$\begin{array}{1 1}(A)\; k+s\\(B)\;\frac{s}{k}\\(C)\;ks\\(D)\;s \end{array} $

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give m is the mean of the observation $x+_1,x_2,x_3,x_4,x_5$

$m =\large\frac{x_1+x_2+x_3+x_4+x_5}{5}$

Let the new mean be n

The observations are $kx_1,kx_2,kx_3,kx_4,kx_5$

$\therefore n= \large\frac{kx_+kx_2+kx_3+kx_4+kx_5}{5}$

=>$k \bigg[\large\frac{x_1+x_2+x_3+x_4+x_5}{5}\bigg]$

$n= km$

Standard deviation $=\sqrt { \large\frac{(x_1-m)^2+(x_3-m)^2+(x_4-m)^2+(x_4 -m)^2}{5}}$

Tjhe new standard deviation will be $S_1 = \sqrt {\large\frac{(kx_1-n)^2 +(kx_2-n)^2 +(kx_3-n)^2 +(kx_4-n)^2+(kx_5-n)^2}{5}}$

we know that

$kx_1 -n =kx_1-mk$

$\qquad= k(x_1-m)$

Similarly $kx_2-n=k(x_2-m)$

$kx_3-n=k(x_3-m)$

$kx_4-n =k(x_4-m)$

$kx_5-n =k(x_5-m)$

Substituting $S_1= \sqrt { \large\frac{k(x_1-m))^2+(k(x_2-m)^2)^2+(k(x_3-m))^2 +(k(x_4-m))^2+(k(x_5-m))^2}{5}}$

$\qquad= ks$

Hence C is the correct answer.

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