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Lesson Example Discussion Quiz: Class Homework

Step-1

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 = 100

6

Step

class intervel

h = 100

7

Clue

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

8

Step

simplification of first quartile

\$Q_1\$ = \$100/4\$ = 25

9

Step

From the c.f \$25^(th)\$ observation is belongs to the class 400 - 500

10

Step

For the first quartile the frequency f = 18,
cumulative frequency of previous class CF = 13
and lower limit L = 400

11

Formula:

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

12

Step

simplification

\$Q_1\$ = \$400 + (100 (25 - 13)) / 18\$

13

Step

simplification

\$Q_1\$ = \$400 + (100 (12)) / 18\$

14

Step

simplification

\$Q_1\$ = \$400 + (1200) / 18\$

15

Step

simplification

\$Q_1\$ = \$400 + 66.67\$

16

Step

After simplification we get

\$Q_1\$ = 466.67

17

Clue

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

18

Step

simplification of second quartile

\$Q_2\$ = \$100/2\$ = 50

19

Step

From the c.f \$50^(th)\$ observation is belongs to the class 500 - 600

20

Step

For the second quartile the frequency f = 35,
cumulative frequency of previous class CF = 31
and lower limit L = 500

21

Formula:

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

22

Step

simplification

\$Q_2\$ = \$500 + (100 (50 - 31)) / 35\$

23

Step

simplification

\$Q_2\$ = \$500 + (100 (19)) / 35\$

24

Step

simplification

\$Q_2\$ = \$500 + (1900) / 35\$

25

Step

simplification

\$Q_2\$ = \$500 + 54.29\$

26

Step

After simplification we get

\$Q_2\$ = 554.29

27

Clue

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

28

Step

simplification of third quartile

\$Q_3\$ = \$(3*100)/4\$ = 75

29

Step

From the c.f \$75^(th)\$ observation is belongs to the class 600 - 700

30

Step

For the third quartile the frequency f = 27,
cumulative frequency of previous class CF = 66
and lower limit L = 600

31

Formula:

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

32

Step

simplification

\$Q_3\$ = \$600 + (100 (75 - 66)) / 27\$

33

Step

simplification

\$Q_3\$ = \$600 + (100 (9)) / 27\$

34

Step

simplification

\$Q_3\$ = \$600 + (900) / 27\$

35

Step

simplification

\$Q_3\$ = \$600 + 33.33\$

36

Step

After simplification we get

\$Q_3\$ = 633.33

37

Formula:

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

38

Step

simplification

sk = \$(633.33 + 466.67 - 2(554.29)) / (633.33 - 466.67)\$

39

Step

simplification

sk = \$(1100 - 1108.58) / 166.67\$

40

Step

After simplification we get

sk = -0.05

41

Answer

A

Tutor: Questions

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1

Problem

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2

Clue

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3

Hint

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4

Step

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5

Step

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