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

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- The formula to calculate the SD is $SD(\sigma) =\sqrt { \large\frac{\sum \limits_{i=1}^{n} (x_i -\bar{x})^2}{n}}$

Step 1:

Given the observations $a,b,c,d,e$

Given mean=m

$m= \large\frac{a+b+c+d+e}{5}$

Let the new mean be n

The new observations are

$a+k,b+k,c+k,d+k,e+k$

$n= \large\frac{a+k+b+k+c+k+d+k+e+k}{5}$

$\quad= \large\frac{a+b+c+d+e+5k}{5}$

$\quad= \large\frac{a+b+c+d+e}{5}+\frac{5k}{5}$

$n=m+k$

The standard deviation $S = \sqrt {\large\frac{(a-m)^2+(b-m)+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$

new standard deviation will be

$S_1= \sqrt {\large\frac{ (a+k-n)^2+(b+k-n)^2+(c+k-n)^2+(d+k-n)^2 +(e+k-m)^2}{5}}$

We know that

$a+k-n=a+k-(m+k)$

$\qquad= a+k-m-k$

$\qquad= a-m$

Similarly, $b+k-n =b-m$

$c+k-n=c-m$

$d+k-n=d-m$

$e+k-n =e-m$

substituting the values ,

$S_1= \sqrt { \large\frac{(a-m)^2+(b-m)^2+(c-m)^2 +(d-m)^2+(e-m)^2}{5}}$

The new standard deviation is s.

Thereby adding or substracting any constant to the observation does not change the value of standard deviation .

Hence A is the correct answer.

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