Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Ex 3.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 3 Skewness Ex 3.1

Question 1.

For a distribution, mean = 100, mode = 127 and S.D. = 60. Find the Pearson coefficient of skewness Sk_{p}.

Solution:

Given, Mean = 100, Mode = 127, S.D. = 60

Question 2.

The mean and variance of a distribution are 60 and 100 respectively. Find the mode and the median of the distribution if Sk_{p} = -0.3.

Solution:

Given, Mean = 60, Variance = 100, Sk_{p} = -0.3

∴ S.D. = √Variance = √100 = 10

Sk_{p} = \(\frac{\text { Mean }-\text { Mode }}{\text { S.D. }}\)

∴ -0.3 = \(\frac{60-\text { Mode }}{10}\)

∴ -3 = 60 – Mode

∴ Mode = 60 + 3 = 63

Mean – Mode = 3 (Mean – Median)

∴ 60 – 63 = 3(60 – Median)

∴ -3 = 180 – 3Median

∴ 3Median = 180 + 3 = 183

∴ Median = \(\frac{183}{3}\)

∴ Median = 61

Question 3.

For a data set, sum of upper and lower quartiles is 100, difference between upper and lower quartiles is 40 and the median is 30. Find the coefficient of skewness.

Solution:

Given, Q_{3} + Q_{1} = 100 ……(i)

Q_{3} – Q_{1} = 40 …..(ii)

Median = Q_{2} = 30

Adding (i) and (ii), we get

2Q_{3} = 140

∴ Q_{3} = 70

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

70 + Q_{1} = 100

∴ Q_{1} = 100 – 70 = 30

Question 4.

For a data set with an upper quartile equal to 55 and median equal to 42, if the distribution is symmetric, find the value of the lower quartile.

Solution:

Upper quartile = Q_{3} = 55

Median = Q_{2} = 42

Since, the distribution is symmetric.

∴ Sk_{b} = 0

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

∴ 0 = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)

∴ 0 = Q_{3} + Q_{1} – 2Q_{2}

∴ Q_{1} = 2Q_{2} – Q_{3}

∴ Q_{1} = 2(42) – 55

∴ Q_{1} = 84 – 55

∴ Q_{1} = 29

Question 5.

Obtain coefficient of skewness by formula and comment on the nature of the distribution.

Solution:

We construct the less than cumulative frequency table as given below.

Q_{1} class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation

∴ \(\frac{\mathrm{N}}{4}=\frac{82}{4}\) = 20.5

Cumulative frequency which is just greater than (or equal) to 20.5 is 30.

∴ Q_{1} lies in the class 60 – 64.

∴ L = 60, h = 4, f = 20, c.f. = 10

Q_{2} class = class containing \(\left(\frac{\mathrm{N}}{2}\right)^{\mathrm{th}}\) observation

∴ \(\frac{\mathrm{N}}{2}=\frac{82}{2}\) = 41

Cumulative frequency which is just greater than (or equal) to 41 is 70.

∴ Q_{2} lies in the class 64 – 68.

∴ L = 64, h = 4, f = 40, c.f. = 30

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

∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 82}{4}\) = 61.5

Cumulative frequency which is just greater than (or equal) to 61.5 is 70.

∴ Q_{3} lies in the class 64 – 68.

∴ L = 64, h = 4, f = 40, c.f. = 30

∴ Sk_{b} = -0.1881

Since, Sk_{b} < 0, the distribution is negatively skewed.

Question 6.

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

17, 17, 21, 14, 15, 20, 19, 16, 13, 17, 18

Solution:

Σx_{i} = 17 + 17 + 21 + 14 + 15 + 20 + 19 + 16 + 13 + 17 + 18 = 187

Mean = \(\frac{\sum x_{i}}{n}=\frac{187}{11}\) = 17

Mode = Observation that occurs most frequently in the data = 17

Question 7.

Calculate Sk_{b} for the following set of observations of the yield of wheat in kg from 13 plots:

4.6, 3.5, 4.8, 5.1, 4.7, 5.5, 4.7, 3.6, 3.5, 4.2, 3.5, 3.6, 5.2

Solution:

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

3.5, 3.5, 3.5, 3.6, 3.6, 4.2, 4.6, 4.7, 4.7, 4.8, 5.1, 5.2, 5.5

Here, n = 13

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

= value of \(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation

= value of (3.50)th observation

= value of 3rd observation + 0.50(value of 4th observation – value of 3rd observation)

= 3.5 + 0.50(3.6 – 3.5)

= 3.5 + 0.50(0.1)

= 3.5 + 0.05

∴ Q_{1} = 3.55

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

= value of 2\(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation

= value of (2 × 3.50)th observation

= value of 7th observation

∴ Q_{2} = 4.6

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

= value of 3\(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 3.50)th observation

= value of (10.50)th observation

= value of 10th observation + 0.50 (value of 11th obseration – value of 10th observation)

= 4.8 + 0.50(5.1 – 4.8)

= 4.8 + 0.50(0.3)

∴ Q_{3} = 4.95

∴ Sk_{b} = -0.5

Question 8.

For a frequency distribution Q_{3} – Q_{2} = 90 and Q_{2} – Q_{1} = 120. Find Sk_{b}.

Solution:

Given, Q_{2} – Q_{1} = 90, Q_{2} – Q_{1} = 120

∴ Sk_{b} = -0.1429