If both the mean and the standard deviation of 50 observations x _...

For observations ${x}_{1}, {x}_{2}, . . . . . . . . . . . {x}_{50}$

We have, mean, $\overset{¯}{x} = \dfrac{∑ {x}_{i}^{}}{50} = 16 . . . . . \left(i\right)$

And, standard deviation, $\sigma = \sqrt{\dfrac{∑ {x}_{i}^{2}}{50} - {\left(\overset{¯}{x}\right)}^{2}} = 16$

$⇒ \dfrac{∑ {x}_{i}^{2}}{50} = {16}^{2} + {\left(\overset{¯}{x}\right)}^{2}$

$⇒ \dfrac{∑ {x}_{i}^{2}}{50} = {16}^{2} + {16}^{2} = 512 . . . . . \left(ii\right)$

Now, the mean value of ${\left({x}_{1} - 4\right)}^{2}, {\left({x}_{2} - 4\right)}^{2}, . . . , {\left({x}_{50} - 4\right)}^{2}$ will be

$= \dfrac{∑ {\left({x}_{i} - 4\right)}^{2}}{50} = \dfrac{∑ {x}_{i}^{2} - 8 ∑ {x}_{i} + 16 \times 50}{50}$

$= \dfrac{∑ {x}_{i}^{2}}{50} - 8\dfrac{∑ {x}_{i}}{50} + 16$

Using \left(i\right)$and \left(ii\right),$

$= 512 - 8 \times 16 + 16 = 400 .$