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(B) First,we calculate the mean $(\bar{X})$ of the distribution: $C$.$I$.$f_i$Mid value $(x_i)$$f_i x_i$$75$-$175$$3$$125$$375$$175$-$275$$2$$225$$450

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$\frac{400 \sqrt{6}}{17}$

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(B) First,we calculate the mean $(\bar{X})$ of the distribution: $C$.$I$.$f_i$Mid value $(x_i)$$f_i x_i$$75$-$175$$3$$125$$375$$175$-$275$$2$$225$$450$$275$-$375$$1$$325$$325$$375$-$475$$0$$425$$0$$475$-$575$$1$$525$$525$$575$-$675$$2$$625$$1250$$675$-$775$$3$$725$$2175$Total$\sum f_i = 12$-$\sum f_i x_i = 5100$The mean is $\bar{X} = \frac{\sum f_i x_i}{\sum f_i} = \frac{5100}{12} = 425$.Given the variance $\sigma^2 = 60000$,the standard deviation is $\sigma = \sqrt{60000} = 100 \sqrt{6}$.The coefficient of variation $(CV)$ is given by the formula: $CV = \frac{\sigma}{\bar{X}} 	imes 100$.$CV = \frac{100 \sqrt{6}}{425} 	imes 100 = \frac{400 \sqrt{6}}{17}$.

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

Consider the following frequency distribution:
$C$.$I$. $75$-$175$ $175$-$275$ $275$-$375$ $375$-$475$ $475$-$575$ $575$-$675$ $675$-$775$
$f_i$ $3$ $2$ $1$ $0$ $1$ $2$ $3$
If the variance of this distribution is $60000$,then the coefficient of variation of the distribution is:

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If the mean and standard deviation of $5$ observations $x_1, x_2, x_3, x_4, x_5$ are $10$ and $3$,respectively,then the variance of $6$ observations $x_1, x_2, x_3, x_4, x_5$ and $-50$ is equal to: (in $.5$)

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$C$.$I$.	$75$-$175$	$175$-$275$	$275$-$375$	$375$-$475$	$475$-$575$	$575$-$675$	$675$-$775$
$f_i$	$3$	$2$	$1$	$0$	$1$	$2$	$3$

Consider the following frequency distribution: $C$.$I$.$75$-$175$$175$-$275$$275$-$375$$375$-$475$$475$-$575$$575$-$675$$675$-$775$$f_i$$3$$2$$1$$0$$1$$2$$3$If the variance of this distribution is $60000$,then the coefficient of variation of the distribution is: