A data consists of n observations: $x_1,x_2,\mathrm{....}{x}_n.$ If $\sum _{i=1}^n{(x_i+1)}^2=9n$ and $\sum _{i=1}^n{(x_i-1)}^2=5n$ then the standard deviation of this data is

$\sum _{i=1}^n{(x_i+1)}^2=9n\Rightarrow \Sigma x_i^2+2\Sigma x_i+n=9n$

$\Sigma x_i^2+2\Sigma x_i=8n \dots .\left(1\right)$

$\sum _{i=1}^n{(x_i-1)}^2=5n\Rightarrow \Sigma x_i^2-2\Sigma x_i+n=5n$

$\Sigma x_i^2-2\Sigma x_i=4n \dots \left(2\right)$

From (1) & (2), $\Sigma x_i^2=6n\&\Sigma x_i=n$

Hence standard deviation $=\sqrt{\frac{\Sigma x_i^2}{h}-{\left(\frac{\Sigma x_i}{h}\right)}^2}=\sqrt{\frac{6n}{h}}-{\left(\frac{n}{n}\right)}^2=\sqrt{5}$