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Q.
If $\displaystyle\sum^{9}_{i =1}\left(x_{i} - 5 \right) = 9$ and $ \displaystyle\sum ^{9}_{i =1}\left(x_{i} - 5 \right)^{2} = 45 ,$ then the standard deviation of the 9 items $x_1 , x_2 , .......,x_9$ is :
Standard deviation of $x_{i}-5$ is
$\sigma=\sqrt{\frac{\displaystyle\sum_{i=1}^{9}\left(x_{i}-5\right)^{2}}{9}-\left(\frac{\displaystyle\sum_{i=1}^{9}\left(x_{i}-5\right)}{9}\right)^{2}}$
$\Rightarrow \sigma=\sqrt{5-1}=2$
As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.
So, $\sigma$ of $x_{i}$ is $2$