If $\sum _{i=1}^{18}(x_i-8)=9$ and $\sum _{i=1}^{18}{(x_i-8)}^2=45$, then standard deviation of $x_1,x_2,...,x_{18}$ is

The correct option is C $\frac{3}{2}$

Given: $\sum _{i=1}^{18}\left(x_i-8\right)=9$

$\sum _{i=1}^{18}{\left(x_i-8\right)}^2=45$

Let $X$ be a random variable taking values $x_1,........x_{18.}$

Then $X-8$ has the values $x_1-8,....,x_{18}-8.$

Now, $E\left(X-8\right)=\frac{\sum _{i=1}^{18}\left(x_i-8\right)}{18}$

$=\frac{9}{18}=\frac{1}{2}$

And $E\left[{\left(X-8\right)}^2\right]=\frac{\sum _{i=1}^{18}{\left(x_i-8\right)}^2}{18}$

$=\frac{45}{18}=\frac{5}{2}$

Thus, $Var\left(X-8\right)=E\left[{\left(X-8\right)}^2\right]-{\left[E\left(X-8\right)\right]}^2$

$=\frac{5}{2}-{\left(\frac{1}{2}\right)}^2$

$=\frac{5}{2}-\frac{1}{4}$

$=\frac{9}{4}$

We know $Var\left(1.X-8\right)=1^{2}Var\left(X\right)$, thus,

$Var\left(X\right)=\frac{9}{4}$

Standard deviation of $X=\sqrt{Var\left(X\right)}$

$=\sqrt{\frac{9}{4}}=\frac{3}{2}$