Solve this following

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

 

  1. $\mathrm{n} \sqrt{\mathrm{a}-1}$

  2. $\sqrt{a-1}$

  3. $a-1$

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


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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