$\sum _{i=1}^n{(x+1)}^2=9n,\sum _{ }^{ }{(x_1-1)}^2=5n$ then standard deviation of these $\text{'}n\text{'}$observation  $\left(x_i\right)$ 

$\begin{array}{l}\sum {(x_i+1)}^2=9n\\ \sum {(x_i-1)}^2=5n\\ \sum x_i^2+2\sum x_i+n=9n\\ \sum x_1^2+2\sum x_i=8n\to \left(1\right)\\ \sum {(x_i-1)}^2=5n\\ \sum (x_1^2-2x_i+1)=5n\\ \sum x_1^2-2\sum x_i+n=5n\\ \sum x_1^2-2\sum x_i=4n\to \left(2\right)\end{array}$

$\begin{array}{l}\left(1\right)+\left(2\right)\\ 2\sum x_i^2=12n\end{array}$

Substitutes $\sum x_i^2=6n$ in eqn (2), we get

$\sum x_i=n$

Standard deviation $=\sqrt{\frac{{\displaystyle \sum x_1^2}}{n}-{\left(\frac{{\displaystyle \sum x_i}}{n}\right)}^2}$

$\Rightarrow \sqrt{\frac{6n}{n}-{\left(\frac{n}{n}\right)}^2}=\sqrt{5}$