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 $\mathrm{n}$ observations
$\mathrm{X}_{1}, \mathrm{X}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}$ is
Correct Option: , 2
Solution:
S.D $=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}-\left(\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}\right)^{2}}$
$=\sqrt{\frac{n a}{n}-\left(\frac{n}{n}\right)^{2}}$
$\left\{\right.$ Given $\left.\sum_{i=1}^{n}\left(x_{i}-a\right)=n \sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\right\}$
$=\sqrt{a-1}$
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