Question

If $\underset {i=1}{\overset{n}{\Sigma}}\left(x_{i}-a\right)=n$ and $\underset {i=1}{\overset{n}{\Sigma}}\left(x_{i}-a\right)^{2}=n a,(n, a>1)$ then the standard deviation of n observations $x _{1}, x _{2}, \ldots ., x _{ n }$ is

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

Answer 1

Answer: (B)

$S.D =\sqrt{\frac{\displaystyle\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}-\left(\frac{\displaystyle\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}\right)^{2}}$
$=\sqrt{\frac{n a}{n}-\left(\frac{n}{n}\right)^{2}}$
$\left\{ \text{Given}\displaystyle\sum_{i=1}^{n}\left(x_{i}-a\right)=n \displaystyle\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\right\}$
$=\sqrt{a-1}$