Discussion on the Measures of Variability/Dispersion under the K to 12 Curriculum Grade 7 Students (Philippine Setting)

1. Measures of
Variability/Dispersion
(Ungrouped Data)
REYNALDO D. SALAYOG, II-LPT-MEd Math

2. MEASURES OF VARIABILITY/DISPERSION
 the “spread” in a set of measurement.
 it refers to how spread out a group of data is
and it measures how much your scores differ
from each other.
Measures of Variation:
1. Range, 2. quartile deviation,
3. mean deviation, 4. variance,
5. standard deviation and the
6. coefficient of variation

3. A. RANGE
- Refers to the difference between the highest and the
lowest observation
𝑟 = 𝐻𝑆 − 𝐿𝑆
Example 1: Find the Range of each set of scores and interpret
the results.
Solution:
Boys: R = 100 – 60
R = 40
Girls: R = 83 – 79
R = 4
Therefore the girls are more
homogeneous than the boys in
their math ability.

4. MERITS and DEMERITS of RANGE:

5. TRY THIS!
Find the Range of each set of scores.
A. 3, 5, 9, 7, 5, 2, 1, 8, 8
B. 13, 17, 11, 16, 19, 20, 31, 17, 19, 11, 15
C. 0.1, 0.3, 0.2, 0.4, 0.5, 0.2, 0.9, 0.6
D. 32, 30, 37, 33, 39, 38, 35, 31
E. 17, 18, 15, 19, 16, 17, 16, 18

6. B. MEAN DEVIATION for Ungrouped
• The average of the summation of the
absolute deviation of each observation/
score from the mean.
𝑀. 𝐷 =
𝑥 − 𝑥
𝑛
𝑥 = 𝑠𝑐𝑜𝑟𝑒/𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠
Example 1: Find the MEAN
DEVIATION of each set of scores
and interpret the results.

7. B. MEAN DEVIATION
Example 1: Find the MEAN
DEVIATION of each set of scores
and interpret the results.
𝑥 − 𝑥 =66
𝑀. 𝐷 =
66
5
=13.2
GIRLS
𝑥 |𝑥 − 𝑥|
Grace 82
Irish 80
Abigail 83
Sherry 81
Kristine 79
𝑥 − 𝑥 =6𝑥 = 81
𝑀. 𝐷 =
6
5
=1.2
Therefore the
girls are more
homogeneous
than the boys
in their math
ability.
BOYS
𝑥 |𝑥 − 𝑥|
Frederick 70
Russel 95
Murphy 60
Jerome 80
Tom 100
𝑥 = 81
11
14
21
1
19
1
1
2
0
2

8. B. MEAN DEVIATION
Example 2: Find the MEAN DEVIATION of the set of scores.
70, 95, 60, 80, 100, 65, 85, 95
𝑥 − 𝑥 = 100
𝑀. 𝐷 =
100
8
=12.5
Solutions
𝑥 |𝑥 − 𝑥 |
70
95
60
80
100
65
85
95
𝑥 = 81.25
11.25
13.75
21.25
1.25
18.75
16.25
3.75
13.75

9. B. MEAN DEVIATION
TRY THIS!: Find the MEAN DEVIATION of the set of scores.
54, 62, 71, 38, 39, 42, 55, 68
𝑥 − 𝑥 = 83.74
𝑀. 𝐷 =
83.74
8
=10.47
Solutions
𝑥 |𝑥 − 𝑥 |
54
62
71
38
39
42
55
68
𝑥 = 53.63
0.37
8.37
17.37
15.63
14.63
11.63
1.37
14.37

10. MERITS and DEMERITS of MEAN
DEVIATION:

11. B. MEAN DEVIATION
Graded Assessment
Find the MEAN DEVIATION of the following set
of scores.
1. 54, 62, 71, 38, 39, 42, 55, 68
2. 75, 68, 69, 70, 71, 78, 76, 69, 71, 68
3. 14, 15, 18, 19, 14, 13, 12, 11, 10
4. 17, 19, 21, 20, 18, 19, 17, 15

12. C. STANDARD DEVIATION for
Ungrouped Data
𝜎 =
𝑥 − 𝑥 2
𝑛
𝑥 = 𝑠𝑐𝑜𝑟𝑒/𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠

13. MERITS and DEMERITS of STANDARD
DEVIATION.

14. C. STANDARD DEVIATION
Example 1: Find the STANDARD DEVIATION of the set of
scores. 70, 95, 60, 80, 100, 65, 85, 95
𝑥 − 𝑥 2
= 1,587.48
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
70
95
60
80
100
65
85
95
𝑥 = 81.25
11.25
13.75
21.25
1.25
18.75
16.25
3.75
13.75
126.56
189.06
451.56
1.56
351.56
264.06
14.06
189.06
𝜎 =
𝑥 − 𝑥 2
𝑛
𝜎 =
1587.48
8
𝜎 = 14.08 𝑜𝑟 14.09

15. C. STANDARD DEVIATION
Example 2: Find the STANDARD DEVIATION of the set of
scores. 54, 62, 71, 38, 39, 42, 55, 68
𝑥 − 𝑥 2
= 1173.9
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
54
62
71
38
39
42
55
68
𝑥 = 53.63
0.37
8.37
17.37
15.63
14.63
11.63
1.37
14.37
0.14
70.06
301.72
244.30
214.04
135.26
1.88
206.50
𝜎 =
𝑥 − 𝑥 2
𝑛
𝜎 =
1173.9
8
𝜎 = 12.11

16. C. STANDARD DEVIATION
TRY THIS: Find the STANDARD DEVIATION of the set of
scores. 23, 21, 19, 18, 18, 15, 12, 10, 10
𝑥 − 𝑥 2
= 179.57
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
23
21
19
18
18
15
12
10
10
𝑥 = 16.22
6.78
4.78
2.78
1.78
1.78
1.22
4.22
6.22
45.97
22.85
7.73
3.17
3.17
1.49
17.81
38.69
𝜎 =
𝑥 − 𝑥 2
𝑛
𝜎 =
179.57
9
𝜎 = 4.46 𝑜𝑟 4.47
6.22 38.69

17. C. STANDARD DEVIATION
Graded Assessment
Find the STANDARD DEVIATION of the following
set of scores.
1. 54, 62, 71, 38, 39, 42, 55, 68
2. 75, 68, 69, 70, 71, 78, 76, 69, 71, 68
3. 14, 15, 18, 19, 14, 13, 12, 11, 10
4. 17, 19, 21, 20, 18, 19, 17, 15

18. D. VARIANCE for Ungrouped Data
𝜎2
=
𝑥 − 𝑥 2
𝑛
𝑥 = 𝑠𝑐𝑜𝑟𝑒/𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠
Variance is the mean or average of
the squares of the deviation of each
measurement from the mean.

19. D. VARIANCE
Example 1: Find the VARIANCE of the set of scores. 70, 95,
60, 80, 100, 65, 85, 95
𝑥 − 𝑥 2
= 1,587.48
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
70
95
60
80
100
65
85
95
𝑥 = 81.25
11.25
13.75
21.25
1.25
18.75
16.25
3.75
13.75
126.56
189.06
451.56
1.56
351.56
264.06
14.06
189.06
𝜎2
=
1587.48
8
𝜎2 = 198.44
𝜎2 =
𝑥 − 𝑥 2
𝑛

20. D. VARIANCE
Example 2: Find the VARIANCE of the set of scores. 54, 62,
71, 38, 39, 42, 55, 68
𝑥 − 𝑥 2
= 1173.9
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
54
62
71
38
39
42
55
68
𝑥 = 53.63
0.37
8.37
17.37
15.63
14.63
11.63
1.37
14.37
0.14
70.06
301.72
244.30
214.04
135.26
1.88
206.50
𝜎2 =
𝑥 − 𝑥 2
𝑛
𝜎2 =
1173.9
8
𝜎2
= 146.74

21. D. VARIANCE
TRY THIS: Find the VARIANCE of the set of scores. 23, 21,
19, 18, 18, 15, 12, 10, 10
𝑥 − 𝑥 2
= 179.57
Solutions
𝒙 𝒙 − 𝒙 𝒙 − 𝒙 𝟐
23
21
19
18
18
15
12
10
10
𝑥 = 16.22
6.78
4.78
2.78
1.78
1.78
1.22
4.22
6.22
45.97
22.85
7.73
3.17
3.17
1.49
17.81
38.69
𝜎2 =
𝑥 − 𝑥 2
𝑛
𝜎2
=
179.57
9
𝜎2
= 19.95
6.22 38.69

22. D. VARIANCE
Graded Assessment
Find the VARIANCE of the following set of scores.
1. 54, 62, 71, 38, 39, 42, 55, 68
2. 75, 68, 69, 70, 71, 78, 76, 69, 71, 68
3. 14, 15, 18, 19, 14, 13, 12, 11, 10
4. 17, 19, 21, 20, 18, 19, 17, 15

23. CHAPTER TEST:
Graded Assessment
Find the RANGE, MEAN DEVIATION, VARIANCE ,
and STANDARD DEVIATION of the given set of
data.

24. Measures of
Variability/Dispersion
(Grouped Data)
REYNALDO D. SALAYOG, II-LPT-MEd Math

25. A. RANGE
- Refers to the difference between the Upper
boundary of the highest class interval and Lower
boundary of the lowest class interval
𝑟 = 𝐻 𝑈𝐵 − 𝐿 𝐿𝐵
Example 1: Find the Range of the distribution.
Solution:
R = 50.5 – 10.5
R = 40

26. GRADED ASSESSMENT:
Find the Range of each grouped data.
1

27. GRADED ASSESSMENT:
2

28. GRADED ASSESSMENT:
3

29. B. MEAN DEVIATION for Grouped
• The average of the summation of the
absolute deviation of each observation/
score from the mean.
𝑀. 𝐷 =
𝑓 𝑚𝑑 − 𝑥
𝑛
𝑚𝑑 = 𝑐𝑙𝑎𝑠𝑠 𝑚𝑎𝑟𝑘/𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛 = 𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

30. B. MEAN DEVIATION
Example 1: Find the MEAN DEVIATION of the grouped data.
𝑓 𝑥 − 𝑥 = ____
𝑀. 𝐷 = =
𝑥 = 24.6
Class Interval f Md 𝑀𝑑 − 𝑥 𝑓(𝑀𝑑 − 𝑥)
36 – 40 1
31 – 35 10
26 – 30 10
21 – 25 16
16 – 20 9
11 – 15 4

31. GRADED ASSESSMENT:
Find the MEAN DEVIATION of each
grouped data.
1

32. GRADED ASSESSMENT:
2

33. GRADED ASSESSMENT:
3

34. C. STANDARD DEVIATION for
Grouped Data
𝜎 =
𝑓 𝑚𝑑 − 𝑥 2
𝑛
𝑚𝑑 = 𝑐𝑙𝑎𝑠𝑠 𝑚𝑎𝑟𝑘
𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠

35. C. STANDARD DEVIATION
Example 1: Find the STANDARD DEVIATION of the grouped
data.
𝑥 = 24.6
Class
Interval
f Md 𝑚𝑑 − 𝑥 𝑚𝑑 − 𝑥 2
𝑓 𝑚𝑑 − 𝑥 2
36 – 40 1
31 – 35 10
26 – 30 10
21 – 25 16
16 – 20 9
11 – 15 4

36. GRADED ASSESSMENT:
Find the STANDARD DEVIATION of each
grouped data.
1

37. GRADED ASSESSMENT:
2

38. GRADED ASSESSMENT:
3

39. D. VARIANCE for Grouped Data
𝜎2
=
𝑓 𝑚𝑑 − 𝑥 2
𝑛
𝑚𝑑 = 𝑐𝑙𝑎𝑠𝑠 𝑚𝑎𝑟𝑘
𝑓 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑥 = 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑎𝑡𝑎
𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑎𝑠𝑒𝑠
Variance is the mean or average of
the squares of the deviation of each
measurement from the mean.

40. D. VARIANCE
Example 1: Find the VARIANCE of the grouped data.
𝑥 = 24.6
Class
Interval
f md 𝑚𝑑 − 𝑥 𝑚𝑑 − 𝑥 2
𝑓 𝑚𝑑 − 𝑥 2
36 – 40 1
31 – 35 10
26 – 30 10
21 – 25 16
16 – 20 9
11 – 15 4

41. GRADED ASSESSMENT:
Find the VARIANCE of each grouped data.
1

42. GRADED ASSESSMENT:
2

43. GRADED ASSESSMENT:
3