Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Miscellaneous Exercise 3 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Skewness Miscellaneous Exercise 3

Question 1.

For u distribution, mean = 100, mode = 80 and S.D. = 20. Find Pearsonian coefficient of skewness Sk_{p}.

Solution:

Given, Mean = 100, Mode = 80, S.D. = 20

Question 2.

For a distribution, mean = 60, median = 75 and variance = 900. Find Pearsonian coefficient of skewness Sk_{p}.

Solution:

Given. Mean = 60, Median = 75, Variance = 900

∴ S.D. = √Variance = √900 = 30

Question 3.

For a distribution, Q_{1} = 25, Q_{2} = 35 and Q_{3} = 50. Find Bowley’s coefficient of skewness Sk_{b}.

Solution:

Given Q_{1} = 25, Q_{2} = 35, Q_{3} = 50

Question 4.

For a distribution Q_{3} – Q_{2} = 40, Q_{2} – Q_{1} = 60. Find Bowlev’s coefficient of skewness Sk_{b}.

Solution:

Given, Q_{3} – Q_{2} = 40, Q_{2} – Q_{1} = 60

Question 5.

For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.

Solution:

Given, Sk_{b} = 0.6, Q_{3} + Q_{1} = 100,

Median = Q_{2} = 38

Skb = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)

∴ 0.6 = \(\frac{100-2(38)}{Q_{3}-Q_{1}}\)

∴ 0.6(Q_{3} – Q_{1}) = 100 – 76 = 24

∴ Q_{3} – Q_{1} = 40 ….(i)

Q_{3} + Q_{1} = 100 …..(ii) (given)

Adding (i) and (ii), we get

2Q_{3} = 140

∴ Q_{3} = 70

Substituting the value of Q_{3} in (ii), we get

70 + Q_{1} = 100

∴ Q_{1} = 100 – 70 = 30

∴ upper quartile = 70 and lower quartile = 30

Question 6.

For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.

Solution:

Mean = \(\bar{x}\) = 200

Coefficient of variation, C.V. = 8%, Skp = 0.3

C.V. = \(\frac{\sigma}{\bar{x}} \times 100\), where σ = standard deviation

∴ 8 = \(\frac{\sigma}{200} \times 100\)

∴ σ = \(\frac{8 \times 200}{100}\) = 16

Now, Sk_{p} = \(\frac{\text { Mean – Mode }}{\text { S.D. }}\)

∴ 0.3 = \(\frac{200-\text { Mode }}{16}\)

∴ 0.3 × 16 = 200 – Mode

∴ Mode = 200 – 4.8 = 195.2

Since, Mean – Mode = 3(Mean – Median)

∴ 200 – 195.2 = 3(200 – Median)

∴ 4.8 = 600 – 3Median

∴ 3Median = 600 – 4.8 = 595.2

∴ Median = 198.4

Question 7.

Calculate Karl Pearsonian’s coefficient of skewness Skp from the follow ing data:

Solution:

The given table is the cumulative frequency table of more than type.

From this table, we have to prepare the frequency distribution table and then calculate the value of Sk_{p}.

Construct the following table:

From the table, N = 120, Σf_{i}x_{i} = 5490 and \(\sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^{2}\) = 284600

Mean = \(\bar{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{5490}{120}\) = 45.75

Maximum frequency 42 is of the class 50 – 60

∴ Mode lies in the class 50 – 60

∴ L = 50, f_{1} = 42, f_{0} = 25, f_{2} = 13, h = 10

Alternate Method:

Let u = \(\frac{x-45}{10}\)

\(\overline{\mathrm{u}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\mathrm{N}}=\frac{9}{120}\) = 0.075

∴ \(\bar{x}\) = 45 + 10(\(\bar{u}\))

= 45 + 10(0.075)

= 45 + 0.75

= 45.75

Var(u) = \(\sigma_{u}^{2}=\frac{\sum f_{i} u_{i}{ }^{2}}{N}-(\bar{u})^{2}\)

= \(\frac{335}{120}\) – (0.075)^{2}

= 2.7917 – 0.0056

= 2.7861

Var(X) = h^{2} × Var(u)

= 100 × 2.7861

= 278.61

S.D. = √278.61 = 16.6916

Maximum frequency 42 is of the class 50 – 60.

∴ Mode lies in the class 50 – 60.

∴ L = 50, f_{1} = 42, f_{0} = 25, f_{2} = 13, h = 10

Question 8.

Calculate Bowley’s coefficient of skewness Skb from the following data.

Solution:

To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Here, N = 120

Q_{1} class = class containing the \(\left(\frac{N}{4}\right)^{t h}\) observation

∴ \(\frac{N}{4}=\frac{120}{4}\) = 30

Cumulative frequency which is just greater than (or equal to) 30 is 35.

∴ Q_{1} lies in the class 30-40.

∴ L = 30, h = 10, f = 13, c.f. = 22

Q_{2} class = class containing the \(\left(\frac{N}{2}\right)^{t h}\) observation

∴ \(\frac{\mathrm{N}}{2}=\frac{120}{2}\) = 60

Cumulative frequency which is just greater than (or equal to) 60 is 60.

∴ Q_{2} lies in the class 40-50.

∴ L = 40, h = 10, f = 25, c.f. = 35

Q_{3} class = class containing the \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation

∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90

Cumulative frequency which is just greater than (or equal to) 90 is 102.

∴ Q_{3} lies in the class 50 – 60

∴ L = 50, h = 10, f = 42, c.f. = 60

Question 9.

Find Sk_{p} for the following set of observations:

18, 27, 10, 25, 31, 13, 28

Solution:

The given data can be arranged in ascending order as follows:

10, 13, 18, 25, 27, 28, 31

Here, n = 7

∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation

= value of \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation

= value of 4th observation

= 25

For finding standard deviation, we construct the following table:

Question 10.

Find Skb for the following set of observations:

18, 27, 10, 25, 31, 13, 28

Solution:

The given data can be arranged in ascending order as follows:

10, 13, 18, 25, 27, 28, 31

Here, n = 7

∴ Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation

= value of 2nd observation

∴ Q_{1} = 13

Q_{2} = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 2\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation

= value of (2 × 2)th observation

= value of 4th observation

∴ Q_{2} = 25

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 2)th observation

= value of 6th observation

∴ Q_{3} = 28

Coefficient of skewness,

∴ Sk_{b} = -0.6