Question

If $\sum _{i=1}^n(x_i-a)$ and $\sum _{i=1}^n(x_i-a{)}^2=na,(n,a>1)$ then the standard deviation of $n$ observations $x_1,x_2,\cdots ,x_n$ is :

• A. $n\sqrt{a-1}$
• B. $\sqrt{na-1}$
• C. $a-1$
• D. $\sqrt{a-1}$

Answer 1

The correct option is D $\sqrt{a-1}$

$S.D=\sqrt{\frac{\sum (x_i-a{)}^2}{n}-{\left(\frac{\sum (x_i-a)}{n}\right)}^2}$
$=\sqrt{(\frac{na}{n}-{\left(\frac{n}{n}\right)}^2)}=\sqrt{a-1}$