Question 1: Two plants A and B  of a factory show following results about the number of workers and the wages paid to them.

$\displaystyle \begin{array} {|l | l | l |} \hline & \text{Plant A} & \text{Plant B} \\ \hline \text{No. of workers } & 5000 & 6000 \\ \hline \text{Average monthly wages } & \text{Rs. } 2500 & \text{Rs. } 2500 \\ \hline \text{Variance of distribution of wages } & 81 & 100 \\ \hline \end{array}$

In which plant A or B is there greater variability in individual wages?

Variance of distribution of wages in plants:

$\displaystyle \text{Plant A} ( \sigma^2 ) = 81 \hspace{2.0cm} \text{Plant B} ( \sigma^2 ) = 100$

Standard deviation of the distribution of wages in plants

$\displaystyle \text{Plant A} ( \sigma ) = 9 \hspace{2.0cm} \text{Plant B} ( \sigma ) = 10$

We observe that the average monthly wages in both the firms is same i.e. Rs. 2500. Therefore the firm with greater variance will have more variability. Thus, firm B has greater variability in individual wages.

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Question 2: The means and standard deviations of heights and weights of 50 students of a class are as follows:

$\displaystyle \begin{array} {|l | l | l |} \hline & \text{Weights} & \text{Heights} \\ \hline \text{Mean} & 63.2 \text{ kg } & 63.2 \text{ inch } \\ \hline \text{Standard deviations } & 5.6 \text{ kg } & 11.5 \text{ inch } \\ \hline \end{array}$

Which shows more variability, heights or weight?

$\displaystyle \text{Coefficient of variation in weights }= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{5.6}{63.2} \times 100 = 8.86$

$\displaystyle \text{Coefficient of variation in heights }= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{11.5}{63.2} \times 100 = 18.19$

$\displaystyle \text{Coefficient of variation in heights } > \text{Coefficient of variation in weights }$

Thus, Height has a greater variability than weights.

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Question 3:  Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

$\displaystyle \text{We know: Coefficient of Variation } (CV) = \frac{\sigma}{\overline{X}} \times 100$

$\displaystyle \overline{X_1} = \frac{21}{60} \times 100 = 35$

$\displaystyle \overline{X_2} = \frac{16}{70} \times 100 = 22.86$

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Question 4: Calculate coefficient of variation from the following data:

$\displaystyle \begin{array} {|l | c | c | c | c| c | c |} \hline \text{Income (in Rs.) } & 1000-1700 & 1700-2400 & 2400 - 3100 & 3100 - 3800 & 3800-4500 & 4500-5200 \\ \hline \text{No. of families: } & 12 & 18 & 20 & 25 & 35 & 10 \\ \hline \end{array}$

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 3450}{700} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 1000-1700 & 12 & 1350 & -3 & -36 & 9 & 108 \\ \hline 1700-2400 & 18 & 2050 & -2 & -36 & 4 & 72 \\ \hline 2400-3100 & 20 & 2750 & -1 & -20 & 1 & 20 \\ \hline 3100 - 3800 & 25 & 3450 & 0 & 0 & 0 & 0 \\ \hline 3800-4500 & 35 & 4150 & 1 & 35 & 1 & 35 \\ \hline 4500 - 5200 & 10 & 4850 & 2 & 20 & 4 & 40 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 120 & & & = -37 & & = 275 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 120, \hspace{0.5cm} \sum f_iu_i = -37, \hspace{0.5cm} \sum f_i{u_i}^2= 275 \hspace{0.5cm} A = 3450, \hspace{0.5cm} h = 700$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 3450 + 700 \Big( \frac{-37}{120} \Big) = 3234.17$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 490000 \Bigg[ \Bigg( \frac{275}{120} \Bigg) - \Bigg( \frac{-37}{120} \Bigg)^2 \Bigg] = 1076332.64 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{1076332.64} = 1037.46$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{1037.46}{3234.17} \times 100 = 32.08$

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Question 5: An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

$\displaystyle \begin{array} {|l | l | l |} \hline & \text{Firm A} & \text{Firm B} \\ \hline \text{No. of wage earners } & 586 & 648 \\ \hline \text{Average weekly wages } & \text{Rs. } 52.5 & \text{Rs. } 47.5 \\ \hline \text{Variance of the distribution of wages } & 100 & 121 \\ \hline \end{array}$

(i) Which firm A or B pays out larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?

(i)

$\displaystyle \text{ Total wages paid by Firm A} \\ \\ = \text{ (Average Wages) } \times \text{ ( No. of wage earners) } = 52.5 \times 586 = 30765$

$\displaystyle \text{ Total wages paid by Firm B} \\ \\ = \text{ (Average Wages) } \times \text{ ( No. of wage earners) } = 47.5 \times 648 = 30780$

Therefore firm  B pays out larger amount as weekly wages.

(ii)

In order to compare the variability of wages among the two firms, we have to calculate their coefficients of variation.

Let $\sigma_1$ and $\sigma_2$ denote the standard deviations of height and weight respectively.

Further let $\overline{X_1}$ and $\overline{X_2}$ be the mean wages in firms A and B respectively.

$\displaystyle \text{We have } \overline{X_1} = 52.5 \hspace{1.0cm} \overline{X_2} = 47.5$

$\displaystyle {\sigma_1}^2 = 100 \text{ and } {\sigma_2}^2 = 121$

$\displaystyle \Rightarrow {\sigma_1} = \sqrt{100} = 10 \text{ and } {\sigma_2} = \sqrt{121} = 11$

$\displaystyle \text{Therefore, coefficient of variation in wages in firm A } \\ \\ { \hspace{1.0cm} = \frac{\sigma_1}{\overline{X_1}} \times 100 = \frac{10}{52.5} \times 100 = 19.05}$

$\displaystyle \text{and coefficient of variation in wages in firm B } \\ \\ { \hspace{1.0cm} = \frac{\sigma_2}{\overline{X_2}} \times 100 = \frac{11}{47.5} \times 100 = 23.05 }$

Clearly, the coefficient of variation in wages is greater for firm B than for firm A. Hence, firm B shows more variability in wages.

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Question 6: The following are some particulars of the distribution of weights of boys and girls in a class:

$\displaystyle \begin{array} {|l | l | l |} \hline & \text{Boys} & \text{Girls} \\ \hline \text{Number} & 100 & 50 \\ \hline \text{Mean weight } & 60 \text{ kg. } & 45 \text{ kg. } \\ \hline \text{Variance } & 9 & 4 \\ \hline \end{array}$

Which of the distributions is more variable?

In order to compare the variability of weight in boys and girls, we have to calculate their coefficients of variation.

Let $\sigma_1$ and $\sigma_2$ denote the standard deviations of boys and girls respectively.

Further let $\overline{X_1}$ and $\overline{X_2}$ be the mean wages in firms A and B respectively.

$\displaystyle \text{We have } \overline{X_1} = 60 \hspace{1.0cm} \overline{X_2} = 45$

$\displaystyle {\sigma_1}^2 = 9 \text{ and } {\sigma_2}^2 = 4$

$\displaystyle \Rightarrow {\sigma_1} = \sqrt{9} = 3 \text{ and } {\sigma_2} = \sqrt{4} = 2$

$\displaystyle \text{Therefore, coefficient of variation in weights in boys } \\ \\ { \hspace{1.0cm} = \frac{\sigma_1}{\overline{X_1}} \times 100 = \frac{3}{60} \times 100 = 5}$

$\displaystyle \text{and coefficient of variation in wages in firm B } \\ \\ { \hspace{1.0cm} = \frac{\sigma_2}{\overline{X_2}} \times 100 = \frac{2}{45} \times 100 =4.44 }$

Clearly, the coefficient of variation in weights is greater for boys than for girls. Hence, boys shows more variability in weights.

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Question 7: The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

$\displaystyle \begin{array} {|l | l | l | l |} \hline \text{Subjects} & \text{Mathematics} & \text{Physics} & \text{Chemistry} \\ \hline \text{Mean } & 42 & 32 & 40.9 \\ \hline \text{Standard Deviation } & 12 & 15 & 20 \\ \hline \end{array}$

Which of the three subjects shows the highest variability in marks and which shows the lowest?

In order to compare the variability of marks in mathematics, physics and chemistry, we have to calculate their coefficients of variation.

Let $\sigma_1$, $\sigma_2$ and $\sigma_3$ denote the standard deviations of mathematics, physics and chemistry respectively.

Further let $\overline{X_1}$, $\overline{X_2}$ and $\overline{X_3}$ be the mean marks in mathematics, physics and chemistry respectively.

$\displaystyle \text{We have } \overline{X_1} = 42 \hspace{1.0cm} \overline{X_2} = 32 \hspace{1.0cm} \overline{X_2} = 40.9$

$\displaystyle {\sigma_1}^2 = 12, \hspace{1.0cm} {\sigma_2}^2 = 15 \text{ and } {\sigma_3}^2 = 20$

$\displaystyle \text{Therefore, coefficient of variation in marks for mathematics } \\ \\ { \hspace{1.0cm} = \frac{\sigma_1}{\overline{X_1}} \times 100 = \frac{12}{42} \times 100 = 28.57}$

$\displaystyle \text{and coefficient of variation in wages in marks for physics } \\ \\ { \hspace{1.0cm} = \frac{\sigma_2}{\overline{X_2}} \times 100 = \frac{15}{32} \times 100 =46.88 }$

$\displaystyle \text{and coefficient of variation in wages in marks for chemistry } \\ \\ { \hspace{1.0cm} = \frac{\sigma_3}{\overline{X_3}} \times 100 = \frac{20}{40.9} \times 100 =48.90 }$

Clearly, the coefficient of variation in marks is greatest in Chemistry and lowest in mathematics.

Hence, marks in chemistry show highest variability and marks in mathematics show lowest variability.

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Question 8: From the data given below state which group is more variable $G_1$ or $G_2$ ?

$\displaystyle \begin{array} {|l | c | c | c | c | c| c | c | } \hline \text{Marks} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Group } G_1 & 9 & 17 & 32 & 33 & 40 & 10 & 9 \\ \hline \text{Group } G_2 & 10 & 20 & 30 & 25 & 43 & 15 & 7 \\ \hline \end{array}$

$\text{For } : G_1$

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 45}{10} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 10-20 & 9 & 15 & -3 & -27 & 9 & 81 \\ \hline 20-30 & 17 & 25 & -2 & -34 & 4 & 68 \\ \hline 30-40 & 32 &35 & - 1& -32 & 1 & 32 \\ \hline 40-50 & 33 &45 & 0 & 0 & 0 & 0 \\ \hline 50-60 & 40 &55 & 1 & 40 & 1 & 40 \\ \hline 60-70 & 10 &65 &2 & 20 &4 & 40 \\ \hline 70-80 & 9 &75 & 3 & 27 & 9 & 81 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 150 & & & = -6 & & = 342 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 150, \hspace{0.5cm} \sum f_iu_i = -6, \hspace{0.5cm} \sum f_i{u_i}^2= 342 \hspace{0.5cm} A = 45, \hspace{0.5cm} h = 10$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 45 + 10 \Big( \frac{-6}{150} \Big) = 44.6$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 100 \Bigg[ \Bigg( \frac{342}{150} \Bigg) - \Bigg( \frac{-6}{150} \Bigg)^2 \Bigg] = 227.84 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{227.84} = 15.09$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{15.09}{44.6} \times 100 = 33.834$

$\text{For } : G_2$

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 45}{10} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 10-20 & 10 & 15 & -3 & -30 & 9 & 90 \\ \hline 20-30 & 20 & 25 & -2 & -40 & 4 & 80 \\ \hline 30-40 & 30 &35 & - 1& -30 & 1 & 30 \\ \hline 40-50 & 25 &45 & 0 & 0 & 0 & 0 \\ \hline 50-60 & 43 &55 & 1 & 43 & 1 & 43 \\ \hline 60-70 & 15 &65 &2 & 30 &4 & 60 \\ \hline 70-80 & 7 &75 & 3 & 21 & 9 & 63 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 150 & & & = -6 & & = 366 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 150, \hspace{0.5cm} \sum f_iu_i = -6, \hspace{0.5cm} \sum f_i{u_i}^2= 366 \hspace{0.5cm} A = 45, \hspace{0.5cm} h = 10$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 45 + 10 \Big( \frac{-6}{150} \Big) = 44.6$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 100 \Bigg[ \Bigg( \frac{366}{150} \Bigg) - \Bigg( \frac{-6}{150} \Bigg)^2 \Bigg] = 243.84 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{243.84} = 15.62$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{15.62}{44.6} \times 100 = 35.02$

Therefore we see that Coefficient of Variation for Group 2 is more than Group 1. Hence the Group 2 is more variable.

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Question 9: Find the coefficient of variation for the following data:

$\displaystyle \begin{array} {|l | c | c | c | c | c| c | } \hline \text{Size (in cm) } & 10-15 & 15-20 & 20-25 & 25-30 & 30-35 & 35-40 \\ \hline \text{No. of Items: } & 2 & 8 & 20 & 35 & 20 & 15 \\ \hline \end{array}$

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 27.5}{5} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 10-15 & 2 & 12.5 & -3 & -6 & 9 & 18 \\ \hline 15-20 & 8 & 17.5 & -2 & -16 & 4 & 32 \\ \hline 20-25 & 20 & 22.5 & -1 & -20 & 1 & 20 \\ \hline 25-30 & 35 & 27.5 & 0 & 0 & 0 & 0 \\ \hline 30-35 & 20 & 32.5 & 1 & 20 & 1 & 20 \\ \hline 35-40 & 15 & 37.5 & 2 & 30 & 4 & 60 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 100 & & & = 8 & & = 150 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 100, \hspace{0.5cm} \sum f_iu_i = 8, \hspace{0.5cm} \sum f_i{u_i}^2= 150 \hspace{0.5cm} A = 27.5, \hspace{0.5cm} h = 5$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 27.5 + 5 \Big( \frac{8}{100} \Big) = 27.9$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 25 \Bigg[ \Bigg( \frac{150}{100} \Bigg) - \Bigg( \frac{8}{100} \Bigg)^2 \Bigg] = 37.34 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{37.34} = 6.11$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{6.11}{27.9} \times 100 = 21.9$

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Question 10: From the prices of shares X and Y given below: find out which is more stable in value:

$\displaystyle \begin{array} {|l | c | c | c | c | c| c | c | c | c| c| } \hline X: & 35 & 54 & 52 & 53 & 56 & 58 & 52 & 50 & 51 & 49 \\ \hline Y: & 108 & 107 & 105 & 105 & 106 & 107 & 104 & 103 & 104 & 101 \\ \hline \end{array}$

Let the assumed mean A be 53.

$\text{For } : X$

$\displaystyle \begin{array} { | c | c| c| } \hline x & d_i = x_i - A & {d_i}^2 \\ \hline 35 & -13 & 169 \\ \hline 24 & -24 & 576 \\ \hline 52 & 4 & 16 \\ \hline 53 & 5 & 25 \\ \hline 56 & 8 & 64 \\ \hline 58 & 10 & 100 \\ \hline 52 & 4 & 16 \\ \hline 50 & 2 & 4 \\ \hline 51 & 3 & 9 \\ \hline 49 & 1 & 1 \\ \hline \sum x_i = 480 & \sum d_i = 0 & \sum {d_i}^2 = 980 \\ \hline \end{array}$

$\displaystyle \text{Here, } n = 10, \sum d_i = 0, \text{ and } \sum {d_i}^2 = 980$

$\displaystyle \text{Mean } = \frac{\sum x_i}{n} = \frac{480}{10} = 48$

$\displaystyle \therefore Var(X) = \frac{1}{n} \Bigg( \sum {d_i}^2 \Bigg) - \Bigg( \frac{1}{n} \sum d_i \Bigg)^2 = \frac{1}{10} \times 980 - \Bigg( \frac{1}{10} \times 0 \Bigg) = 98$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{98} = 9.9$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{9.9}{48} \times 100 = 20.6$

Let the assumed mean A be 105.

$\text{For } : Y$

$\displaystyle \begin{array} { | c | c| c| } \hline y & d_i = y_i - A & {d_i}^2 \\ \hline 108 & 3 & 9 \\ \hline 107 & 2 & 4 \\ \hline 105 & 0 & 0 \\ \hline 105 & 0 & 0 \\ \hline 106 & 1 & 1 \\ \hline 107 & 2 & 4 \\ \hline 104 & -1 & 1 \\ \hline 103 & -2 & 4 \\ \hline 104 & -1 & 1 \\ \hline 101 & -4 & 16 \\ \hline \sum x_i = 1050 & \sum d_i = 0 & \sum {d_i}^2 = 40 \\ \hline \end{array}$

$\displaystyle \text{Here, } n = 10, \sum d_i = 0, \text{ and } \sum {d_i}^2 = 40$

$\displaystyle \text{Mean } = \frac{\sum x_i}{n} = \frac{1050}{10} = 105$

$\displaystyle \therefore Var(X) = \frac{1}{n} \Bigg( \sum {d_i}^2 \Bigg) - \Bigg( \frac{1}{n} \sum d_i \Bigg)^2 = \frac{1}{10} \times 40 - \Bigg( \frac{1}{10} \times 0 \Bigg) = 4$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{4} = 2$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{2}{105} \times 100 = 1.9$

Coefficient of variation of Y  is lesser than that of X, hence price of Y is stable.

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Question 11: Life of bulbs produced by two factories A and B are given below:

$\displaystyle \begin{array} {|l | c | c | c | c | c| } \hline \text{Length of life (in hours):} & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{Factory A (number of bulbs) } & 10 & 22 & 52 & 20 & 16 \\ \hline \text{Factory B (number of bulbs) } & 8 & 60 & 24 & 16 & 12 \\ \hline \end{array}$

The bulbs of which factory are more consistent from the point of view of length of life?

$\text{For : Factory A}$

Let assumed mean A = 800 and h = 100

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 800}{100} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 550-650 & 10 & 600 & -2 & -20 & 4 & 40 \\ \hline 650-750 & 22 & 700 & -1 & -22 &1 & 22 \\ \hline 750-850 & 52 & 800 & 0 & 0 & 0 & 0 \\ \hline 850-950 & 20 & 900 & 1 & 20 & 1 & 20 \\ \hline 950-1050 &16 & 1000 & 2 & 32 & 4 & 64 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 120 & & & = 10 & & = 146 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 120, \hspace{0.5cm} \sum f_iu_i = 10, \hspace{0.5cm} \sum f_i{u_i}^2= 146 \hspace{0.5cm} A = 800, \hspace{0.5cm} h = 100$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 800 + 100 \Big( \frac{10}{120} \Big) = 808.33$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 10000 \Bigg[ \Bigg( \frac{146}{120} \Bigg) - \Bigg( \frac{10}{120} \Bigg)^2 \Bigg] = 12097.2 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{12097.2} = 109.98$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{109.98}{808.33} \times 100 = 13.61$

$\text{For : Factory B}$

Let assumed mean A = 800 and h = 100

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 800}{100} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 550-650 & 8 & 600 & -2 & -16 & 4 & 32 \\ \hline 650-750 & 60 & 700 & -1 & -60 &1 & 60 \\ \hline 750-850 & 24 & 800 & 0 & 0 & 0 & 0 \\ \hline 850-950 & 16 & 900 & 1 & 16 & 1 & 16 \\ \hline 950-1050 &12 & 1000 & 2 & 24 & 4 & 48 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 120 & & & = -36 & & = 156 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 120, \hspace{0.5cm} \sum f_iu_i = -36, \hspace{0.5cm} \sum f_i{u_i}^2= 156 \hspace{0.5cm} A = 800, \hspace{0.5cm} h = 100$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 800 + 100 \Big( \frac{-36}{120} \Big) = 800-30=770$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 10000 \Bigg[ \Bigg( \frac{156}{120} \Bigg) - \Bigg( \frac{-36}{120} \Bigg)^2 \Bigg] = 12100 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{12100} = 110$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{110}{770} \times 100 = 14.29$

Since, the coefficient of variation of factor B is greater than the coefficient of variation of factory A, therefore factory B has more variability than factory A.

This means that the bulbs from factory A are more consistent from the point of views of length of life.

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Question 12: Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests :

$\displaystyle \begin{array} {|l | c | c | c | c | c| c | c | c | c| c| } \hline \text{Ravi: } & 25 & 50 & 45 & 30 & 70 & 42 & 36 & 48 & 35 & 60 \\ \hline \text{Hashina: } & 10 & 70 & 50 & 20 & 95 & 55 & 42 & 60 & 48 & 85 \\ \hline \end{array}$

Who is more intelligent and who is more consistent?

Let the assumed mean A be 45.

$\text{For Ravi} :$

$\displaystyle \begin{array} { | c | c| c| } \hline x & d_i = x_i - 45 & {d_i}^2 \\ \hline 25 & -20 & 400 \\ \hline 50 & -5 & 25 \\ \hline 45 & 0 & 0 \\ \hline 30 & -15 & 225 \\ \hline 70 & 25 & 625 \\ \hline 42 & -3 & 9 \\ \hline 36 & -9 & 81 \\ \hline 48 & 3 & 9 \\ \hline 35 & -10 & 100 \\ \hline 60 & 15 & 225 \\ \hline \sum x_i = 770 & \sum d_i = -9 & \sum {d_i}^2 = 1699 \\ \hline \end{array}$

$\displaystyle \text{Here, } n = 10, \sum d_i = -9, \text{ and } \sum {d_i}^2 = 1699$

$\displaystyle \text{Mean } = \frac{\sum x_i}{n} = \frac{770}{10} = 77$

$\displaystyle Var(X) = \frac{1}{n} \Bigg( \sum {d_i}^2 \Bigg) - \Bigg( \frac{1}{n} \sum d_i \Bigg)^2 = \frac{1}{10} \times 1699 - \Bigg( \frac{1}{10} \times (-9) \Bigg)^2 = 169.09$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{169.09} = 13.003$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{110}{770} \times 100 = 14.29$

Let the assumed mean A be 55.

$\text{For Hashina} :$

$\displaystyle \begin{array} { | c | c| c| } \hline x & d_i = x_i - 55 & {d_i}^2 \\ \hline 10 & -45 & 2025 \\ \hline 70 & 15 & 625 \\ \hline 50 & -5 & 25 \\ \hline 20 & -35 & 1225 \\ \hline 95 & 40 & 1600 \\ \hline 55 & 0 & 0 \\ \hline 42 & -13 & 169 \\ \hline 60 & 5 & 25 \\ \hline 48 & -7 & 49 \\ \hline 80 & 25 & 625 \\ \hline \sum x_i = 530 & \sum d_i = -20 & \sum {d_i}^2 = 6368 \\ \hline \end{array}$

$\displaystyle \text{Here, } n = 10, \sum d_i = -20, \text{ and } \sum {d_i}^2 = 6368$

$\displaystyle \text{Mean } = \frac{\sum x_i}{n} = \frac{530}{10} = 53$

$\displaystyle \therefore Var(X) = \frac{1}{n} \Bigg( \sum {d_i}^2 \Bigg) - \Bigg( \frac{1}{n} \sum d_i \Bigg)^2 = \frac{1}{10} \times 6368 - \Bigg( \frac{1}{10} \times (-20) \Bigg)^2 = 632.8$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{632.8} = 25.16$

$\displaystyle \text{Coefficient of variation}= \frac{\text{S.D}}{\text{Mean}} \times 100 = \frac{25.16}{53} \times 100 = 47.47$

Since the coefficient of variation in mark obtained by Hashina is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.