Answer 1

For Ravi

 

Marks   \[\left( x_i \right)\]	\[d_i = x_i - 45\]	\[d_i^2\]
25	−20	400
50	5	25
45	0	0
30	−15	225
70	25	625
42	−3	9
36	−9	81
48	3	9
35	−10	100
60	15	225
	\[\sum_{} d_i = - 9\]	\[\sum_{} d_i^2 = 1699\]

\[X_R = A + \frac{\sum_{} d_i}{10} = 45 + \frac{\left( - 9 \right)}{10} = 44 . 1\]

Standard deviation,

\[\sigma_R = \sqrt{\frac{\sum_{} d_i^2}{10} - \left( \frac{\sum_{} d_i}{10} \right)^2} = \sqrt{\frac{1699}{10} - \left( \frac{- 9}{10} \right)^2} = \sqrt{169 . 09} = 13 . 003\]

Coefficicent of variation = \[\frac{\sigma_B}{X_B} \times 100 = \frac{110}{770} \times 100 = 14 . 29\]

For Hashina

Marks \[\left( x_i \right)\]	\[d_i = x_i - 55\]	\[d_i^2\]
10	−45	2025
70	15	625
50	−5	25
20	−35	1225
95	40	1600
55	0	0
42	−13	169
60	5	25
48	−7	49
80	25	625
	\[\sum_{} d_i = - 20\]	\[\sum_{} d_i^2 = 6368\]

Mean,

\[X_H = A + \frac{\sum_{} d_i}{10} = 55 + \frac{\left( - 20 \right)}{10} = 53\]

Standard deviation,

\[\sigma_H = \sqrt{\frac{\sum_{} d_i^2}{10} - \left( \frac{\sum_{} d_i}{10} \right)^2} = \sqrt{\frac{6368}{10} - \left( \frac{- 20}{10} \right)^2} = \sqrt{632 . 8} = 25 . 16\]

Coefficicent of variation = \[\frac{\sigma_H}{X_H} \times 100 = \frac{25 . 16}{53} \times 100 = 47 . 47\]

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