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

Given:

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

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

And, $\sum _{i=1}^n{\left(x_i-1\right)}^2=5n$

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

Form $\left(1\right)+\left(2\right)$, we get

$\sum _{i=1}^n\left(2x_i^2+2\right)=14n$ 

$\Rightarrow \sum _{i=1}^n{x}_i^2=6n ...\left(3\right)$

From $\left(1\right)-\left(2\right)$, we get 

$\Rightarrow \sum _{i=1}^n{x}_i=n ...\left(4\right)$

$\therefore$ Variance, ${\sigma }^2=E\left(x^2\right)-{\left(E\left(x\right)\right)}^2$

$\Rightarrow {\sigma }^2=\frac{\sum x^2}{n}-\frac{\sum x}{n}$

$\Rightarrow {\sigma }^2=\frac{6n}{n}-{\left(\frac{n}{n}\right)}^2=6-1=5$

$\therefore$ Standard deviation, $\sigma =\sqrt{5}$.