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BULACAN STATE UNIVERSITY
COLLEGE OF SCIENCE

MMW 101
MATHEMATICS IN THE MODERN WORLD

Module 16
Measures of Variability
“Statistics: Our Life Saver
and Influencer”

Measures of Variability
Objectives of the module:
At the end of the module, you should be able to:
1
...
Solve for the range, the mean deviation, the quartile deviation,
the variance, and the standard deviation of given datasets
3
...
Some
measures are usually needed to supplement these measures in describing and
comparing the sets of data
...
How will you
compare the scores in Set A and set B?
Set A
Set B
12
6
14
15
15
15
15
19
19
20
Computing the mean, the median, and the mode of the sets of data, we will get the
following values:
̅ = 𝚺𝑿 =75 = 15
̅ = 𝚺𝑿 = 75 = 15
𝑿
𝑿
𝒏
5
𝒏
5
md = 15
md = 15
mo = 15
mo = 15
The two sets of scores have the same mean, median, and mode
...
This shows that computing the measures of central
tendency will not give us all the features or characteristics of a given set of data
...

The measures of variability tell us how the data are spread out or dispersed
around the center
...
On the other hand, a high measure of variability
indicates that the values fall farther from the center
...
The most common
measures of variability are the following: 1
...
) quartile deviation or semiinterquartile range, 3
...
) variance, and 5
...

The range is the simplest measure of variability but the most unstable because
its value quickly fluctuates when there is a change in either the lowest or highest value
...
It does not
give the dispersion or the spread of the values between the highest and the lowest
value
...
The
formula is:
Range = Highest Value - Lowest Value
R = HV – LV

Formula 12

A measure of variability that is not influenced by the presence of outliers in the
data is the interquartile range
...
It gives the spread of the scores
around the median (Q2)
...

The formula is:
Interquartile Range = 3rd Quartile - 1st Quartile
IQR = Q3 - Q1 or IQR = P75 - P25

Formula 13

To illustrate, we have:
Q3
Q2

IQR

Q1

The quartile deviation is also called the semi-interquartile range
...
It tells something about how data is
dispersed around the median
...
In
the formula, we have

Quartile Deviation =
QD =

𝑸𝟑− 𝑸𝟏
𝟐

=

3𝑟𝑑 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒−1𝑠𝑡 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒

𝑰𝑸𝑹
𝟐

2

or

QD =

𝑃75− 𝑃25
2

=

𝐼𝑄𝑅

Formula 14

2

The mean deviation (MD) or the average deviation (AD) is the sum of the
absolute deviations of each value from the mean divided by the total number of
observations in the distribution
...

2

The variance is defined as the average of the squared deviations from the
mean
...
These are the deviation
method and the raw score method
...
A low standard deviation indicates that the values are closer to the
mean
...

The standard deviation is the positive square root of the variance
...
The Range
Example: Consider the following scores of students in a test:
Sets of observation:
A: 10, 8, 6, 5, 12, 11, 13, 7
B: 9, 4, 8, 6, 10, 9, 10, 17
Solving for the range:
FOR SETA: R = HV - LV
R = 13 - 5
R=8

FOR SETB: R = HV - LV
R = 17 - 4
R = 13

If you noticed, the range is computed using only the lowest and the highest
values and does not include the other values
...
This shows that the scores in set B are more spread than the scores in set A
...


2
...
75

6th 7th
11 12
𝑃𝑖 (𝑛+1)

Position:

100

8th
13

=

25(8+1)
100

= 2
...
75 is between the 6th The location of 2
...

2nd value and the 3rd value
...
7th value –6th value = 12 -11 = 1
2
...
75) = 0
...
0
...
75 + 11 =
11
...
3rd value - 2nd value = 7 – 6 = 1
2
...
25) = 0
...
0
...
25 + 6 = 6
...
75

Q1=P25 = 6
...
75 - 6
...
5
QD =

𝑸𝟑− 𝑸𝟏
𝟐

=

𝑰𝑸𝑹
𝟐

=

𝟓
...
75

Interpretation: We can say that half of the values are between 6
...
75
...
5
...

P75= 75 and n = 8
1st
4

Find: Q1 = P25
P25= 25 and n = 8
2nd
6

3rd
8

4th
9

5th
9

6th
10

7th
10

8th
17

4

𝑃𝑖 (𝑛+1)

Position:

100

=

75(8+1)
100

= 6
...
25

The location of 6
...


The location of 2
...


Since the values are the same, there is
no need to interpolate
...
3rd value - 2nd value = 8 - 6 = 2
2
...
25) = 0
...
0
...
5 + 6 = 6
...


Q1 = P25 = 6
...
5
IQR = 3
...
𝟓
𝟐

= 1
...
5 and 10
...
5
...
5
2
...
5
1
...
25

10

P75 = 10

9

QD = 1
...
75
10

IQR =
5
...
75

8

7

P25 =
6
5

6
...
5

9
QR = 1
...
5
6
4

5

Interpretation (for QD): Comparing the results obtained from the two
sets of values, we can say that the values in set B are less dispersed
around the median as compared to the values in set A
...
The Mean Deviation or the Average Deviation

𝑴𝑫 =

̅|
𝚺|𝑿 − 𝑿
𝒏

Formula 15

where: MD = mean deviation
X = individual value
̅ = mean of data
𝑿
n = total number of items or observations
̅ | = deviations from the mean regardless of algebraic signs
|𝑿 − 𝑿

Example: Mean Deviation for Ungrouped Data
Determine the mean deviation of the two sets of data and follow the steps below
...
Arrange the values in ascending or descending order
...
Find the sum of the values
...
Compute the value of the mean
...
Find the value of the deviation of each score from the mean
...
Find the absolute value of each deviation obtained in Step 4
...
Find the sum of the absolute values in Step 5
...
Substitute the values in the formula and solve
...
𝟓

Interpretation: This means that on the average, the scores of
the students in set A deviated from the mean by 2
...

Note: The lower the mean deviation, the closer the values are around the
mean
...


4
...
Arrange the values in ascending
or descending order
...
Find the sum of the values
...
Compute the value of the mean
...
Obtain the individual deviations
from the mean
...
Square each deviation and write
the results under column 3
...
Find the sum of the squared
deviations
...
Substitute the values in the
formula and solve
...
Arrange the values in
ascending or descending
order
...
Find the sum of the
values
...
Square each value and
write the results under
column 2
...
Get the sum of the
squared values in step 3
...
Substitute the values in
the formula and solve
...
𝟓𝟕

s2

=

𝚺 𝑥2

−

(𝚺𝑥)2
𝑛

𝑛−1

=

708−

(72)2
8

8−1

= 8
...
The square root of the variance will
give us a measure that has the same unit as the data, and this is the standard
deviation
...


5
...

For example, when the unit of measurements of the data are meters, inches,
feet, the standard deviation and the mean are expressed in the same units
...

Example 4
...

Sets of observations:
A: 10, 8, 6, 5, 12, 11, 13, 7
B: 9, 4, 8, 6, 10, 9, 10, 17
To solve the standard deviation, follow the procedures in computing the
variance either by deviation method or by raw score method
...


Standard
Deviation

s=

√𝑠 2

SET A
= √8
...
𝟗𝟑

s=

√𝑠 2

SET B
= √14
...
𝟖𝟎

Interpretation: The computed standard deviations of the two
sets of values show that the values in set A are less dispersed
from the mean since its standard deviation is smaller than the
standard deviation of set B
...


9

Let us have another example
...
You cannot choose which brand to buy
...

Take five one-kilo packs of each brand, weigh them, and list the results
...
By just comparing the values, it would be difficult for you to choose a better
brand
...
87 kg
0
...
03 kg
0
...
04 kg
1
...
01 kg
1
...
14 kg
1
...
018
1
...
0968
s = 0
...
Since the standard deviation of brand A is
smaller than the standard deviation of brand B, then it means that the contents of the
packs of sugar of brand A are more uniform than brand B
...

Measures of Variability for Grouped Data
For the computation of the measures of variability for grouped data, let us
consider the frequency distribution of two sections in Math 323 in a 60-item
examination
...
The Range
Range = Highest UPPER
R = HUCB - LLCB

CLASS BOUNDARY

- Lowest LOWER CLASS BOUNDARY

FOR SET A
HUCB = 59
...
5
R = HUCB - LLCB
R = 59
...
5
R = 30
The range of 30 shows the width of the distribution
...

What conclusions can you make between the two sections?

2
...
5 - 34
...
5 - 39
...
5 - 44
...
5 - 49
...
5 - 54
...
5 - 59
...
25 is found here)
26( 37(33
...

FOR SET A
Solving for Q3 = P75
...


i = 75 and n = 45
𝑖𝑛

=
100

75(45)
100

i = 25 and n = 45
𝑖𝑛

=
100

= 33
...
5 + (
P75 = 48
...
25
𝒊𝝁

𝑃25 = 𝓵𝒊 + (𝟏𝟎𝟎

)𝒄

33
...
5 + (
P25 = 39
...
02−39
...
94
2

− <𝑐𝑓𝒃𝒊
𝒇𝒊

11
...
𝟒𝟕

Interpretation: The results show that the middle 50% of the scores of
the forty-five students in Math 323 lie between 39
...
02
...
47
...
Interpret the result and make a
conclusion
...
The Mean Deviation, Variance, and the Standard Deviation
In computing the mean deviation, you have to follow the steps below
...

Step 2
...

Step 4
...

Step 6
...

Multiply each frequency (f) by its corresponding class mark(𝑿𝒊 )
...

̅ )of the distribution
...

Subtract the mean (𝐗
Write the absolute values of the results obtained in Step 5 under the
̅ | column
...
Find the products of items under column (f) and items under |𝑿𝒊 − 𝐗
column
...

Step 8
...
Compute the mean deviation
...

Step 10
...

Step 11
...

Step 12
...

Method 1
...

Step 13
...


𝒔= √

̅) 𝟐
𝚺𝒇(𝑿𝒊 − 𝐗
𝒏−𝟏

Formula 20a

13

Table 1 shows how you will continue constructing the additional columns for calculating the mean deviation, the variance, and the standard
deviation of SET A
...
56 = -11
...
56 = - 6
...
56 = - 1
...
56 =
3
...
56 =
8
...
56 = 13
...
56
6
...
56
3
...
44
13
...
562 133
...
562
43
...
562
2
...
83
71
...
63

̅)2
f(𝐗 − 𝐗
(Step 11)
3x133
...
89
9x 43
...
27
14 x 2
...
02
130
...
38
361
...
95
Σf(𝐗 − 𝐗
(Step 12)

̅ = 𝚺fX = 1960 = 𝟒𝟑
...
56 34
...
56 59
...
56 21
...
84
50
...
88
̅| =230
...
92

45
𝟏𝟕𝟒𝟎
...
13

= 𝟑𝟗
...
𝟐𝟗

TABLE 3
...
08

Q3
48
...
94

QD
4
...
56

MD
5
...
57

s
6
...

Suppose you want to solve the variance of grouped data by using Method 2
(Short Method)
...

Step 1
...

1
...

1
...
1, assign negative values starting from -1
...
3 Class intervals above
...

Step 2
...

Step 3
...

Step 4
...

Step 5
...

Step 6
...

Step 7
...

s2 = c2⌈

𝐧 𝚺 𝐟𝐝𝟐 − (𝚺 𝐟𝐝)𝟐
𝐧 (𝐧−𝟏)

⌉

Formula 19b
s2 = variance

where:
f = frequencies
d = coded values
n = number of observations

In determining the standard deviation, follow the step below
...
Substitute the computed value of the variance obtained in Step 7 in the
standard deviation Method 2 (Short Method) formula and solve
...
SHORT METHOD (FOR SET A)
LL - UL

f

30 -34
35 -39
40 - 44
45 - 49
50 - 54
55 - 59

3
9
14
11
6
2

c=5

n = 45

d
(Step 1)
-2
-1
0
1
2
3

d2
(Step 2)
(-2)2 4
(-1)2 1
02 0
1
4
9

fd
(Step 3)
3 x -2
-6
9 x -1 -9 (-15)
14x 0
0
11
12
6 (29)
Σfd = (15+29)=14
(Step 4)

fd2
(Step 5)
3 x 4 12
9x1
9
14x 0
0
11
24
18
2
Σfd = 74
(Step 6)

Note: Σfd can be negative
...
𝟓𝟕

𝐧 𝚺 𝐟𝐝𝟐 − (𝚺 𝐟𝐝)𝟐

s=c√

𝐧 (𝐧−𝟏)

s =√𝑠 2 = √39
...
29
Summary of Values:
c=5
n = 45
Σ𝑓𝑑 2 = 74
(Σ𝑓𝑑)2 = (14)2

Let us assume that there is another section C that took the same test
...
03 and a standard deviation of 7
...

Which section has more variability in their scores?
Comparing section A with section C:
Section A has a mean of 43
...
29
...
03, with a standard deviation of 7
...
Since the standard
deviation of section C is bigger than the standard deviation of section A, we say
that the scores in section C are more variable than in section A
...

Note: Disperse means scatter or spread out in different directions or over a
wider area
...
Compare
the results between section A and section B and interpret them

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Title: ASSESSMENT OF LEARNING
Description: this will serve as your compilation of notes in assessment of learning 1.

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