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Lesson Example Discussion Quiz: Class Homework

Step-2

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 30 - 40

10

Step

For the first quartile the frequency f = 13,
cumulative frequency of previous class CF = 22
and lower limit L = 30

11

Formula:

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

12

Step

simplification

\$Q_1\$ = \$30 + (10 (30 - 22)) / 13\$

13

Step

simplification

\$Q_1\$ = \$30 + (10 (8)) / 13\$

14

Step

simplification

\$Q_1\$ = \$30 + (80) / 13\$

15

Step

simplification

\$Q_1\$ = \$30 + 6.15\$

16

Step

After simplification we get

\$Q_1\$ = 36.15

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 40 - 50

20

Step

For the second quartile the frequency f = 25,
cumulative frequency of previous class CF = 35
and lower limit L = 40

21

Formula:

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

22

Step

simplification

\$Q_2\$ = \$40 + (10 (60 - 35)) / 25\$

23

Step

simplification

\$Q_2\$ = \$40 + (10 (25)) / 25\$

24

Step

simplification

\$Q_2\$ = \$40 + (250) / 25\$

25

Step

simplification

\$Q_2\$ = \$40 + 10\$

26

Step

After simplification we get

\$Q_2\$ = 50

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 50 - 60

30

Step

For the third quartile the frequency f = 42,
cumulative frequency of previous class CF = 60
and lower limit L = 50

31

Formula:

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

32

Step

simplification

\$Q_3\$ = \$50 + (10 (90 - 60)) / 42\$

33

Step

simplification

\$Q_3\$ = \$50 + (10 (30)) / 42\$

34

Step

simplification

\$Q_3\$ = \$50 + (300) / 42\$

35

Step

simplification

\$Q_3\$ = \$50 + 7.14\$

36

Step

After simplification we get

\$Q_3\$ = 57.14

37

Formula:

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

38

Step

simplification

sk = \$(57.14 + 36.15 - 2(50)) / (57.14 - 36.15)\$

39

Step

simplification

sk = \$(93.29 - 100) / 20.99\$

40

Step

After simplification we get

sk = -0.31

41

Answer

B

Tutor: Questions

Seq Type Question Audio

1

Problem

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2

Clue

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3

Hint

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4

Step

What did you learn from the Steps?

5

Step

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