Class 11

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Correlation Miscellaneous Exercise 5

Question 1.
Two series of x and y with 50 items each have standard deviations of 4.8 and 3.5 respectively. If the sum of products of deviations of x and y series from respective arithmetic means is 420, then find the correlation coefficient between x and y.
Solution:

Question 2.
Find the number of pairs of observations from the following data,
r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40.
Solution:
Given, r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40

Question 3.
Given that r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900. Find the number of pairs of observations.
Solution:
Given, r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900

Question 4.
Given the following information, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5, where x_i and y_i are the deviations from their respective means, find the number of items.
Solution:
Here, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5
Here, x_i and y_i are the deviations from their respective means.
∴ If X_i, Y_i are elements of x and y series respectively, then
X_i – \(\bar{x}\) = x_i and Y_i – \(\bar{y}\) = y_i
∴ Σx_iy_i = \(\Sigma\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)\left(\mathrm{Y}_{\mathrm{i}}-\bar{y}\right)\) = 60, \(\sum x_{\mathrm{i}}^{2}=\sum\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)^{2}\) = 90

Question 5.
A sample of 5 items is taken from the production of a firm. The length and weight of 5 items are given below. [Given: √0.8823 = 0.9393]

Calculate the correlation coefficient between length and weight and interpret the result.
Solution:
Let length = x_i (in cm), Weight = y_i (in gm)


∴ the value of r indicates a high degree of positive correlation between length and weight of items.

Question 6.
Calculate the correlation coefficient from the following data and interpret it.

Solution:


∴ the value of r indicates a perfect negative correlation between x and y.

Question 7.
Calculate the correlation coefficient from the following data and interpret it.

Solution:



∴ the value of r indicates a perfect positive correlation between x and y.

Question 8.
If the correlation coefficient between X and Y is 0.8, what is the correlation coefficient between
(i) 2X and Y
(ii) \(\frac{X}{2}\) and Y
(iii) X and 3Y
(iv) X – 5 and Y – 3
(v) X + 7 and Y + 9
(vi) \(\frac{X-5}{7}\) and \(\frac{Y-3}{8}\)?
Solution:
The correlation coefficient remains unaffected by the change of origin and scale.
i.e., if u_i = \(\frac{x_{i}-\mathrm{a}}{\mathrm{h}}\) and v_i = \(\frac{y_{i}-\mathrm{b}}{\mathrm{k}}\), then Corr(U, V) = ±Corr(X, Y).
according to the same or opposite signs of h and k.
(i) u_i = \(\frac{2\left(x_{i}-0\right)}{1}\), v_i = \(\frac{y_{i}-0}{1}\)
Here, h = 1 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(ii) u_i = \(\frac{x_{i}-0}{2}\), v_i = \(\frac{y_{\mathrm{i}}-0}{1}\)
Here, h = 2 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(iii) Corr (X, 3Y) = Corr (X, Y) = 0.8

(iv) Corr (X – 5, Y – 3) = Corr(X, Y) = 0.8

(v) Corr (X + 7, Y + 9) = Corr(X, Y) = 0.8

(vi) Corr(\(\frac{X-5}{7}, \frac{Y-3}{8}\)) = Corr(X, Y) = 0.8

Question 9.
In the calculation of the correlation coefficient between the height and weight of a group of students of a college, one investigator took the measurements in inches and pounds while the other investigator took the measurements in cm. and kg. Will they get the same value of the correlation coefficient or different values? Justify your answer.
Solution:
The coefficient of correlation is a ratio of covariance and standard deviations.
Since covariance and standard deviations are independent of units of measurement.
∴ coefficient of correlation is also independent of units of measurement.
∴ values of coefficient of correlation obtained by first and second investigators are the same.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Correlation Ex 5.1

Question 1.
Draw a scatter diagram for the data given below and interpret it.

Solution:

Since all the points lie in a band rising from left to right.
Therefore, there is a positive correlation between the values of X and Y respectively.

Question 2.
For the following data of marks of 7 students in Physics (x) and Mathematics (y), draw scatter diagram and state the type of correlation.

Solution:
We take marks in Physics on X-axis and marks in Mathematics on Y-axis and plot the points as below.

We get a band of points rising from left to right. This indicates the positive correlation between marks in Physics and marks in Mathematics.

Question 3.
Draw a scatter diagram for the data given below. Is there any correlation between Aptitude score and Grade points?

Solution:

The points are completely scattered i.e., no trend is observed.
∴ there is no correlation between Aptitude score (X) and Grade point (Y).

Question 4.
Find correlation coefficient between x andy series for the following data:
n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122
Solution:
Here, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122

Question 5.
The correlation coefficient between two variables x and y are 0.48. The covariance is 36 and the variance of x is 16. Find the standard deviation of y.
Solution:
Given, r = 0.48, Cov(X, Y) = 36
Since \(\sigma_{X}^{2}\) = 16
∴ σ_x = 4

∴ the standard deviation of y is 18.75.

Question 6.
In the following data, one of the values of y is missing. Arithmetic means of x and y series are 6 and 8 respectively. (√2 = 1.4142)

(i) Estimate missing observation.
(ii) Calculate correlation coefficient.
Solution:
(i) Let X = x_i, Y = y_i and missing observation be ‘a’.
Given, \(\bar{x}\) = 6, \(\bar{y}\) = 8, n = 5
∴ 8 = \(\frac{35+a}{5}\)
∴ 40 = 35 + a
∴ a = 5
(ii) We construct the following table:

Question 7.
Find correlation coefficient from the following data. [Given: √3 = 1.732]

Solution:

Question 8.
The correlation coefficient between x and y is 0.3 and their covariance is 12. The variance of x is 9, find the standard deviation of y.
Solution:
Given, r = 0.3, Cov(X, Y) = 12,
\(\sigma_{X}^{2}\) = 9
∴ \(\sigma_{\mathrm{X}}\) = 3

∴ the standard deviation of y is 13.33.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4

Question 1.
Following data gives the coded price (X) and demand (Y) of a commodity.

Classify the data by taking classes 0 – 4, 5 – 9, etc. for X and 5 – 8, 9 – 12, etc. for Y.
Also find
(i) marginal frequency distribution of X and Y.
(ii) conditional frequency distribution of Y when X is less than 10.
Solution:
Given, X = coded price
Y = demand
Bivariate frequency table can be prepared by taking class intervals 0 – 4, 5 – 9,… etc for X and 5 – 8, 9 – 12,… etc for Y.
Bivariate frequency distribution is as follows.

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X < 10:

Question 2.
Following data gives the age in years and marks obtained by 30 students in an intelligence test.


Prepare a bivariate frequency distribution by taking class intervals 16 – 18, 18 – 20,…,etc. for age and 10 – 20, 20 – 30,…, etc. for marks.
Find
(i) marginal frequency distributions.
(ii) conditional frequency distribution of marks obtained when age of students is between 20 – 22.
Solution:
Let X = Age in years
Y = Marks
Bivariate frequency table can be prepared by taking class intervals 16 – 18, 18 – 20,…, etc for X and 10 – 20, 20 – 30,…, etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X is between 20 – 22:

Question 3.
Following data gives Sales (in Lakh ?) and Advertisement Expenditure (in Thousand ₹) of 20 firms.
(115, 61) (120, 60) (128, 61) (121, 63) (137, 62) (139, 62) (143, 63) (117, 65) (126, 64) (141, 65) (140, 65) (153, 64) (129, 67) (130, 66) (150, 67) (148, 66) (130, 69) (138, 68) (155, 69) (172, 68)
(i) Construct a bivariate frequency distribution table for the above data by taking classes 115 – 125, 125 – 135, ….etc. for sales and 60 – 62, 62 – 64, …etc. for advertisement expenditure.
(ii) Find marginal frequency distributions
(iii) Conditional frequency distribution of Sales when the advertisement expenditure is between 64 – 66 (Thousand ₹)
(iv) Conditional frequency distribution of advertisement expenditure when the sales are between 125 – 135 (lakh ₹)
Solution:
(i) Let X = Sales (in lakh ₹)
Y = Advertisement Expenditure (in Thousand ₹)
Bivariate frequency table can be prepared by taking class intervals 115 – 125, 125 – 135, …. etc for X and 60 – 62, 62 – 64, ….etc for Y.
Bivariate frequency distribution is as follows:

(ii) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of X when Y is between 64 – 66:

(iii) Conditional frequency distribution of Y when X is between 125 – 135:

Question 4.
Prepare a bivariate frequency distribution for the following data, taking class intervals for X as 35 – 45, 45 – 55, …. etc and for Y as 115 – 130, 130 – 145, … etc where, X denotes the age in years and Y denotes blood pressure for a group of 24 persons.
(55, 151) (36, 140) (72, 160) (38, 124) (65, 148) (46, 130) (58, 152) (50, 149) (38, 115) (42, 145) (41, 163) (47, 161) (69, 159) (60, 161) (58, 131) (57, 136) (43, 141) (52, 164) (59, 161) (44, 128) (35, 118) (62, 142) (67, 157) (70, 162)
Also find
(i) Marginal frequency distribution of X.
(ii) Conditional frequency distribution of Y when X < 45.
Solution:
Given X = Age in years
Y = Blood pressure
Bivariate frequency table can be prepared by taking class intervals 35 – 45, 45 – 55, …, etc for X and 115 – 130, 130 – 145, ….., etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

(ii) Conditional frequency distribution of Y when X < 45:

Question 5.
Thirty pairs of values of two variables X and Y are given below. Form a bivariate frequency table. Also find marginal frequency distributions of X and Y.

Solution:
Bivariate frequency table can be prepared by taking class intervals 80 – 90, 90 – 100, etc for X and 500 – 600, 600 – 700, …., etc for Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y

Question 6.
The following table shows how the samples of Mathematics and Economics scores of 25 students are distributed:

Find the value of ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{35 \times 25}{50}\) = 17.5
E₁₂ = \(\frac{35 \times 25}{50}\) = 17.5
E₂₁ = \(\frac{15 \times 25}{50}\) = 7.5
E₂₂ = \(\frac{15 \times 25}{50}\) = 7.5
Table of expected frequencies.

Question 7.
Compute ϰ² statistic from the following data:

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 60}{100}\) = 30
E₁₂ = \(\frac{50 \times 40}{100}\) = 20
E₂₁ = \(\frac{50 \times 60}{100}\) = 30
E₂₂ = \(\frac{50 \times 40}{100}\) = 20
Table of expected frequencies.

Question 8.
The attitude of 250 employees towards a proposed policy of the company is as observed in the following table. Calculate ϰ² statistic.

Solution:
Table of observed frequencies

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{150 \times 95}{250}\) = 57
E₁₂ = \(\frac{150 \times 95}{250}\) = 57
E₁₃ = \(\frac{150 \times 60}{250}\) = 36
E₂₁ = \(\frac{100 \times 95}{250}\) = 38
E₂₂ = \(\frac{100 \times 95}{250}\) = 38
E₂₃ = \(\frac{100 \times 60}{250}\) = 24
Table of observed frequencies.

Question 9.
In a certain sample of 1000 families, 450 families are consumers of tea. Out of 600 Hindu families, 286 families consume tea. Calculate ϰ² statistic.
Solution:
The given data can be arranged in the following table.

Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{600 \times 450}{1000}\) = 270
E₁₂ = \(\frac{600 \times 550}{1000}\) = 330
E₂₁ = \(\frac{400 \times 450}{1000}\) = 180
E₂₂ = \(\frac{400 \times 550}{1000}\) = 220
Table of expected frequencies.

Question 10.
A sample of boys and girls were asked to choose their favourite sport, with the following results. Find the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{200 \times 126}{300}\) = 84
E₁₂ = \(\frac{200 \times 90}{300}\) = 60
E₁₃ = \(\frac{200 \times 69}{300}\) = 46
E₁₄ = \(\frac{200 \times 15}{300}\) = 10
E₂₁ = \(\frac{100 \times 126}{300}\) = 42
E₂₂ = \(\frac{100 \times 90}{300}\) = 30
E₂₃ = \(\frac{100 \times 69}{300}\) = 23
E₂₄ = \(\frac{100 \times 15}{300}\) = 5
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2

Question 1.
The following table shows the classification of applications for secretarial and for sales positions according to gender. Calculate the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{225 \times 100}{300}\) = 75
E₁₂ = \(\frac{225 \times 200}{300}\) = 150
E₂₁ = \(\frac{75 \times 100}{300}\) = 25
E₂₂ = \(\frac{75 \times 200}{300}\) = 50
Table of expected frequencies.

Question 2.
200 teenagers were asked which takeaway food do they prefer – French fries, burgers, or pizza. The results were-

Compute ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 24}{200}\) = 6
E₁₂ = \(\frac{50 \times 60}{200}\) = 15
E₁₃ = \(\frac{50 \times 116}{200}\) = 29
E₂₁ = \(\frac{150 \times 24}{200}\) = 18
E₂₂ = \(\frac{150 \times 60}{200}\) = 45
E₂₃ = \(\frac{150 \times 116}{200}\) = 87
Table of expected frequencies.

Question 3.
A sample of men and women who had passed their driving test either in 1st attempt or in 2nd attempt
were surveyed. Compute ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{60 \times 40}{80}\) = 30
E₁₂ = \(\frac{60 \times 40}{80}\) = 30
E₂₁ = \(\frac{20 \times 40}{80}\) = 10
E₂₂ = \(\frac{20 \times 40}{80}\) = 10
Table of expected frequencies.

Question 4.
800 people were asked whether they wear glasses for reading with the following results.

Compute the ϰ² square statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{400 \times 600}{800}\) = 300
E₁₂ = \(\frac{400 \times 200}{800}\) = 100
E₂₁ = \(\frac{400 \times 600}{800}\) = 300
E₂₂ = \(\frac{400 \times 200}{800}\) = 100
Table of expected frequencies.

Question 5.
Out of a sample of 120 persons in a village, 80 were administered a new drug for preventing influenza, and out of the 18 were attacked by influenza. Out of those who are not administered the new drug, 10 persons were not attacked by influenza:
(i) Prepare a two-way table showing frequencies.
(ii) Compute the ϰ² square statistic.
Solution:
(i) The given data can be arranged in the following table.

The observed frequency table can be prepared as follows:

(ii) Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{48 \times 80}{120}\) = 32
E₁₂ = \(\frac{48 \times 40}{120}\) = 16
E₂₁ = \(\frac{72 \times 80}{120}\) = 48
E₂₂ = \(\frac{72 \times 40}{120}\) = 24
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1

Question 1.
The following table gives income (X) and expenditure (Y) of 25 families:

Find
(i) Marginal frequency distributions of income and expenditure.
(ii) Conditional frequency distribution of X when Y is between 300 – 400.
(iii) Conditional frequency distribution of Y when X is between 200 – 300.
(iv) How many families have their income ₹ 300 and more and expenses ₹ 400 and less?
Solution:
The bivariate frequency distribution table for Income (X) and Expenditure (Y) is as follows:

(i) Marginal frequency distribution of income (X):

Marginal frequency distribution of expenditure ( Y):

(ii) Conditional frequency distribution of X when Y is between 300 – 400:

(iii) Conditional frequency distribution of Y when X is between 200 – 300:

(iv) The cells 300 – 400 and 400 – 500 are having income ₹ 300 and more and the cells 200 – 300 and 300 – 400 are having expenditure ₹ 400 and less.
Now, the following table indicates the number of families satisfying the above condition.

∴ There are 17 families with income ₹ 300 and more and expenditure ₹ 400 and less.

Question 2.
Two dice are thrown simultaneously 25 times. The following pairs of observations are obtained.
(2, 3) (2, 5) (5, 5) (4, 5) (6, 4) (3, 2) (5, 2) (4, 1) (2, 5) (6, 1) (3, 1) (3, 3) (4, 3) (4, 5) (2, 5) (3, 4) (2, 5) (3, 4) (2, 5) (4, 3) (5, 2) (4, 5) (4, 3) (2, 3) (4, 1)
Prepare a bivariate frequency distribution table for the above data. Also, obtain the marginal distributions.
Solution:
Let X = Observation on 1st die
Y = Observation on 2nd die
Now, the minimum value of X is 1 and the maximum value is 6.
Also, the minimum value of Y is 1 and the maximum value is 6.
A bivariate frequency distribution can be prepared by taking X as row and Y as a column.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Question 3.
Following data gives the age of husbands (X) and age of wives (Y) in years. Construct a bivariate frequency distribution table and find the marginal distributions.

Find conditional frequency distribution of age of husbands when the age of wife is 23 years.
Solution:
Given, X = Age of Husbands (in years)
Y = Age of Wives (in years)
Now, the minimum value of X is 25 and the maximum value is 29.
Also, the minimum value of Y is 19 and the maximum value is 23.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of X when Y is 23:

Question 4.
Construct a bivariate frequency distribution table of the marks obtained by students in Statistics (X) and English (Y).

Construct a bivariate frequency distribution table for the above data by taking class intervals 20 – 30, 30 – 40, …. etc. for both X and Y. Also find the marginal distributions and conditional frequency distribution of Y when X lies between 30 – 40.
Solution:
Given, X = Marks in Statistics
Y = Marks in English
A bivariate frequency table can be prepared by taking class intervals 20 – 30, 30 – 40,…, etc for both X and Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when X lies between 30 – 40:

Question 5.
Following data gives height in cms (X) and weight in kgs (Y) of 20 boys. Prepare a bivariate frequency table taking class intervals 150 – 154, 155 – 159,…etc. for X and 35 – 39, 40 – 44,…, etc for Y. Also, find
(i) Marginal frequency distributions.
(ii) Conditional frequency distribution of Y when 155 ≤ X ≤ 159.
(152,40) (160,54) (163,52) (150,35) (154,36) (160,49) (166,54) (157,38)
(159,43) (153,48) (152,41) (158,51) (155,44) (156,47) (156,43) (166,53)
(160,50) (151,39) (153,50) (158,46)
Solution:
Given X = Height in cms.
Y = Weight in kgs.
Bivariate frequency table can be prepared by taking class intervals 150 – 154, 155 – 159, …, etc for X and 35 – 39, 40 – 44,…etc for Y.
The bivariate frequency distribution table is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when 155 ≤ X ≤ 159:

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, Q₁ = 25, Q₂ = 35 and Q₃ = 50. Find Bowley’s coefficient of skewness Sk_b.
Solution:
Given Q₁ = 25, Q₂ = 35, Q₃ = 50

Question 4.
For a distribution Q₃ – Q₂ = 40, Q₂ – Q₁ = 60. Find Bowlev’s coefficient of skewness Sk_b.
Solution:
Given, Q₃ – Q₂ = 40, Q₂ – Q₁ = 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, Q₃ + Q₁ = 100,
Median = Q₂ = 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(Q₃ – Q₁) = 100 – 76 = 24
∴ Q₃ – Q₁ = 40 ….(i)
Q₃ + Q₁ = 100 …..(ii) (given)
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (ii), we get
70 + Q₁ = 100
∴ Q₁ = 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_ix_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, f₁ = 42, f₀ = 25, f₂ = 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) = h² × 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, f₁ = 42, f₀ = 25, f₂ = 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
Q₁ 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.
∴ Q₁ lies in the class 30-40.
∴ L = 30, h = 10, f = 13, c.f. = 22

Q₂ 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.
∴ Q₂ lies in the class 40-50.
∴ L = 40, h = 10, f = 25, c.f. = 35

Q₃ 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.
∴ Q₃ 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
∴ Q₁ = 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
∴ Q₁ = 13
Q₂ = 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
∴ Q₂ = 25
Q₃ = 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
∴ Q₃ = 28
Coefficient of skewness,

∴ Sk_b = -0.6

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, Q₃ + Q₁ = 100 ……(i)
Q₃ – Q₁ = 40 …..(ii)
Median = Q₂ = 30
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (i), we get
70 + Q₁ = 100
∴ Q₁ = 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 = Q₃ = 55
Median = Q₂ = 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 = Q₃ + Q₁ – 2Q₂
∴ Q₁ = 2Q₂ – Q₃
∴ Q₁ = 2(42) – 55
∴ Q₁ = 84 – 55
∴ Q₁ = 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.

Q₁ 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.
∴ Q₁ lies in the class 60 – 64.
∴ L = 60, h = 4, f = 20, c.f. = 10

Q₂ 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.
∴ Q₂ lies in the class 64 – 68.
∴ L = 64, h = 4, f = 40, c.f. = 30

Q₃ 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.
∴ Q₃ 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
Q₁ = 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
∴ Q₁ = 3.55
Q₂ = 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
∴ Q₂ = 4.6
Q₃ = 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)
∴ Q₃ = 4.95

∴ Sk_b = -0.5

Question 8.
For a frequency distribution Q₃ – Q₂ = 90 and Q₂ – Q₁ = 120. Find Sk_b.
Solution:
Given, Q₂ – Q₁ = 90, Q₂ – Q₁ = 120

∴ Sk_b = -0.1429

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2

Question 1.
Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Solution:
Here, largest value (L) = 203, smallest value (S) = 114
∴ Range = L – S
= 203 – 114
= 89

Question 2.
Given below the frequency distribution of weekly w ages of 400 workers. Find the range.

Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Question 3.
Find the range of the following data.

Solution:
Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115
∴ Range = L – S
= 175 – 115
= 60

Question 4.
The city traffic police issued challans for not observing the traffic rules:

Find Q.D.
Solution:
The given data can be arranged in ascending order as follows:
24, 36, 40, 58, 62, 80
Here, n = 6
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (1.75)th observation
= value of 1st observation + 0.75(value of 2nd observation – value of 1st observation)
= 24 + 0.75(36 – 24)
= 24 + 0.75(12)
= 24 + 9
∴ Q₁ = 33
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 1.75)th observation
= value of (5.25)th observation
= value of 5th observation + 0.25(value of 6th observation – value of 5th observation)
= 62 + 0.25(80 – 62)
= 62 + 0.25(18)
= 62 + 4.5
= 66.5
∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}=\frac{66.5-33}{2}=\frac{33.5}{2}\) = 16.75

Question 5.
Calculate Q.D. from the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 35
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{35}{4}\) = 8.75
Cumulative frequency which is just greater than (or equal to) 8.75 is 15.
∴ Q₁ lies in the class 20-30.
∴ L = 20, c.f. = 8, f = 7, h = 10

Question 6.
Calculate the appropriate measure of dispersion for the following data.

Solution:
Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.
We construct the less than cumulative frequency table as follows:

Here N = 250
Q₁ class class containing \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{250}{4}\) = 62.5
Cumulative frequency which is just greater than (or equal to) 62.5 is 65.
∴ Q₁ lies in the class 35-40.
∴ L = 35, c.f. = 15, f = 50, h = 5

The cumulative frequency which is just greater than (or equal to) 187.5 is 190.
∴ Q₃ lies in the class 45-50.
∴ L = 45, c.f. = 150, f = 40, h = 5

Question 7.
Calculate Q.D. of the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 120
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 4-6.
∴ L = 4, c.f. = 15, f = 20, h = 2

Cumulative frequency which is just greater than (or equal to) 90 is 90.
∴ Q₃ lies in the class 10-12.
∴ L = 10, c.f. = 72, f = 18, h = 2

Question 8.
Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 27, 25 (Given √6.6 = 2.57)
Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 9.
Following data gives no. of goals scored by a team in 90 matches.

Compute the variance and standard deviation for the above data.
Solution:
We prepare the following table for the calculation of variance and S.D:

Question 10.
Compute the arithmetic mean and S.D. and C.V. (Given √296 = 17.20)

Solution:
We prepare the following table for calculation of arithmetic mean and S.D.:

Question 11.
The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.
Solution:
Here, n = 200, \(\bar{x}\) = Mean = 60, S.D. = 20
Wrongly taken items are 3 and 67.
Correct items are 13 and 17.
Now, \(\bar{x}\) = 60

Correct value of \(\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i}\) (sum of wrongly taken items) + (sum of correct items)
= 12000 – (3 + 67) + (13 + 17)
= 12000 – 70 + 30
= 11960
Correct value of mean = \(\frac{1}{n}\) × correct value of \(\sum_{i=1}^{n} x_{i}\)
= \(\frac{1}{200}\) × 11960
= 59.8
Now, S.D. = 20
Variance = (S.D.)² = 20²
∴ Variance = 400
∴ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^{2}-(\bar{x})^{2}=400\)

Correct value of \(\sum_{i=1}^{n} x_{i}^{2}\)
= \(\sum_{i=1}^{n} x_{i}^{2}\) – (Sum of squares of wrongly taken items) + (Sum of squares of correct items)
= 800000 – (3² + 67²) + (13² + 17²)
= 800000 – (9 + 4489) + (169 + 289)
= 800000 – 4498 + 458
= 795960
∴ Correct value of Variance = (\(\frac{1}{n}\) × \(\sum_{i=1}^{n} x_{i}^{2}\)) – (correct value of \(\bar{x}\))²
= \(\frac{1}{200}\) × 795960 – (59.8)²
= 3979.8 – 3576.04
= 403.76
∴ The correct mean is 59.8 and correct variance is 403.76.

Question 12.
The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.
Solution:

Question 13.
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Solution:
Let n₁ and n₂ be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{\mathrm{c}}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{1}\) = 62, σ₁ = 8, σ₂ = 10
Here, n₁ + n₂ = n
∴ n₁ + n₂ = 200 …….(i)
Combined mean is given by

∴ 70n₁ + 62n₂ = 13000
∴ 35n₁ + 31n₂ = 6500 ……..(ii)
Solving (i) and (ii), we get
n₁ = 75, n₂ = 125
Combined standard deviation is given by,

Question 14.
From the following data available for 5 pairs of observations of two variables x and y, obtain the combined S.D. for all 10 observations,
where \(\sum_{i=1}^{n} x_{i}=30, \sum_{i=1}^{n} y_{i}=40, \sum_{i=1}^{n} x_{i}^{2}=225, \sum_{i=1}^{n} y_{i}^{2}=340\)
Solution:

Question 15.
The mean and standard deviations of two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.
Solution:

Since C.V. (I) > C.V. (II)
∴ the brand I is more variable.

Question 16.
Calculate the coefficient of variation for the data given below. [Given √3.3 = 1.8166]

Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Ex 2.3

Question 1.
The mean and standard deviation of two distributions of 100 and 150 items are 50, 5, and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:


∴ The mean and standard deviation of all 250 items taken together are 44 and √55.6 respectively.

Question 2.
For certain bivariate data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are: 2, 4, 6, 8, 10. (Given: √8 = 2.8284)
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:
Given, \(\bar{x}\) = 25, σ = 5
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
= 100 × \(\frac{5}{25}\)
= 20%

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variation of 2.5%. What is the standard deviation of their height?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
∴ 2.5 = 100 × \(\frac{\sigma}{150.4}\)
∴ \(\frac{2.5 \times 150.4}{100}\) = σ
∴ σ = 3.76
∴ the standard deviation of students’ height is 3.76.

Question 6.
Two workers on the same job show1 the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since, C.V. (P) < C.V.(Q)
∴ Worker P is more consistent regarding the time required to complete the job.

(ii) Since, \(\bar{p}\) > \(\bar{q}\)
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since, \(\bar{x}_{1}>\bar{x}_{2}\)
i.e., average weekly wages are more for the first department.
∴ the first department has a larger bill.
(ii) Since, C.V. (1) < C.V. (2)
∴ the second department is less consistent.
∴ the second department has larger variability in wages.

Question 8.
The following table gives the weights of the students of class A. Calculate the coefficient of variation (Given √8 = 0.8944)

Solution:

Question 9.
Compute coefficient of variation for team A and team B. (Given: √2.5162 = 1.5863, √2.244 = 1.4980)

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.


Since C.V. of team A > C.V. of team B.
∴ team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M)
∴ The subject statistics show higher variability in marks.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Ex 2.2

Question 1.
Find the variance and S.D. for the following set of numbers.
7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
Solution:
Given data: 7, 11, 2, 4, 9, 6, 3, 7, 11, 2, 5, 8, 3, 6, 8, 8, 2, 6
The tabulated form of the above data is as given below:

We prepare the following table for the calculation of variance and S. D.

Question 2.
Find the variance and S.D. for the following set of numbers.
65, 77, 81, 98, 100, 80, 129
Solution:

Question 3.
Compute the variance and standard deviation for the following data:

Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 4.
Compute the variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 5.
The following data gives the age of 100 students in a school. Calculate variance and S.D.

Solution:
We prepare the following table for the calculation of variance and S.D:

Question 6.
The mean and variance of 5 observations are 3 and 2 respectively. If three of the five observations are 1, 3, and 5, find the values of the other two observations.
Solution:


∴ (x₄ – 4)(x₄ – 2) = 0
∴ x₄ = 4 or x₄ = 2
From (i), we get
x₅ = 2 or x₅ = 4
∴ The two numbers are 2 and 4.

Question 7.
Obtain standard deviation for the following data:

Solution:
We prepare the following table for the calculation of standard deviation.

Question 8.
The following distribution was obtained by change of origin and scale of variable X.

If it is given that mean and variance are 59.5 and 413 respectively, determine actual class intervals.
Solution:
Here, Mean = \(\bar{x}\) = 59.5, and
Var(X) = σ² = 413
Let x_i be a mid value of class and
d = \(\frac{x-a}{h}\), where a is assumed mean and h is class width.
We prepare the following table for calculation of mean and variance of d_i.


Now, Var(X) = h². Var(D)
∴ 413 = h² × 4.13
∴ h² = 100
∴ h = 10
Substituting h = 10 in (i), we get
-0.1 × 10 + a = 59.5
∴ -1 + a = 59.5
∴ a = 59.5 + 1
∴ a = 60.5
We prepare the following table to determine actual class intervals for corresponding values of d_i.

∴ The actual class intervals are 15.5 – 25.5, 25.5 – 35.5, …….., 95.5 – 105.5