The document discusses several statistical measures of variation including range, average deviation, variance, and semi-interquartile range. It provides formulas and step-by-step examples for calculating each measure using both ungrouped and grouped data. Key measures include the range being the difference between highest and lowest values, average deviation being the mean of absolute deviations from the mean, and variance being the mean of squared deviations from the mean.

1. Measure of Variation
● Measure of Variation - it is used to determine the scatter of values
in a distribution.
● Range it is the difference between the highest and the lowest
value in the distribution.
Formula:
R = H – L
Where: R is range
H is highest values
L is the lowest value

2. Steps:
1. Determined the lowest value.
2. Determined the highest value.
3. Substitute the given values in the formula.
4. Solve the problem.

3. Example:
Statistics classes with 60 students were given an examination and the results are the following:
48 73 57 57 69 88 11 80 82 47
46 70 49 45 75 81 33 65 38 59
94 59 62 36 58 69 45 55 58 65
30 49 73 29 41 53 37 35 61 48
22 51 56 55 60 37 56 59 57 36
12 36 50 63 68 30 56 70 53 28

4. Solution:
1. Determined the lowest value.
L = 11
2. Determined the highest value.
H = 94
3. Substitute the given values in the formula.
R = H – L
= 94 – 11
4. Solve the problem.
R = H – L
= 94 – 11
= 83

5. Sometimes in finding the range of group data we used the upper class boundary and
the lower class boundary. To determine the upper class boundary just add point five
and for lower boundary just subtract point five.
Example:
H = 94 (94 +.5 =94.5) 94.5 is the upper boundary or the highest value
L = 11 (11 - .5 = 10.5) 10.5 is the lower boundary or the lowest value.
R = H – L
= 94.5 – 10.5
= 84

6. Semi-inter quartile range or quartile Deviation
Another measure of deviation is the semi-inter quartile
range or quartile deviation. This value is obtained by getting
one half the differences between the third and the first
quartiles.
Formula:
𝑄 =
𝑄3 − 𝑄1
2

7. Example:
The examination score of 50 students in a statistics class resulted to the
following values: Q3 = 75. 43 and Q1 = 54.24. Determine the value of the semi-
inter quartile range.
Given:
Q3 = 75. 43
Q1 = 54.24
Q = ?

8. Solution:
𝑄 =
𝑄3 − 𝑄1
2
𝑄 =
75.43 − 54.24
2
𝑄 =
21.19
2
𝑄 = 10.60

9. Another Example:
Suppose the performance rating of 100 faculty members of a certain college
were taken and are presented in a frequency distribution as follows:
Classes f
71-74 3
75-78 10
79-82 13
83-86 18
87-90 25
91-94 19
95-98 12
n = 100

10. Solution:
We will first compute the value of Q3 and Q1 since only the frequency
distribution is given.
Classes f <cumf
71-74 3 3
75-78 10 13
79-82 13 26 (1st Quartile class)
83-86 18 44
87-90 25 69
91-94 19 88 (3rd Quartile class)
95-98 12 100

11. Formula:
𝑄1 = 𝑥𝑙𝑏 + (
n
4
−𝑐𝑢𝑚𝑓𝑏
𝑓𝑄1
)𝑐
Solution:
𝑄1 = 𝑥𝑙𝑏 + (
n
4
−𝑐𝑢𝑚𝑓𝑏
𝑓𝑄1
)𝑐
𝑄1 = 78.5 + (
25−13
13
)4
𝑄1 = 78.5 + (
12
13
)4
𝑄1 = 78.5 + (0.923076923)4
𝑄1 = 78.5 + 3.692307692
𝑄1 = 82.19230769 or 82.19
Formula:
𝑄3 = 𝑥𝑙𝑏 + (
3n
4
−𝑐𝑢𝑚𝑓𝑏
𝑓𝑄3
)𝑐
Solution:
𝑄3 = 𝑥𝑙𝑏 + (
3n
4
−𝑐𝑢𝑚𝑓𝑏
𝑓𝑄3
)𝑐
𝑄3 = 90.5 + (
75−69
19
)4
𝑄3 = 90.5 + (
6
19
)4
𝑄3 = 90.5 + (0.315789473)4
𝑄3 = 90.5 + 1.263157895
𝑄3 = 91.7631579 𝑜𝑟 91.76

12. The value of Q can now be obtained. Thus,
𝑄 =
𝑄3−𝑄1
2
𝑄 =
91.76−82.19
2
𝑄 =
9.57
2
𝑄 = 4.78

13. Average Deviation
● The average deviation refers to the
arithmetic mean of the absolute deviations
of the values from the mean of the
distribution. This measure in sometimes
referred to as the mean absolute deviation.

14. Average Deviation for Ungroup Data
AD =
|x − 𝒙|
𝒏
where: x represents the individual values
𝒙 is the mean of the distribution

15. Average Deviation of Ungrouped Data
Steps:
1. Arrange the values in column according to magnitude.
2. Computed the value of the mean (𝒙).
3. Determine the deviations (𝒙 − 𝒙)
4. Convert the deviations in step 3 into positive deviations. Use the absolute
values sign |x− 𝒙|
5. Get the sum of the absolute deviations in step 4.
6. Divide the sum in step 5 by n.

16. Example:
Consider the following values.
x: 13, 16, 9, 6, 15, 7, 11
Determined the value of the average deviation.
Solution:
First we arrange the values in vertical column and then we compute the
value of the mean.

17. Step 1: Step 2: Step 3:
x x x − 𝒙
6 𝒙 =
𝒙
𝒏
6 6 – 11 = -5
7 7 7 – 11 = -4
9 =
𝟕𝟕
𝟕
9 9 – 11 = -2
11 11 11 – 11 = 0
13 = 11 13 13 – 11 = 2
15 15 15 – 11 = 4
16 16 16 – 11 = 5
𝒙 = 𝟕𝟕

18. Step 4:
x x − 𝒙 |x − 𝒙|
6 6 – 11 = -5 5
7 7 – 11 = -4 4
9 9 – 11 = -2 2
11 11 – 11 = 0 0
13 13 – 11 = 2 2
15 15 – 11 = 4 4
16 16 – 11 = 5 5
Notice that the some of the deviations from negative. Hence, we make an
assumption that all deviations are positive deviations by introducing the absolute
value sign. Adding all these absolute deviations.

19. Step 5: Step 6:
x x − 𝒙 |x − 𝒙| AD =
|x − 𝒙|
𝒏
6 6 – 11 = -5 5
7 7 – 11 = -4 4 =
𝟐𝟐
𝟕
9 9 – 11 = -2 2
11 11 – 11 = 0 0 = 3.14
13 13 – 11 = 2 2
15 15 – 11 = 4 4
16 16 – 11 = 5 5
|x − 𝒙| = 𝟐𝟐

20. Average Deviation for Group Data
AD =
𝒇|x − 𝒙|
𝒏
Where:
f – represent the frequency of each class
x – the midpoint of each class
𝒙- The mean of the distribution
n – the total number of frequency.

21. Average Deviation for Group Data
Steps:
1. Compute the value of mean.
2. Get the deviation by using the expression 𝒙 − 𝒙.
3. Multiply the deviation by its corresponding frequency.
4. Add the results in step 3
5. Divide the sum in step 4 by n.

22. Example:
Compute the value of the average deviation of the frequency distribution as follows:
Classes f
11-22 3
23-34 5
35-46 11
47-58 19
59-70 14
71-82 6
83-94 2
n = 60

23. Solution:
Step 1: compute the value of the mean
𝑥 =
𝑓𝑥
𝑛
=
3174
60
= 52.90
Classes f x fx
11-22 3 16.5 49.5
23-34 5 28.5 142.5
35-46 11 40.5 445.5
47-58 19 52.5 997.5
59-70 14 64.5 903.0
71-82 6 76.5 459.0
83-94 2 88.5 177.0
n = 60
𝒇𝒙 = 𝟑, 𝟏𝟕𝟒

24. Step 2: Construct the deviation column 𝒙 − 𝒙
𝒙 − 𝒙
16.5 – 52.90 = -36.4
28.5 – 52.90 = -24.4
40.5 – 52.90 = -12.4
52.5 – 52.90 = -0.4
64.5 – 52.90 = 11.6
76.5 – 52.90 = 23.6
88.5 – 52.90 = 35.6
Classes f x fx 𝒙 − 𝒙
11-22 3 16.5 49.5 -36.4
23-34 5 28.5 142.5 -24.4
35-46 11 40.5 445.5 -12.4
47-58 19 52.5 997.5 -0.4
59-70 14 64.5 903.0 11.6
71-82 6 76.5 459.0 23.6
83-94 2 88.5 177.0 35.6

25. Step 3: Convert the deviations to positive deviations
Classes f x fx 𝒙 − 𝒙 |x− 𝒙|
11-22 3 16.5 49.5 -36.4 36.4
23-34 5 28.5 142.5 -24.4 24.4
35-46 11 40.5 445.5 -12.4 12.4
47-58 19 52.5 997.5 -0.4 0.4
59-70 14 64.5 903.0 11.6 11.6
71-82 6 76.5 459.0 23.6 23.6
83-94 2 88.5 177.0 35.6 35.6

26. Step 4: multiply the positive deviations by their corresponding frequencies.
Classes f x fx 𝒙 − 𝒙 |x− 𝒙| f |x− 𝒙|
11-22 3 16.5 49.5 -36.4 36.4 109.2
23-34 5 28.5 142.5 -24.4 24.4 122.0
35-46 11 40.5 445.5 -12.4 12.4 136.4
47-58 19 52.5 997.5 -0.4 0.4 7.6
59-70 14 64.5 903.0 11.6 11.6 162.4
71-82 6 76.5 459.0 23.6 23.6 141.6
83-94 2 88.5 177.0 35.6 35.6 71.2
n = 60 ∑fx = 3,174 ∑f |x− 𝒙| = 750.4

27. Step 5: Divide the sum in step 4 by n
AD =
𝒇|x − 𝒙|
𝒏
=
750.4
60
= 12.51

28. Variance
● In the preceding sections, we noticed that the sum of the
deviations from the mean equals 0 if the signs are
considered. If the signs are ignored, that is, assuming
that all the deviations are positive numbers, then the
sum will also be a positive number. In these sections,
we shall consider another procedure of handling signed
numbers. If the deviations 𝑥 − 𝑥 are squared, then the
sum of the squared deviations will also be equal to a
positive number. If this sum is divided by the sample
size n, then we will be able to come up with the mean of
squared deviations. We shall call this measure the
variance.

29. Variance for Ungrouped Data
𝑺𝟐 =
|x − 𝒙|𝟐
𝒏
where: x represents the individual values in the distribution
𝒙 is the mean of the distribution
n is the sample size

30. Steps:
1. Computed the value of the mean (𝒙).
2. Get the deviation of each value from mean
3. Square the deviations
4. Calculate the sum of the squared deviations
5. Divide the sum by the total number of values.
Example:
Compute the value of variance of the following measurements
x: 13, 5, 7, 9, 10, 17, 15, 12

31. Step 1: Compute the Value of Mean
x
5
7
9
10
12
13
15
17
𝒙 = 88
𝒙 =
𝒙
𝒏
= 88/8
= 11

32. Step 2: Get the deviation of each value from mean
x 𝒙 x − 𝒙
5 11 -6
7 11 -4
9 11 -2
10 11 -1
12 11 1
13 11 2
15 11 4
17 11 6
𝐱 = 88

33. Step 3: Square the deviations
x x − 𝒙 (x − 𝒙) 2
5 -6 36
7 -4 16
9 -2 4
10 -1 1
12 1 1
13 2 4
15 4 16
17 6 36
𝐱 = 88

34. Step 4: Calculate the sum of squared deviation
x x − 𝒙 (x − 𝒙) 2
5 -6 36
7 -4 16
9 -2 4
10 -1 1
12 1 1
13 2 4
15 4 16
17 6 36
𝒙 = 88 = 114

35. Step 5:
𝑺𝟐 =
|x − 𝒙|
𝟐
𝒏
=
𝟏𝟏𝟒
𝟖
= 14.25

36. Variance for Grouped Data
The computing formulas for the variance as in 𝑠2 =
𝑓 𝑥−𝐱 2
𝑛
is based on
ungrouped data. When the data are presented in a frequency distribution,
however, we shall have another computing formula.
𝒔𝟐 =
𝒇 𝒙−𝒙 𝟐
𝒏
Where: x is midpoint of each class interval
𝒙 is the mean
n is the sample size

37. Steps:
1. Compute the value of the mean.
2. Determined the deviation x− x
̄ by subtracting the mean
from the midpoint of each class interval.
3. Square the deviations obtained in step 2.
4. Multiply the frequencies by their corresponding squared
deviations.
5. Add the results in step 4.
6. Divide the result in step 5 by the sample size.

38. Example:
Calculate the variance of the distribution of the following:
Classes f
11-22 3
23-34 5
35-46 11
47-58 19
59-70 14
71-82 6
83-94 2
𝒇 = 𝟔𝟎

39. Step 1: Compute the value of the mean
Classes f x fx
11-22 3 16.5 49.5
23-34 5 28.5 142.5
35-46 11 40.5 445.5
47-58 19 52.5 997.5
59-70 14 64.5 903
71-82 6 76.5 459
83-94 2 88.5 177
𝒇 = 𝟔𝟎 𝒇𝒙 = 𝟑𝟏𝟕𝟒
𝒙 =
𝒇𝒙
𝒏
=
𝟑𝟏𝟕𝟒
𝟔𝟎
= 52.90

40. Step 2: Construct the deviation 𝑥 − 𝑥
Classes f x fx 𝒙 − 𝒙
11-22 3 16.5 49.5 -36.4
23-34 5 28.5 142.5 -24.4
35-46 11 40.5 445.5 -12.4
47-58 19 52.5 997.5 -0.4
59-70 14 64.5 903 11.6
71-82 6 76.5 459 23.6
83-94 2 88.5 177 35.6

41. Step 3: square the deviation
Classes f x fx 𝒙 − 𝒙 |x − 𝒙|2
11-22 3 16.5 49.5 -36.4 1324.96
23-34 5 28.5 142.5 -24.4 595.36
35-46 11 40.5 445.5 -12.4 153.76
47-58 19 52.5 997.5 -0.4 0.16
59-70 14 64.5 903 11.6 134.56
71-82 6 76.5 459 23.6 556.96
83-94 2 88.5 177 35.6 1267.36

42. Step 4: Multiply the deviations by their corresponding
frequencies
Classes f x fx 𝒙 − 𝒙 |x − 𝒙|2 f|x − 𝒙|2
11-22 3 16.5 49.5 -36.4 1324.96 3974.88
23-34 5 28.5 142.5 -24.4 595.36 2976.8
35-46 11 40.5 445.5 -12.4 153.76 1691.36
47-58 19 52.5 997.5 -0.4 0.16 3.04
59-70 14 64.5 903 11.6 134.56 1883.84
71-82 6 76.5 459 23.6 556.96 3341.76
83-94 2 88.5 177 35.6 1267.36 2535.92
𝒇 = 𝟔𝟎 𝒇𝒙 = 𝟑𝟏𝟕𝟒 𝒇|x−𝒙| =
𝟏𝟔𝟒𝟎𝟕. 𝟔

43. Step 5: Divide the sum of Step 4 by n
𝒔𝟐
=
𝒇 𝒙−𝒙 𝟐
𝒏
=
𝟏𝟔𝟒𝟎𝟕.𝟔
𝟔𝟎
= 273.46