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Let $x_,x_2,x_3,x_4,x_5$ be the observations with mean m and standard deviation s. The standard deviation of the observation $kx_1,kx_2,kx_3,kx_4,kx_5$ is

$\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 $kx_2-n=k(x_2-m)$
$kx_4-n =k(x_4-m)$
$kx_5-n =k(x_5-m)$
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.
answered Jul 8, 2014 by meena.p

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