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?

Answer:

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?

Answer:

\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?

Answer:

\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}

Answer:

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?

Answer:

(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?

Answer:

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?

Answer:

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}

Answer:

\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}

Answer:

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}

Answer:

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?

Answer:

\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?

Answer:

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.