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# Let $a,b,c,d,e$ be the observations with mean m and standard deviations $S$. The standard deviation of the observations $a+k,b+k,c+k,d+k,e+k$ is

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