A data consists of n observations: x _ 1 , x _ 2 , , x _ n . If ∑ _...

Given:

$\mathop{∑}_{i = 1}^{n}{\left({x}_{i} + 1\right)}^{2} = 9n$

$⇒ \mathop{∑}_{i = 1}^{n}\left({x}_{i}^{2} + 2{x}_{i} + 1\right) = 9n . . . \left(1\right)$

And, $\mathop{∑}_{i = 1}^{n}{\left({x}_{i} - 1\right)}^{2} = 5n$

$⇒ \mathop{∑}_{i = 1}^{n}\left({x}_{i}^{2} - 2{x}_{i} + 1\right) = 5n . . . \left(2\right)$

Form \left(1\right) + \left(2\right)$, we get</p><p>$\mathop{∑}_{i = 1}^{n}\left(2{x}_{i}^{2} + 2\right) = 14n$</p><p>$⇒ \mathop{∑}_{i = 1}^{n}{x}_{i}^{2} = 6n . . . \left(3\right)$</p><p>From \left(1\right) - \left(2\right)$, we get

$⇒ \mathop{∑}_{i = 1}^{n}{x}_{i} = n . . . \left(4\right)$

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

$⇒ {\sigma}^{2} = \dfrac{∑ {x}^{2}}{n} - \dfrac{∑ x}{n}$

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

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