\displaystyle \textbf{Question 1: }\text{Two plants A and B of a factory show the following results about the}
\displaystyle \text{number of workers and the wages paid to them.}
\displaystyle \begin{array}{|l|c|c|}\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}
\displaystyle \text{Answer:}
\displaystyle \text{Variance of distribution of wages:}
\displaystyle \sigma_A^2=81,\hspace{2cm}\sigma_B^2=100.
\displaystyle \text{Therefore, the standard deviations are}
\displaystyle \sigma_A=\sqrt{81}=9,\hspace{2cm}\sigma_B=\sqrt{100}=10.
\displaystyle \text{The average monthly wages in both the plants are the same, i.e., Rs. }2500.
\displaystyle \text{Hence, the plant having the greater variance (or standard deviation) will have greater variability}
\displaystyle \text{in individual wages.}
\displaystyle \text{Since }100>81,\text{ Plant B has greater variability in individual wages.}
\displaystyle \therefore \text{Plant B has greater variability in individual wages.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The means and standard deviations of heights and weights of }50\text{ students}
\displaystyle \text{of a class are as follows:}
\displaystyle \begin{array}{|l|c|c|}\hline &\text{Weights}&\text{Heights}\\ \hline \text{Mean}&63.2\text{ kg}&63.2\text{ inch}\\ \hline \text{Standard deviation}&5.6\text{ kg}&11.5\text{ inch}\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Coefficient of variation of weights}=\frac{\sigma}{\overline{X}}\times100
\displaystyle =\frac{5.6}{63.2}\times100=8.86\%.
\displaystyle \text{Coefficient of variation of heights}=\frac{\sigma}{\overline{X}}\times100
\displaystyle =\frac{11.5}{63.2}\times100=18.20\%.
\displaystyle \text{Since the coefficient of variation of heights is greater than that of weights,}
\displaystyle \text{heights show greater variability than weights.}
\displaystyle \therefore \text{Heights show greater variability than weights.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Coefficient of variation of two distributions are }60\%\text{ and }70\%
\displaystyle \text{and their standard deviations are }21\text{ and }16\text{ respectively. What are their}
\displaystyle \text{arithmetic means?}
\displaystyle \text{Answer:}
\displaystyle \text{We know that}
\displaystyle \mathrm{C.V.}=\frac{\sigma}{\overline{X}}\times100.
\displaystyle \therefore \overline{X}=\frac{\sigma\times100}{\mathrm{C.V.}}.
\displaystyle \text{For the first distribution,}
\displaystyle \overline{X_1}=\frac{21\times100}{60}=35.
\displaystyle \text{For the second distribution,}
\displaystyle \overline{X_2}=\frac{16\times100}{70}=\frac{1600}{70}=22.8571\approx22.86.
\displaystyle \therefore \text{The arithmetic means of the two distributions are }35\text{ and }22.86\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Calculate the 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}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=3450\text{ and class width }h=700.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-3450}{700}&f_iu_i&u_i^2&f_iu_i^2\\ \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=120&&&\sum f_iu_i=-37&&\sum f_iu_i^2=275\\ \hline \end{array}
\displaystyle \text{Clearly, }N=120,\quad \sum f_iu_i=-37,\quad \sum f_iu_i^2=275,\quad A=3450\text{ and }h=700.
\displaystyle \overline X=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =3450+700\left(\frac{-37}{120}\right)
\displaystyle =3234.1667\approx3234.17.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =700^2\left[\frac{275}{120}-\left(\frac{-37}{120}\right)^2\right]
\displaystyle =1076332.6389\approx1076332.64.
\displaystyle \sigma=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{1076332.6389}=1037.4645\approx1037.46.
\displaystyle \mathrm{C.V.}=\frac{\sigma}{\overline X}\times100
\displaystyle =\frac{1037.46}{3234.17}\times100
\displaystyle =32.08\%.
\displaystyle \therefore \text{The coefficient of variation is }32.08\%\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{An analysis of the weekly wages paid to workers in two firms A and B,}
\displaystyle \text{belonging to the same industry, gives the following results:}
\displaystyle \begin{array}{|l|c|c|}\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}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Total weekly wages paid by Firm A}
\displaystyle =(\text{Average weekly wages})\times(\text{Number of wage earners})
\displaystyle =52.5\times586=30765.
\displaystyle \text{Total weekly wages paid by Firm B}
\displaystyle =(\text{Average weekly wages})\times(\text{Number of wage earners})
\displaystyle =47.5\times648=30780.
\displaystyle \text{Since }30780>30765,\text{ Firm B pays out the larger amount as weekly wages.}
\displaystyle \text{(ii) To compare the variability of wages, we calculate the coefficients of variation.}
\displaystyle \overline{X_1}=52.5,\qquad \overline{X_2}=47.5.
\displaystyle \sigma_1^2=100,\qquad \sigma_2^2=121.
\displaystyle \therefore \sigma_1=\sqrt{100}=10,\qquad \sigma_2=\sqrt{121}=11.
\displaystyle \mathrm{C.V.}\text{ of Firm A}=\frac{\sigma_1}{\overline{X_1}}\times100
\displaystyle =\frac{10}{52.5}\times100=19.05\%.
\displaystyle \mathrm{C.V.}\text{ of Firm B}=\frac{\sigma_2}{\overline{X_2}}\times100
\displaystyle =\frac{11}{47.5}\times100=23.16\%.
\displaystyle \text{Since }23.16\%>19.05\%,\text{ Firm B has greater variability in individual wages.}
\displaystyle \therefore \text{(i) Firm B pays out the larger amount as weekly wages.}
\displaystyle \therefore \text{(ii) Firm B has greater variability in individual wages.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The following are some particulars of the distribution of weights of}
\displaystyle \text{boys and girls in a class:}
\displaystyle \begin{array}{|l|c|c|}\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}
\displaystyle \text{Answer:}
\displaystyle \text{To compare the variability of weights of boys and girls, we calculate their coefficients of variation.}
\displaystyle \overline{X_1}=60,\qquad \overline{X_2}=45.
\displaystyle \sigma_1^2=9,\qquad \sigma_2^2=4.
\displaystyle \therefore \sigma_1=\sqrt{9}=3,\qquad \sigma_2=\sqrt{4}=2.
\displaystyle \mathrm{C.V.}\text{ of boys}=\frac{\sigma_1}{\overline{X_1}}\times100
\displaystyle =\frac{3}{60}\times100=5\%.
\displaystyle \mathrm{C.V.}\text{ of girls}=\frac{\sigma_2}{\overline{X_2}}\times100
\displaystyle =\frac{2}{45}\times100=4.44\%.
\displaystyle \text{Since the coefficient of variation of boys is greater than that of girls,}
\displaystyle \text{the distribution of boys' weights is more variable than that of girls.}
\displaystyle \therefore \text{The distribution of boys' weights is more variable.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The mean and standard deviation of marks obtained by }50\text{ students of a}
\displaystyle \text{class in three subjects, Mathematics, Physics and Chemistry, are given below:}
\displaystyle \begin{array}{|l|c|c|c|}\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}
\displaystyle \text{Answer:}
\displaystyle \text{To compare the variability of marks in the three subjects, we calculate their coefficients of variation.}
\displaystyle \overline{X_1}=42,\qquad \overline{X_2}=32,\qquad \overline{X_3}=40.9.
\displaystyle \sigma_1=12,\qquad \sigma_2=15,\qquad \sigma_3=20.
\displaystyle \mathrm{C.V.}\text{ for Mathematics}=\frac{\sigma_1}{\overline{X_1}}\times100
\displaystyle =\frac{12}{42}\times100=28.57\%.
\displaystyle \mathrm{C.V.}\text{ for Physics}=\frac{\sigma_2}{\overline{X_2}}\times100
\displaystyle =\frac{15}{32}\times100=46.88\%.
\displaystyle \mathrm{C.V.}\text{ for Chemistry}=\frac{\sigma_3}{\overline{X_3}}\times100
\displaystyle =\frac{20}{40.9}\times100=48.90\%.
\displaystyle \text{Since the coefficient of variation is greatest for Chemistry, it shows the highest variability.}
\displaystyle \text{Since the coefficient of variation is least for Mathematics, it shows the lowest variability.}
\displaystyle \therefore \text{Chemistry shows the highest variability in marks, while Mathematics shows the lowest variability.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{From the data given below, state which group, }G_1\text{ or }G_2,\text{ is more variable.}
\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}
\displaystyle \text{Answer:}
\displaystyle \text{For Group }G_1:
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-45}{10}&f_iu_i&u_i^2&f_iu_i^2\\ \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=150&&&\sum f_iu_i=-6&&\sum f_iu_i^2=342\\ \hline \end{array}
\displaystyle \text{Clearly, }N=150,\quad \sum f_iu_i=-6,\quad \sum f_iu_i^2=342,\quad A=45\text{ and }h=10.
\displaystyle \overline X_1=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =45+10\left(\frac{-6}{150}\right)=44.6.
\displaystyle \mathrm{Var}(G_1)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =10^2\left[\frac{342}{150}-\left(\frac{-6}{150}\right)^2\right]
\displaystyle =227.84.
\displaystyle \sigma_1=\sqrt{227.84}=15.09\text{ (approx.).}
\displaystyle \text{For Group }G_2:
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-45}{10}&f_iu_i&u_i^2&f_iu_i^2\\ \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=150&&&\sum f_iu_i=-6&&\sum f_iu_i^2=366\\ \hline \end{array}
\displaystyle \text{Clearly, }N=150,\quad \sum f_iu_i=-6,\quad \sum f_iu_i^2=366,\quad A=45\text{ and }h=10.
\displaystyle \overline X_2=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =45+10\left(\frac{-6}{150}\right)=44.6.
\displaystyle \mathrm{Var}(G_2)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =10^2\left[\frac{366}{150}-\left(\frac{-6}{150}\right)^2\right]
\displaystyle =243.84.
\displaystyle \sigma_2=\sqrt{243.84}=15.62\text{ (approx.).}
\displaystyle \text{Since }\overline X_1=\overline X_2,\text{ the variances or standard deviations may be compared directly.}
\displaystyle \text{Since }243.84>227.84,\text{ Group }G_2\text{ has greater variability.}
\displaystyle \therefore \text{Group }G_2\text{ is more variable.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{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}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=27.5\text{ and class width }h=5.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-27.5}{5}&f_iu_i&u_i^2&f_iu_i^2\\ \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=100&&&\sum f_iu_i=8&&\sum f_iu_i^2=150\\ \hline \end{array}
\displaystyle \text{Clearly, }N=100,\quad \sum f_iu_i=8,\quad \sum f_iu_i^2=150,\quad A=27.5\text{ and }h=5.
\displaystyle \overline X=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =27.5+5\left(\frac{8}{100}\right)=27.9.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =5^2\left[\frac{150}{100}-\left(\frac{8}{100}\right)^2\right]
\displaystyle =37.34.
\displaystyle \sigma=\sqrt{\mathrm{Var}(X)}=\sqrt{37.34}=6.11\text{ (approx.).}
\displaystyle \mathrm{C.V.}=\frac{\sigma}{\overline X}\times100
\displaystyle =\frac{6.11}{27.9}\times100=21.90\%.
\displaystyle \therefore \text{The coefficient of variation is }21.90\%.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{From the prices of shares }X\text{ and }Y\text{ given below, find 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}
\displaystyle \text{Answer:}
\displaystyle \text{For share }X,\text{ let the assumed mean be }A=53.
\displaystyle \begin{array}{|c|c|c|}\hline x_i&d_i=x_i-A&d_i^2\\ \hline 35&-18&324\\ \hline 54&1&1\\ \hline 52&-1&1\\ \hline 53&0&0\\ \hline 56&3&9\\ \hline 58&5&25\\ \hline 52&-1&1\\ \hline 50&-3&9\\ \hline 51&-2&4\\ \hline 49&-4&16\\ \hline \sum x_i=510&\sum d_i=-20&\sum d_i^2=390\\ \hline \end{array}
\displaystyle \text{Clearly, }n=10,\quad \sum x_i=510,\quad \sum d_i=-20\text{ and }\sum d_i^2=390.
\displaystyle \overline X=\frac{\sum x_i}{n}=\frac{510}{10}=51.
\displaystyle \mathrm{Var}(X)=\frac{\sum d_i^2}{n}-\left(\frac{\sum d_i}{n}\right)^2
\displaystyle =\frac{390}{10}-\left(\frac{-20}{10}\right)^2=39-4=35.
\displaystyle \sigma_X=\sqrt{\mathrm{Var}(X)}=\sqrt{35}=5.92\text{ (approx.).}
\displaystyle \mathrm{C.V.}\text{ of }X=\frac{\sigma_X}{\overline X}\times100
\displaystyle =\frac{5.92}{51}\times100=11.60\%\text{ (approx.).}
\displaystyle \text{For share }Y,\text{ let the assumed mean be }A=105.
\displaystyle \begin{array}{|c|c|c|}\hline y_i&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 y_i=1050&\sum d_i=0&\sum d_i^2=40\\ \hline \end{array}
\displaystyle \text{Clearly, }n=10,\quad \sum y_i=1050,\quad \sum d_i=0\text{ and }\sum d_i^2=40.
\displaystyle \overline Y=\frac{\sum y_i}{n}=\frac{1050}{10}=105.
\displaystyle \mathrm{Var}(Y)=\frac{\sum d_i^2}{n}-\left(\frac{\sum d_i}{n}\right)^2
\displaystyle =\frac{40}{10}-\left(\frac{0}{10}\right)^2=4.
\displaystyle \sigma_Y=\sqrt{\mathrm{Var}(Y)}=\sqrt{4}=2.
\displaystyle \mathrm{C.V.}\text{ of }Y=\frac{\sigma_Y}{\overline Y}\times100
\displaystyle =\frac{2}{105}\times100=1.90\%\text{ (approx.).}
\displaystyle \text{Since the coefficient of variation of }Y\text{ is less than that of }X,
\displaystyle \text{the price of share }Y\text{ is more stable in value.}
\displaystyle \therefore \text{Share }Y\text{ is more stable in value.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{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)}&550-650&650-750&750-850&850-950&950-1050\\ \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}
\displaystyle \text{Answer:}
\displaystyle \text{For Factory A, let }A=800\text{ and }h=100.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-800}{100}&f_iu_i&u_i^2&f_iu_i^2\\ \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=120&&&\sum f_iu_i=10&&\sum f_iu_i^2=146\\ \hline \end{array}
\displaystyle \text{Clearly, }N=120,\quad \sum f_iu_i=10,\quad \sum f_iu_i^2=146,\quad A=800\text{ and }h=100.
\displaystyle \overline X=800+100\left(\frac{10}{120}\right)=808.33.
\displaystyle \mathrm{Var}(X)=100^2\left[\frac{146}{120}-\left(\frac{10}{120}\right)^2\right]=12097.22.
\displaystyle \sigma=\sqrt{12097.22}=109.99\text{ (approx.).}
\displaystyle \mathrm{C.V.}=\frac{109.99}{808.33}\times100=13.61\%.
\displaystyle \text{For Factory B, let }A=800\text{ and }h=100.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-800}{100}&f_iu_i&u_i^2&f_iu_i^2\\ \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=120&&&\sum f_iu_i=-36&&\sum f_iu_i^2=156\\ \hline \end{array}
\displaystyle \text{Clearly, }N=120,\quad \sum f_iu_i=-36,\quad \sum f_iu_i^2=156,\quad A=800\text{ and }h=100.
\displaystyle \overline X=800+100\left(\frac{-36}{120}\right)=770.
\displaystyle \mathrm{Var}(X)=100^2\left[\frac{156}{120}-\left(\frac{-36}{120}\right)^2\right]=12100.
\displaystyle \sigma=\sqrt{12100}=110.
\displaystyle \mathrm{C.V.}=\frac{110}{770}\times100=14.29\%.
\displaystyle \text{Since the coefficient of variation of Factory A is less than that of Factory B,}
\displaystyle \text{Factory A has less variability in the life of its bulbs.}
\displaystyle \therefore \text{The bulbs produced by Factory A are more consistent from the point of view of length of life.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Following are the marks obtained, out of }100,\text{ by two students Ravi and Hashina}
\displaystyle \text{in }10\text{ 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}
\displaystyle \text{Answer:}
\displaystyle \text{For Ravi, let the assumed mean be }A=45.
\displaystyle \begin{array}{|c|c|c|}\hline x_i&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=441&\sum d_i=-9&\sum d_i^2=1699\\ \hline \end{array}
\displaystyle \text{Clearly, }n=10,\quad \sum x_i=441,\quad \sum d_i=-9\text{ and }\sum d_i^2=1699.
\displaystyle \overline X_R=\frac{\sum x_i}{n}=\frac{441}{10}=44.1.
\displaystyle \mathrm{Var}(X_R)=\frac{\sum d_i^2}{n}-\left(\frac{\sum d_i}{n}\right)^2
\displaystyle =\frac{1699}{10}-\left(\frac{-9}{10}\right)^2=169.09.
\displaystyle \sigma_R=\sqrt{169.09}=13.003\text{ (approx.).}
\displaystyle \mathrm{C.V.}\text{ of Ravi}=\frac{\sigma_R}{\overline X_R}\times100
\displaystyle =\frac{13.003}{44.1}\times100=29.49\%\text{ (approx.).}
\displaystyle \text{For Hashina, let the assumed mean be }A=55.
\displaystyle \begin{array}{|c|c|c|}\hline x_i&d_i=x_i-55&d_i^2\\ \hline 10&-45&2025\\ \hline 70&15&225\\ \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 85&30&900\\ \hline \sum x_i=535&\sum d_i=-15&\sum d_i^2=6245\\ \hline \end{array}
\displaystyle \text{Clearly, }n=10,\quad \sum x_i=535,\quad \sum d_i=-15\text{ and }\sum d_i^2=6245.
\displaystyle \overline X_H=\frac{\sum x_i}{n}=\frac{535}{10}=53.5.
\displaystyle \mathrm{Var}(X_H)=\frac{\sum d_i^2}{n}-\left(\frac{\sum d_i}{n}\right)^2
\displaystyle =\frac{6245}{10}-\left(\frac{-15}{10}\right)^2=622.05.
\displaystyle \sigma_H=\sqrt{622.05}=24.94\text{ (approx.).}
\displaystyle \mathrm{C.V.}\text{ of Hashina}=\frac{\sigma_H}{\overline X_H}\times100
\displaystyle =\frac{24.94}{53.5}\times100=46.62\%\text{ (approx.).}
\displaystyle \text{Since Hashina has the greater mean marks, she is more intelligent on the basis of average performance.}
\displaystyle \text{Since Ravi has the smaller coefficient of variation, he is more consistent.}
\displaystyle \therefore \text{Hashina is more intelligent and Ravi is more consistent.}
\displaystyle \\

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