If both the mean and the standard deviation of $50$ observations $x_1,\mathrm{ }{x}_2,\mathrm{ }\dots ,\mathrm{ }{x}_{50}$ are equal to $16$, then the mean of ${\left(x_1-4\right)}^2,\mathrm{ }{\left(x_2-4\right)}^2,\mathrm{ }\dots ,{\left(x_{50}-4\right)}^2$ is

For observations $x_1,\hspace{0.17em}{x}_2,\hspace{0.17em}...........x_{50}$

We have, mean, $\overline{x}=\frac{\sum x_i^{ }}{50}=16 .....\left(i\right)$

And, standard deviation, $\sigma =\sqrt{\frac{\sum x_i^2}{50}-{\left(\overline{x}\right)}^2}=16$

$\Rightarrow \frac{\sum x_i^2}{50}={16}^2+{\left(\overline{x}\right)}^2$

$\Rightarrow \frac{\sum x_i^2}{50}={16}^2+{16}^2=512 .....\left(ii\right)$

Now, the mean value of ${\left(x_1-4\right)}^2,\hspace{0.17em}{\left(x_2-4\right)}^2,\hspace{0.17em}..., {\left(x_{50}-4\right)}^2$ will be

$=\frac{\sum {\left(x_i-4\right)}^2}{50}=\frac{\sum x_i^2-8 \sum x_i+16\times 50}{50}$

$=\frac{\sum x_i^2}{50}-8\frac{\sum x_i}{50}+16$

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

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