The marks obtained by students A and B in 3 e | vedclass

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(D) For student $A$: Marks are $30, 20, 40$.Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.Standard deviation $\sigma_A = \sqrt{\frac{(30-3

Vedclass · Accepted Answer

$5 : 3 \sqrt{61}$

Vedclass · Answer

(D) For student $A$: Marks are $30, 20, 40$.Mean $\bar{x}_A = \frac{30+20+40}{3} = \frac{90}{3} = 30$.Standard deviation $\sigma_A = \sqrt{\frac{(30-30)^2 + (20-30)^2 + (40-30)^2}{3}} = \sqrt{\frac{0 + 100 + 100}{3}} = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}}$.Coefficient of variation $(CV)_A = \frac{\sigma_A}{\bar{x}_A} 	imes 100 = \frac{10 \sqrt{2}}{\sqrt{3} 	imes 30} 	imes 100 = \frac{\sqrt{2}}{3 \sqrt{3}} 	imes 100$.For student $B$: Marks are $70, 0, 5$.Mean $\bar{x}_B = \frac{70+0+5}{3} = \frac{75}{3} = 25$.Standard deviation $\sigma_B = \sqrt{\frac{(70-25)^2 + (0-25)^2 + (5-25)^2}{3}} = \sqrt{\frac{45^2 + (-25)^2 + (-20)^2}{3}} = \sqrt{\frac{2025 + 625 + 400}{3}} = \sqrt{\frac{3050}{3}} = \sqrt{\frac{25 	imes 122}{3}} = 5 \sqrt{\frac{122}{3}}$.Coefficient of variation $(CV)_B = \frac{\sigma_B}{\bar{x}_B} 	imes 100 = \frac{5 \sqrt{122}}{\sqrt{3} 	imes 25} 	imes 100 = \frac{\sqrt{122}}{5 \sqrt{3}} 	imes 100$.Ratio $\frac{(CV)_A}{(CV)_B} = \frac{\sqrt{2}}{3 \sqrt{3}} \div \frac{\sqrt{122}}{5 \sqrt{3}} = \frac{\sqrt{2}}{3 \sqrt{3}} 	imes \frac{5 \sqrt{3}}{\sqrt{122}} = \frac{5 \sqrt{2}}{3 \sqrt{2} \sqrt{61}} = \frac{5}{3 \sqrt{61}}$.Thus,the ratio is $5 : 3 \sqrt{61}$.

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