Prove the following


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 :

  1. (1) $a-1$

  2. (2) $n \sqrt{a-1}$

  3. (3) $\sqrt{n(a-1)}$

  4. (4) $\sqrt{a-1}$

Correct Option: , 4


Standard deviation

$=\sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)^{2}}{n}-\left(\frac{\sum_{i=1}^{n}\left(x_{i}-a\right)}{n}\right)^{2}} \quad[\because n, a>1]$

$=\sqrt{\frac{n a}{n}-\left(\frac{n}{n}\right)^{2}}=\sqrt{a-1}$

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