Question:
If $\sum_{i=1}^{n}\left(x_{i}-a\right)=n$ and $\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a,(n, a>1)$, then
the standard deviation of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is :
Correct Option: , 4
Solution:
Standard deviation
$=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}}{n}-\left(\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}\right)^{2}} \quad[\because n, a>1]$
$=\sqrt{\frac{n a}{n}-\left(\frac{n}{n}\right)^{2}}=\sqrt{a-1}$