences between the mean, median and mode;
♦
compute for the mean, median and mode of a given set of data; and
♦
use the mean, median and mode to analyze and interpret data to solve problems in
daily life.
In a fishing village, Mario, Gimo, Oscar and Caloy are comparing their catches for the day.
Mario has caught 7.6 kilograms of fish while Gimo has caught only 6 kilograms. Oscar has 10.2
kilograms of fish while Caloy has 11.5 kilograms. What is their average fish catch for the day?
3
The average fish catch can also be referred to as the mean fish catch. The mean is the
arithmetic average of a set of data or given numbers. In this case, the set of data is the number of
kilograms of fish each of the fishermen caught. These given numbers, or what is referred to as the
scores are added up. When we add all the kilograms of fish caught for the day, we have 35.3
kilograms of fish. Then we count the number of scores, and we get 4. When the sum of the
scores is divided by the number of scores, we get an average of 8.825 kilograms of fish. This
means that the mean of the group of data is 8.825.
Let’s Learn
The mean, also known as the arithmetic mean, is the average of a set of data, or scores.
This is the number or the score that best represents a group of scores. It is one of the measures of
central tendency used when you have interval or ratio scores. Do you still remember the terms
“interval” and “ratio”? These are 2 of the 4 scales of measurement. If you want to review them,
you may read the module entitled “Graphs and Frequency Distributions. This topic is discussed in
lesson one of that module.
The computation for the mean may be depicted using symbols. Let us use x to represent a
single score in a given data. In the example above, Mario has caught 7.6 kilograms of fish. This
score would be represented as: x = 7.6. This score can also be referred to as a raw score if it is
untreated in any way.
The sum of a set of scores is represented by the capital Greek letter sigma (Σ). This
symbols represents the summation of a set of scores. Whenever this symbol appears, it means
that whatever follows it must be summed or added up. For example, the notation Σx indicates
that all the values or scores represented by x should be added up.
The symbol N is also useful in statistical computations. This represents the number of scores
in the given set. In the previous example, the number of scores given is 4. Therefore, N = 4.
Combining these three symbols (x, Σ, N), we have the formula to be used in calculating for
the mean of a given set of data. If we let the mean be x, the formula is:
x=
Σx
N
where x = the mean
Σ =" the sum of"
N = number of scores
4
Let’s Try This
Read the following problems carefully. Solve for the mean.
EXAMPLE 1
A company makes electrical circuits. The quality control manager decides to
find out the average number of defective circuits made per week. He
examines each electrical circuit produced per day, and records the number
of electrical circuits with defects. He collected data for 7 days. These data
are listed below:
Day 1
25 defective circuits
Day 2
22 defective circuits
Day 3
30 defective circuits
Day 4
32 defective circuits
Day 5
31 defective circuits
Day 6
27 defective circuits
Day 7
16 defective circuits
Compute for the mean failure rate or the mean number of defective electrical
circuits. Let’s do it together.
In order to get the mean, we use the formula
x=
Σx
N
In the formula, Σx indicates that we sum up all the given scores. Thus,
25 + 22 + 30 + 32 + 31 + 27 + 16 = 183
We then divide the sum with the number of scores N. Here, N = 7. Thus,
x=
Σx
183
= 26.14 = 26
=
7
N
Therefore, the mean failure rate is 26 defective electrical circuits.
Try to solve the following on your own.
5
EXAMPLE 2
You are an employee of a certain company in your province. In your
company, there are 20 employees in all. You would want to determine the
average monthly salary that the company gives. Listed below are the salaries
of these 20 employees (including your own salary) in thousands of pesos.
Employee 1
P 4,000
Employee 11
P 3,000
Employee 2
P 2,500
Employee 12
P 3,000
Employee 3
P 5,200
Employee 13
P 3,000
Employee 4
P 2,800
Employee 14
P 4,200
Employee 5
P 1,200
Employee 15
P 2,100
Employee 6
P 2,100
Employee 16
P 2,100
Employee 7
P 2,100
Employee 17
P 1,800
Employee 8
P 2,100
Employee 18
P 1,500
Employee 9
P 2,000
Employee 19
P 2,100
Employee 10
P 3,200
Employee 20
P 4,000
Compare your answers with those found in the Answer Key on page 37.
Let’s Learn
When scores are in a grouped data or are presented in a frequency distribution, such as in
the table below, we may need to use a more convenient formula to compute for the mean.
The following table shows the test scores in a class composed of 40 students.
Table 1
Scores in a test
Interval of scores
Frequency (f )
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
15 – 19
10 – 14
1
2
5
10
9
6
4
3
In the case of a frequency distribution table, we take note of how frequent an interval
appears in the data. This is represented by f. We also take note of the midpoint in an interval.
This is represented by xc. For example, the midpoint xc of the interval 45 – 49 is 47. For the
interval 40 – 44, what do you think is the midpoint? _____. In order to get the mean using the
new formula, the midpoint is multiplied by the frequency. Let us present the information above in a
new table.
6
Table 2
Scores in a test
Interval of
scores
45
40
35
30
25
20
15
10
–
–
–
–
–
–
–
–
Midpoint x c
Frequency f
47
42
37
32
27
22
17
12
1
2
5
10
9
6
4
__3__
N = 40
49
44
39
34
29
24
19
14
To compute the mean for this frequency distribution, we use the formula below:
x=
Σfx c
N
Using the data in Table 2, we now solve for the mean,
1115
40
= 27.875
x=
= 28
Therefore, the mean is 28.
Let’s Learn
When data is presented in a frequency distribution, we compute for the mean by using the
following formula:
Σfx c
N
where x = the mean
x=
Σ = “the sum of”
x c = midpoint of an interval
f = frequency of cases in an interval
N = number of scores
7
Given a frequency distribution, we follow the following step-by-step procedure:
STEP 1
Determine the number of scores N.
STEP 2
Determine the midpoint xc for each class interval.
STEP 3
Determine the fxc for each class interval.
STEP 4
Determine the sum Σfxc.
STEP 5
Substitute the values in the formula.
Let’s Try This
EXAMPLE 3
Given the following frequency distribution table of the ages of 30
participants in a workshop training, compute for the mean.
Interval of ages
Frequency
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
2
5
4
3
12
4
Let us use the step-by-step procedure. Try to fill in the blanks.
STEP 1
Determine the number of scores N.
The total number of scores is 30 since there are 30 participants in the
workshop training. Therefore, N = 30.
STEP 2
Determine the midpoint xc for each class interval.
Interval of ages
Midpoint Xc
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
47
42
37
___
___
___
8
Frequ
1
___
N=
STEP 3
Determine the fxc for each class interval.
Interval of ages
Midpoint xc
Frequency f
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
47
42
37
32
27
22
2
5
4
3
12
__4__
N = 30
STEP 4
Determine the sum Σfxc.
Interval of ages
Midpoint xc
Frequency f
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
47
42
37
32
27
22
2
5
4
3
12
__4__
N = 30
STEP 5
Substitute the values in the formula.
Σfx c
N
960
=
30
= 32
x=
Therefore, the mean is 32.
EXAMPLE 4
Given the following frequency distribution table of 40 scores in a Science
test, compute for the mean.
Interval of scores
Frequency
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
1
4
5
7
6
5
3
5
2
2
9
STEP 1
Determine the number of scores N.
N = _______
STEP 2
STEP 3
STEP 4
Determine the midpoint xc for each class interval.
Interval of ages
Midpoint xc
Frequency f
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
____
____
____
____
____
____
____
____
____
____
1
4
5
7
6
5
3
5
2
2
N = ____
Determine the fxc for each class interval.
Interval of ages
Midpoint xc
Frequency f
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
____
____
____
____
____
____
____
____
____
____
1
4
5
7
6
5
3
5
2
2
N = ____
Interval of ages
Midpoint xc
Frequency f
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
____
____
____
____
____
____
____
____
____
____
1
4
5
7
6
5
3
5
2
2
N = ____
Determine the sum Sfxc.
10
Σ
STEP 5
Substitute the values in the formula.
Σfx c
N
= ______
= ______
x=
Therefore, the mean is_______.
Compare your answers with those found in the Answer Key on pages 37–39.
Let’s Study and Analyze
Suppose you are watching a basketball game. The following are the final scores of the
players that you took note of:
Player
David
Limpot
Ong
Meneses
Crisano
Score/Number or Points
45
10
9
8
8
11
If we are to look for the middle score but used the method for computing the mean, the
score that we will get will be higher than the true value we are looking for. The computed middle
score will be higher because one score is too high compared to the other scores. The scores of
Limpot, Ong, Meneses and Crisano are close to each other. However, there is one value that is
too high, compared to the other scores – the score of David. We say that the score of David is an
extreme score or an outlier.
So instead of computing for the mean, just look for the middle score. This score is called the
median. The median is the midpoint in a set of scores.
When looking for the median, arrange the scores from the highest to the lowest score. This
arrangement of scores is called an array. Then, look for the score that is exactly in the middle of
the array such as the scores above and below it equal number. In the example, there are five
scores, and the middle number is 9. There are 2 scores below it and also 2 scores above it.
Therefore, the score 9 is the median. This is quite easy to determine because the number of
scores N is an odd number.
For a set of scores with an even number of scores N, the median is computed by getting the
average of the two middle scores. For example, in the set of scores below,
5 9 12 13 15 21 22 30
the middle scores are 13 and 15. The median is computed by getting the average of these
two scores. The average of 13 and 15 is 14. Therefore, we say that the median is 14. Note that
there is no 14 in the given set of data, which means that the median is not a score but a point that
divides a distribution in the middle.
Let’s Learn
The median is the score or value that divides a given data set or a distribution into two
equal halves wherein 50% of the scores are above it, while 50% are below it.
The median is not greatly affected by extremely high or low scores. If one or more scores in
the data are too high or too low compared to most of the other scores, then the median is a better
measure of central tendency than the mean. These extreme scores are called outliers.
The median is oftentimes denoted as “Md”. There are two different cases considered in
finding the median of a given set of data.
CASE 1
The number of scores N is odd.
In the array of scores, the median is the middle score.
CASE 2
The number of scores N is even.
In the array of scores, the median is the average of the two middle scores.
12
Let’s Try This
Read the following problems carefully. Find the median of the given set of data
EXAMPLE 1
The following are the amount of pastilyas Ria sold for 5 days.
Day 1
P 1,620
Day 2
P 1,560
Day 3
P 2,220
Day 4
P 5,560
Day 5
P 2,300
It is important that the given data is in the form of an array. In the example,
th scores are not in an array, so we still have to arrange it from the highest to
the lowest value. Rearranging the given data, we’ll have
Day 4
P 5,560
Day 5
P 2,300
Day 3
P 2,220
Day 1
P 1,620
Day 2
P 1,560
Which value may be considered an outlier in Ria’s data? Notice that the
amount she got in Day 4 is much higher than in the other days. Therefore,
the outlier in the data is P 5,560.
How do we solve for the median? Notice that the number of scores given is
5, an odd value. Therefore, Case 1 applies to this example. To get the
median, we simply look for the middle score.
The middle score in the given data is the amount corresponding to Day 3. It
is so because there are 2 scores above and below it. From the definition,
the median divides a distribution of data into two equal parts.
Therefore, the median is P 2,220.
13
EXAMPLE 2
The following are the monthly salaries of 10 employees from a small
company.
Employee 1
Employee 2
Employee 3
Employee 4
Employee 5
Employee 6
Employee 7
Employee 8
Employee 9
Employee 10
P 15,400
P 3,500
P 3,400
P 3,300
P 3,300
P 3,100
P 2,700
P 2,600
P 2,500
P 2,200
Check first if the data is already presented in an array.
Which salary may be considered an outlier in the data? _________.
Take note of the number of scores given. Which case applies to this
example? ________.
Case 2 applies to data with an even number of scores. It states that the
median is the average of the two middle scores. What are the middle scores
in this example? ______ and P 3,100. Notice that there are 4 values both
above and below the middle scores. We can now determine the median by
getting the average of the middle scores.
Md = (_____ + P 3,100) ÷ 2
= ___________
Therefore, the median is _________.
Try to solve the following example on your own.
EXAMPLE 3
The following are the number of boxes of soap sold by a certain company in
the past year.
January
February
March
April
May
June
July
August
September
October
November
December
3,214
2,459
2,330
2,422
2,354
2,575
2,132
2,079
1,991
2,560
2,438
1,857
14
Compute for the median.
Compare your answers with those found in the Answer Key on page 39.
Let’s Learn
Now you know how to find the median of a given set of data. But what if you are given data
presented in a frequency distribution table? How will you solve for the median in this case?
Let’s look at the following example.
The table below shows the test scores in a class composed of 30 students.
Table 3
Scores in a test
Interval of scores
Midpoint xc
Frequency f
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
____
____
____
____
____
____
____
____
____
____
1
4
5
7
6
5
3
5
2
2
N = ____
Let us solve for the median using a step-by-step procedure.
STEP 1
Determine the number of cases N.
The total number of cases is 30 since there are 30 students in the class.
Therefore, N = 30.
Note that we use the term “case” because we are referring to how frequent
a “score” is used.
STEP 2
Solve for N/2 or half the number of cases in the distribution.
The number of cases N is 30, therefore, N/2 = 30/2 = 15.
15
STEP 3
Count up the number of cases until the interval containing the N/2 case is
reached.
From the bottom, we count up until we reach the N/2 case, which is 15. So,
2 + 3 + 6 + 8 = 19
This is 4 cases more than 15 and falls in the interval 25 – 29.
STEP 4
Determine how many cases were needed out of all the cases in the interval
to reach N/2. Divide this by the number of cases in the interval.
Out of the 8 cases in the interval 25 – 29, only 4 are used to reach 15.
Therefore, only 4/8 or ½ of the interval was needed.
STEP 5
Multiply this by the size of the interval.
The size of the class interval 25 – 29 is 5.
STEP 6
Add this to the lower limit of the interval containing the median.
The interval containing the median is 25 – 29. The lowest number in this
interval 25, but instead, we will be using 24.5 since this is really the limit
before the next interval.
If we do Steps 4 to 6 simultaneously, we will have:
4/8 × 5 + 24.5 = 27
Therefore, we say that 27 is the median.
Let’s Learn
When data is presented in a frequency distribution, we compute for the median by using the
following step-by-step procedure:
STEP 1
Determine the number of cases N. Get the sum of the given frequencies.
STEP 2
Solve for N/2 or half the number of cases in the distribution.
STEP 3
Count up the number of cases until the interval containing the N/2 case is
reached.
STEP 4
Determine how many cases were needed out of all the cases in the interval
to reach N/2. Divide this by the number of cases in the interval.
STEP 5
Multiply this by the size of the interval.
STEP 6
Add this to the lower limit of the interval containing the median.
16
Let’s Try This
EXAMPLE 4
The table shows the daily earnings of employees in a certain barangay.
Solve for the median using the step-by-step procedure.
STEP 1
Interval of scores
(earnings in P)
Frequency
(number of employees)
235 – 239
230 – 234
225 – 229
220 – 224
215 – 219
210 – 214
205 – 209
200 – 204
3
5
11
8
4
5
3
1
Determine the number of cases N.
N = ______
STEP 2
Solve for N/2 or half the number of cases in the distribution.
N/2 = _____ /2 = _____.
STEP 3
Count up the number of cases until the interval containing the N/2 case is
reached.
From the bottom, we count up until we reach the N/2 case, which is ____.
So,
___+___+___+___+___+___ = ______
This is _____ case/s more than ______ and falls in the interval
__________.
STEP 4
Determine how many cases were needed out of all the cases in the interval
to reach N/2. Divide this by the number of cases in the interval.
Out of the _____ cases in the interval 220 – 224, only ______ are used to
reach 20. Therefore, only ___(4)___ of the interval was needed.
STEP 5
Multiply this by the size of the interval.
The size of the class interval 220 – 224 is ___(5)__.
STEP 6
Add this to the lower limit of the interval containing the median.
The lower limit to be used = __(6)___.
If we do Steps 4 to 6 simultaneously, we will have:
__(4)___ × __(5)___ + ___(6)___ = _________
Therefore, we say that _______ is the median.
17
By now you should be able to solve for the median without having to write
down the step-by-step procedure. Try to do the following example on your
own.
EXAMPLE 5
The table shows the age of the students in a dance class of 35. Find the
median height.
Interval of scores
(ages in years)
Frequency
(number of students)
33 – 35
30 – 32
27 – 29
24 – 26
21 – 23
18 – 20
15 – 17
1
1
3
3
6
12
9
Compare your answers with those found in the Answer Key on page 40.
Let’s Study and Analyze
Suppose your family wants to open up a sari-sari store and you want to know which brand
of bath soap sells most among the families in your barangay in order to help you decide which
brand you should buy. You went over and collected data from one of the bigger sari-sari stores in
your barangay by asking the storeowner the number of bars of different bath soaps sold daily. He
got the following data on the next page:
18
Brand of Bath Soap
Average Number of
_________________________Bars Sold per Day
Cascade
Buoy
Satin
Palms
20
10
15
32
The brand of bath soap with the most number of sales is the Palms bath soap, with an
average of 32 bars sold for a day. The Palms bath soap is called the mode or modal category of
the data collected.
Let’s Learn
The mode is the score or category that occurs with the highest frequency. By the word
“category”, we simply mean a brand, a name or a group of scores. This means that among all the
categories in your data, this score or category is the one which occurred the most number of
times.
The modal category of the data refers to the most frequently occurring cardinal data while
mode refers to the most frequently occurring nominal data.
In the case of data presented in a frequency distribution table, the mode is the midpoint of
the class interval with the highest frequency of occurrence.
Let’s Try This
EXAMPLE 1
A survey of favorite books among teens is given below. Find the modal
book.
Title of Book
Frequency
Heather Potter
Sweet Dale
Mushroom Soup
Teen Blood
EXAMPLE 2
54
37
42
33
Find the mode for the following set of scores.
Interval of scores
Midpoint
Frequenc
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
15 – 19
10 – 14
37
42
37
32
27
22
17
12
3
1
6
12
9
5
2
2
Compare your answers with those found in the Answer Key on page 40.
19
Let’s See What You Have Learned
A.
Solve what is asked for.
1.
The expenses of a household for four weeks are as follows:
First week
Second week
Third week
Fourth week
–
–
–
–
P 1,900
P 1,800
P 2,100
P 2,100
What is the weekly mean expenses of the household?
2.
The duration in minutes of telephone calls made in a certain pay phone for one
day are as follows:
8, 9, 20, 4, 29, 15, 2, 4, 3, 12, 10
Find the median.
3.
A survey of favorite TV channels in a class gives the following data:
Channel Z
Cinema TV
Cartoon Channel
Sports Stars
Nature Show
–
–
–
–
–
10
9
13
8
5
Find the modal program.
20
B.
The following is the frequency distribution of 30 scores in a test.
Interval of scores
Frequency
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
1
3
4
6
5
4
2
3
1
1
a.
Find the mean.
b.
Find the median.
c.
Find the mode.
Compare your answers with those found in the Answer Key on page 41.
Did you get all the answers right? If yes, congratulations! You really understood the
concepts discussed in this lesson. If you did not get all the correct answers, just go over the parts
you did not understand very well, then answer the exercises again.
21
Let’s Remember
The mean is the average of a set of scores. It is computed by adding all the scores in your
data set and dividing the sum by the number of scores. In formula form,
x=
Σx
N
where x = the mean
Σ =" the sum of"
N = number of scores
When data is presented in a frequency distribution, we compute for the mean by using the
following formula:
Σfxc
N
where x = the mean
x=
∑ = “the sumof”
x c = midpoint of an int erval
f = frequencyof casesin an int erval
N = number ofscores
♦
To compute for the mean of a frequency distribution, we have the following steps:
STEP 1
Determine the number of scores N.
STEP 2
Determine the midpoint xc for each class interval.
STEP 3
Determine the fxc for each class interval.
STEP 4
Determine the sum Σfxc.
STEP 5
Substitute the values in the formula.
♦
The median is the score or value that divides a given data set or a distribution into two
equal halves wherein 50% of the scores are above it, while 50% are below it.
♦
The median is oftentimes denoted as “Md”. There are two different cases considered
in finding the median of a given set of data.
CASE 1
The number of scores N is odd.
In the array of scores, the median is the middle score.
CASE 2
The number of scores N is even.
In the array of scores, the median is the average of the two middle scores.
22
♦
When data is presented in a frequency distribution, we compute for the median by
using the following step-by-step procedure:
STEP 1
Determine the number of cases N.
STEP 2
Solve for N/2 or half the number of cases in the distribution.
STEP 3
Count up the number of cases until the interval containing the N/2 case is
reached.
STEP 4
Determine how many cases were needed out of all the cases in the interval
to reach N/2. Divide this by the number of cases in the interval.
STEP 5
Multiply this by the size of the interval.
STEP 6
Add this to the lower limit of the interval containing the median.
♦
The mode is the score or category that occurs with the highest frequency. By the word
“category”, we simply mean a brand, a name or a group of scores. This means that
among all the categories in your data, this score or category is the one which occurred
the most number of times.
♦
The modal category of the data refers to the most frequently occurring cardinal data
while mode refers to the most frequently occurring nominal data.
♦
In the case of data presented in a frequency distribution table, the mode is the midpoint
of the class interval with the highest frequency of occurrence.
23
LESSON 2
Measures of Dispersion
How do we know how close or far apart are the scores in a given data? In Statistics, it is
measured by values called measures of dispersion or measures of variability. These measures
tell us how close to each other are the scores in the data. It should be remembered that the
smaller the value of these measures, the closer together are the scores in the data.
Measures of dispersion provide information about the variations in a given distribution of
data. They reflect the way in which data are distributed or dispersed in either directions from the
center of the distribution.
This lesson will introduce you to the three measures of dispersion called the range, mean
deviation and standard deviation; how they are computed and how they are used in everyday life.
After studying this lesson, you should be able to:
♦
describe the differences between the range, mean deviation and standard deviation;
♦
compute for the range, mean deviation and standard deviation of a given set of data;
♦
use the range, mean deviation and standard deviation to analyze and interpret data to
solve problems in daily life.
Let’s Study and Analyze
Hey, how did you 3 do in the
basketball tournament last year?
24
Oh, I did not get scores as high as
you did. I got 9 for the first game, 7
for the second game and 8 for the
last game.
Well, these are my scores during the 3
games of the basketball tournament: 15
for the first game, 10 for the second
game and 15 again for the last game.
How about you, Noel? What were your
scores?
I am not as lucky as you are but I
think I am the most consistent among
all of us. I got 7 in all 3 games.
You know what? I got 20 for the
first game, 21 for the second game
and 24 for the last game. I think I
got lucky.
List down below the scores that each person got for the 3 games of the basketball
tournament. Then answer the questions that follow.
Mario
____
____ ____
Noel
____
____ ____
Paul
____
____ ____
Oscar
____
____ ____
1.
Whose scores are closest to each other?
________
2.
Who is the highest scoring player?
________
3.
Who is the most consistent player?
________
Compare your answers with those found in the Answer Key on page 41.
25
Let’s Learn
The most basic measure of variability is the range. To compute for the range of a given
data, we find the highest and lowest scores in the data, and then find the difference between these
two values.
R = XH − XL
Where
R is the range;
XH is the highest score; and
XL is the lowest score.
Let us take as an example Mario’s scores in the 3 basketball games. Here are the steps in
computing for the range:
STEP 1
Look for the highest score and the lowest score.
In the example above, the highest score is 15 and the lowest score is 10.
STEP 2
Subtract the lowest score from the highest score.
R = 15 − 10 = 5
Therefore, the range Mario’s scores is 5.
Let’s Try This
Compute for the range of Noel, Paul and Oscar’s scores.
1.
2.
3
Noel’s scores
____ ____
____
highest score
XH = _____
lowest score
XL = _____
range
R = _____
Paul’s scores
____ ____ ____
highest score
XH = _____
lowest score
XL = _____
range
R = _____
Oscar’s scores
____ ____ ____
highest score
XH = _____
lowest score
XL = _____
range
R = _____
Compare your answers to those found in the Answer Key on page 41.
26
Let’s Learn
The range is used to measure the variability of a set of data. However, if the data has an
outlier, or the distribution of most of the scores are located close to the mean but has one
extremely low or high score, the range will show a much larger spread of scores than what really
exists. In this case, we would have to use other measures of variability.
The variance is another measure of variability. It may also be considered as a mean. It is
the average of the squared differences from the mean of the data. It is used for data in the interval
or ratio scales.
In computing for the variance, the distance of the scores from the mean will be used. We
can represent this deviation from the mean as x. This is obtained by subtracting the mean from the
raw score,
x=X–x
where x = deviation from the mean
X = raw score
x = the mean
We can now use the deviation from the mean in the formula for variance. Representing the
variance as S2,
Σx 2
S =
N
2
where S2 = the variance
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
Recall that the S symbol tells you to add the values indicated beside this symbol.. So what
are the steps again in getting the variance?
Let’s take Mario’s score in the basketball game as an example. Here are the steps in finding
the variance.
STEP 1
Compute for the mean.
x=
15 + 10 + 15
40
= 11.67 =
= 13.33
3
3
The mean of the scores of Mario is 13.33 points. His average score for the 3
games is 13.33 points.
27
STEP 2
Subtract each of the raw scores from the mean.
From the formula of computing for the deviation from the mean,
x=X–x
we have,
(15 – 33.33) = –1.67
(10 – 33.33) = –3.33
(15 – 33.33) = –1.67
Did you get the same answers? You might be wonder that you got negative
values. You should not worry about this. In the next step, you will square these
values, so the result shall still be positive.
STEP 3
Square each value.
(1.67) 2 = 2.79
(–3.33) 2 = 11.09
(1.67) 2 = 2.79
STEP 4
Add all the squared values.
2.79 + 11.09 + 2.79 = 16.67
STEP 5
Divide the sum of the squared values by N or the total number of scores.
16.67 ÷ 3 = 5.56
Therefore, the variance of Mario’s scores is 5.56.
Let’s Try This
Compute for the variance of Noel and Paul’s scores.
1.
Noel’s scores 9 7 8
STEP 1
Compute for the mean.
The mean of the Noel’s scores is _____ points.
STEP 2
Subtract each of the raw scores from the mean.
From the formula of computing for the deviation from the mean,
x=X–x
we have, ______
______
______
28
STEP 3
Square each value.
_______
_______
_______
STEP 4
Add all the squared values.
______ + ______ + ______ = ______
STEP 5
Divide the sum of the squared values by N or the total number of scores.
______ ÷ ______ = ______
Therefore, the variance of Noel’s scores is _______.
2.
Paul’s scores 20 21 24
Try to solve this one without writing down the step-by-step procedure.
Compare your answers with those found in the Answer Key on page 42.
29
Let’s Study and Analyze
The standard deviation is the most statistically useful measure of variability. The mean and
the standard deviation are related. While the mean is the average of the scores in a set, the
standard deviation is the average of how far the scores are from the mean.
Recall the formula for the variance. You can see that it is a squared value. But, the original
scores in your data set are not squared values. To convert these squared values into the unit of
the original scores, we get the square root of the variance. This is what the standard deviation is:
the square root of the variance.
Look at the formula for standard deviation below. It has the same formula as the variance,
except for the square root sign.
S2 =
Σx 2
N
where S2 = the standard deviation
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
Simply, we can say that the standard deviation is the square root of the variance.
S = S2
Of course, if you have already computed for the variance of a set of scores you do not have
to compute the standard deviation using the long process. You simply find the square root of the
variance.
Let’s go back to Mario’s scores in the basketball game. You have already computed for the
variance of these scores, right? Then it will be very easy to find the standard deviation. Here are
the steps in finding the standard deviation:
STEP 1
Compute for the variance.
We have already done that for Mario’s scores.
S2 = 5.56
STEP 2
Get the square root of the variance.
5.56 = 2.36
Therefore, the standard deviation of Mario’s scores is 2.36.
30
Let’s Try This
Compute for the standard deviation of Noel and Paul’s scores.
1.
Variance of Noel’s scores = ________
STEP 1
Compute for the variance.
S2 = ______
STEP 2
Get the square root of the variance.
___ = _____
Therefore, the standard deviation of Noel’s scores is _____.
2.
Variance of Paul’s scores = ______
Standard deviation of Paul’s scores = _______
Compare your answers with those found in the Answer Key on page 43.
Let’s See What You Have Learned
1.
The following are the test scores of Maria, Rachel and Mike for four of their subjects.
Math
Science
Literature
History
Maria
82
85
90
91
Rachel
80
95
88
87
Mike
96
91
80
81
Compute for the range, variance and standard deviation. Complete the table below.
Range
Variance
Maria
Rachel
Mike
Compare your answers with those found in the Answer Key on page 43. If you got all the 9
answers correct, you’re doing great! However if you got a score of less than 5, you need to read
the lesson again, then answers some more exercises, for you to fully grasp the concepts
discussed.
31
Let’s Remember
♦
The range is the difference of the highest and the lowest scores in a given data. The
formula for computing the range is
R = XH − XL
Where
R is the range;
XH is the highest score; and
XL is the lowest score.
♦
In computing for the variance and standard deviation, we look for the deviation from
the mean by using the following formula,
x=X–x
where x = deviation from the mean
X = raw score
x = the mean
♦
The variance is the average of the squared differences from the mean of the data. The
formula is
S2 =
Σx 2
N
where S2 = the variance
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
♦
The standard deviation is the average of how far the scores are from the mean. It is
the square root of the variance. The formula is
S2 =
Σx 2
N
where S2 = the standard deviation
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
32
What Have You Learned
A.
C.
The following is the frequency distribution of 40 scores in a Math test.
Interval of scores
Frequency
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
2
4
5
7
6
5
3
4
2
2
4.
Find the mean.
5.
Find the median.
6.
Find the mode.
The following are the basketball scores of Jack and Kevin for four games in a
basketball tournament.
Game 1
Jack
Kevin
Game 2
Game 3
Game 4
23
18
32
30
26
25
19
27
Complete the table below.
Jack
Kevin
Range
Variance
(1)
(4)
(2)
(5)
Compare your answers with those found in the Answer Key on page 43.
If you got a score of:
8–9
Very good! You learned a lot from this module. You are now ready to move on to
the next module.
6–7
Very satisfactory. Just review the items that you missed.
4–5
Satisfactory. Review the parts of the module you did not understand very well.
1–3
You should study the whole module again.
33
Let’s Sum Up
♦
The mean is the average of a set of scores. It is computed by adding all the scores in
your data set and dividing the sum by the number of scores. In formula form,
x=
Σx
N
where x = the mean
Σ =" the sum of"
N = number of scores
♦
When data is presented in a frequency distribution, we compute for the mean by using
the following formula:
Σfxc
N
where x = the mean
x=
∑ = “the sumof”
x c = midpoint of an int erval
f = frequencyof casesin an int erval
N = number ofscores
♦
To compute for the mean of a frequency distribution, we have the following steps:
STEP 1
Determine the number of scores N.
STEP 2
Determine the midpoint xc for each class interval.
STEP 3
Determine the fxc for each class interval.
STEP 4
Determine the sum Σfxc.
STEP 5
Substitute the values in the formula.
♦
The median is the score or value that divides a given data set or a distribution into two
equal halves wherein 50% of the scores are above it, while 50% are below it.
♦
The median is oftentimes denoted as “Md”. There are two different cases considered
in finding the median of a given set of data.
CASE 1
The number of scores N is odd.
In the array of scores, the median is the middle score.
CASE 2
The number of scores N is even.
In the array of scores, the median is the average of the two middle scores.
34
♦
When data is presented in a frequency distribution, we compute for the median by
using the following step-by-step procedure:
STEP 1
Determine the number of cases N.
STEP 2
Solve for N/2 or half the number of cases in the distribution.
STEP 3
Count up the number of cases until the interval containing the N/2 case is
reached.
STEP 4
Determine how many cases were needed out of all the cases in the interval
to reach N/2. Divide this by the number of cases in the interval.
STEP 5
Multiply this by the size of the interval.
STEP 6
Add this to the lower limit of the interval containing the median.
♦
The mode is the score or category that occurs with the highest frequency. By the word
“category”, we simply mean a brand, a name or a group of scores. This means that
among all the categories in your data, this score or category is the one which occurred
the most number of times.
♦
The modal category of the data refers to the most frequently occurring cardinal data
while mode refers to the most frequently occurring nominal data.
♦
In the case of data presented in a frequency distribution table, the mode is the midpoint
of the class interval with the highest frequency of occurrence.
♦
The range is the difference of the highest and the lowest scores in a given data. The
formula for computing the range is
R = XH − XL
Where
R is the range;
XH is the highest score; and
XL is the lowest score.
♦
In computing for the variance and standard deviation, we look for the deviation from
the mean by using the following formula,
x=X–x
where x = deviation from the mean
X = raw score
x = the mean
35
♦
The variance is the average of the squared differences from the mean of the data. The
formula is
S2 =
Σx 2
N
where S2 = the variance
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
♦
The standard deviation is the average of how far the scores are from the mean. It is
the square root of the variance. The formula is
Σx 2
S =
N
2
where S2 = the standard deviation
Σx 2 = sum of the squared deviations from the mean
N = number of scores in the data
36
Answer Key
A.
Let’s See What You Already Know (page 2)
A.
Mean
Median
Mode
=
=
=
76
77.36 = 77
82
B.
Jeremy
Theresa
B.
Range
Variance
28
26
105.50
87.75
Lesson 1
Let’s Try This (pages 5–6)
EXAMPLE 2
54,000
= 2,700
20
Let’s Try This (pages 8–11 )
EXAMPLE 3
STEP 2
STEP 3
Determine the midpoint xc for each class interval.
Interval of ages
Midpoint xc
Frequency f
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
47
42
37
32
27
22
2
5
4
3
12
__4__
N = 30
Determine the fxc for each class interval.
Interval of ages
Midpoint xc
45 – 49
40 – 44
35 – 39
30 – 34
25 – 29
20 – 24
47
42
37
32
27
22
Frequency
2
5
4
3
12
__4__
N = 30
37
EXAMPLE 1
STEP 1
Determine the number of scores N.
N = 40
STEP 2
STEP 3
Determine the midpoint xc for each class interval.
Interval of ages
Midpoint xc
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
97
92
87
82
77
72
67
62
57
52
Frequenc
1
4
5
7
6
5
3
5
2
__2__
N = 40
Determine the fxc for each class interval.
Interval of
ages
Midpoint xc
Frequency f
fx
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
97
92
87
82
77
72
67
62
57
52
1
4
5
7
6
5
3
5
2
__2__
N = 40
9
36
43
57
46
36
20
31
11
10
38
STEP 4
STEP 5
Determine the sum Sfxc.
Interval of
ages
Midpoint xc
Frequency f
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
97
92
87
82
77
72
67
62
57
52
1
4
5
7
6
5
3
5
2
__2__
N = 40
Substitute the values in the formula.
Σfx c
N
3025
=
40
= 75.625
x=
Therefore, the mean is 76.
Let’s Try This (13–15)
EXAMPLE 2
Employee 1’s salary
Case 2
P 3,300
Md = (P 3,300 + P 3,100) ÷ 2
= P 3,200
Therefore, the median is P 3,200.
EXAMPLE 3
Md = (P 2,422 + P 2,354) ÷ 2
= P 2,388
Therefore, the median is P 2,388.
39
___
Σfxc
Let’s Try This (ages 17–18)
EXAMPLE 4
STEP 1
Determine the number of cases N.
N = 40
STEP 2
Solve for N/2 or half the number of cases in the distribution.
N/2 = 40 /2 = 20.
STEP 3
Count up the number of cases until the interval containing the N/2 case
is reached.
From the bottom, we count up until we reach the N/2 case, which is
20. So,
1 + 3 + 5 + 4 + 8 = 21
This is 1 case/s more than 20 and falls in the interval 220 – 224.
STEP 4
Determine how many cases were needed out of all the cases in the
interval to reach N/2. Divide this by the number of cases in the interval.
Out of the 8 cases in the interval 220 – 224, only 7 are used to reach
20. Therefore, only 7/8 of the interval was needed.
STEP 5
Multiply this by the size of the interval.
The size of the class interval 220 – 224 is 5.
STEP 6
Add this to the lower limit of the interval containing the median.
The lower limit to be used = 219.5.
If we do Steps 4 to 6 simultaneously, we will have:
7/8 × 5 + 219.5 = 223.875
Therefore, we say that 224 is the median.
EXAMPLE 5
Md = 9/12 × 3 + 17.5 = 19.75
= 20
Let’s Try This (page 19)
EXAMPLE 1
Heather Potter
EXAMPLE 2
32
40
Let’s See What You Have Learned (pages 20–21)
C.
A.
1.
2.
3.
1,975
9
Cartoon Channel
B.
1.
2.
3.
77
78.5
82
Lesson 2
Let’s Study and Analyze (pages 24–25)
1.
2.
3.
Mario
15
10
15
Noel
9
7
8
Paul
20
21
24
Oscar
7
7
7
Mario
Paul
Oscar
Let’s Try This (page 26)
1.
2.
3.
Noel
9
highest score
lowest score
range
XH = 9
XL = 7
R = 2
Paul
21
20
7
8
24
highest score
XH = 24
lowest score
XL = 20
range
R
= 4
Oscar
7
7
highest score
XH = 7
lowest score
XL = 7
range
R
7
= 0
41
Let’s Try This (pages 28–29)
1.
Noel’s scores 9 7 8
STEP 1
Compute for the mean.
The mean of the Noel’s scores is 8.
STEP 2
Subtract each of the raw scores from the mean.
From the formula of computing for the deviation from the mean,
x=X–x
we have,
9–8=1
8–8=0
7 – 8 = –1
STEP 3
Square each value.
12 = 1
02 = 0
(–1)2 = 1
STEP 4
Add all the squared values.
1+ 0+1 = 2
STEP 5
Divide the sum of the squared values by N or the total number of
scores.
2 ÷ 3 = 0.67
Therefore, the variance of Noel’s scores is 0.67.
2.
Σx 2
S =
N
= 8.67/3
= 2.89
2
42
Let’s Try This (page 31)
1.
Variance of Noel’s scores = 0.67
STEP 1
Compute for the variance.
S2 = 0.67
STEP 2
Get the square root of the variance.
0.67 = 0.82
Therefore, the standard deviation of Noel’s scores is 0.82.
2.
Variance of Paul’s scores = 2.89
Standard deviation of Paul’s scores = 1.70
Let’s See What You Have Learned (page 31)
1.
Maria
Rachel
Mike
D.
Range
Variance
9
15
16
13.50
28.19
45.50
What Have You Learned? (page 33)
A.
mean. 76.5
median. 77.83 = 78
mode. 82
B.
Jack
Kevin
Range
Variance
13
12
22.50
19.50
References
Mathematics III. SEDP Series. Quezon City: IMC. 1991.
Ho, Ju Se T., et al. 21st Century Mathematics (Third Year). Quezon City: Phoenix
Publishing. 1996.
Lacuesta, Debbie P. Basic Statistical Concepts. Quezon City: SEAMEO INNOTECH.
1998.
Downie, N.M. and Heath R.W. Basic Statistical Methods. 4th ed. 1974.
43