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Lesson Example Discussion Quiz: Class Homework

Step-4

Title: Bowley’s Coefficient of skewness

Grade: 9-a Lesson: S2-L8

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step Type Explanation Answer

1

Problem

Find Bowley’s coefficient of skewness for the following data

2

Hint

To know the Bowley’s coefficient of skewness, we need to calculate three quartiles they are \$N / 4 , N / 2 ,(3N) / 4\$

3

Formula:

Bowley’s coefficient of skewness formula sk = \$(Q_3 + Q_1 - 2Q_2) / (Q_3 - Q_1)\$

4

Step

1

5

Step

Total frequency

N = 120

6

Step

class intervel

h = 10

7

Clue

First quartile Q_1 = \$"value of" (N/4)^(th) "observation"\$

8

Step

simplification of first quartile

\$Q_1\$ = \$120/4\$ = 30

9

Step

From the c.f \$30^(th)\$ observation is belongs to the class 80 - 90

10

Step

For the first quartile the frequency f = 18,
cumulative frequency of previous class CF = 23
and lower limit L = 80

11

Formula:

Formula for first quartile \$Q_1\$ = \$L + (h(N / 4 - CF)) / f\$

12

Step

simplification

\$Q_1\$ = \$80 + (10 (30 - 23)) / 18\$

13

Step

simplification

\$Q_1\$ = \$80 + (10 (7)) / 18\$

14

Step

simplification

\$Q_1\$ = \$80 + (70) / 18\$

15

Step

simplification

\$Q_1\$ = \$80 + 3.88\$

16

Step

After simplification we get

\$Q_1\$ = 83.89

17

Clue

Second quartile \$Q_2\$ = \$"value of" (N/2)^(th) "observation"\$

18

Step

simplification of second quartile

\$Q_2\$ = \$120/2\$ = 60

19

Step

From the c.f \$60^(th)\$ observation is belongs to the class 90 - 100

20

Step

For the second quartile the frequency f = 25,
cumulative frequency of previous class CF = 41
and lower limit L = 90

21

Formula:

Formula for second quartile \$Q_2\$ = \$L + (h(N / 2 - CF)) / f\$

22

Step

simplification

\$Q_2\$ = \$90 + (10 (60 - 41)) / 25\$

23

Step

simplification

\$Q_2\$ = \$90 + (10 (19)) / 25\$

24

Step

simplification

\$Q_2\$ = \$90 + (190) / 25\$

25

Step

simplification

\$Q_2\$ = \$90 + 7.6\$

26

Step

After simplification we get

\$Q_2\$ = 97.6

27

Clue

Third quartile \$Q_3\$ = \$"value of" ((3N)/4)^(th) "observation"\$

28

Step

simplification of third quartile

\$Q_3\$ = \$(3*120)/4\$ = 90

29

Step

From the c.f \$90^(th)\$ observation is belongs to the class 100 - 110

30

Step

For the third quartile the frequency f = 19,
cumulative frequency of previous class CF = 18
and lower limit L = 100

31

Formula:

Formula for third quartile \$Q_3\$ = \$L + (h((3N) / 4 - CF)) / f\$

32

Step

simplification

\$Q_3\$ = \$100 + (10 (90 - 18)) / 19\$

33

Step

simplification

\$Q_3\$ = \$100 + (10 (72)) / 19\$

34

Step

simplification

\$Q_3\$ = \$100 + (720) / 19\$

35

Step

simplification

\$Q_3\$ = \$100 + 37.89\$

36

Step

After simplification we get

\$Q_3\$ = 137.89

37

Formula:

Bowley’s coefficient of skewness formula sk = \$(Q_3 + Q_1 - 2Q_2) / (Q_3 - Q_1)\$

38

Step

simplification

sk = \$(137.89 + 83.89 - 2(97.6)) / (137.89 - 83.89)\$

39

Step

simplification

sk = \$(221.78 - 195.2) / 54\$

40

Step

After simplification we get

sk = 0.49

41

Answer

D

Tutor: Questions

Seq Type Question Audio

1

Problem

What did you learn from this problem?

2

Clue

What did you learn from the clues?

3

Hint

What did you learn from the Hints?

4

Step

What did you learn from the Steps?

5

Step

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