Solution:
It is given that mean and standard deviation of 100 items are 50 and 4, respectively
We now need to find the sum of all items and the sum of the squares of the items
As per the criteria given,
No. of items, $\mathrm{n}=100$
Given items mean, $\bar{x}=50$
But it is known that,
$\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$
Substitute the corresponding values,
$50=\frac{\sum x_{i}}{100}$
$\Rightarrow \sum x_{\mathrm{i}}=50 \times 100=5000$
As a result, the sum of all the 100 items $=5000$
It is also given that the standard deviation of the 100 items is 4
i.e., $\sigma=4$
But it is known that
$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$
Substitute the corresponding values,
$4=\sqrt{\frac{\sum x_{i}^{2}}{100}}-\left(\frac{5000}{100}\right)^{2}$
Taking square on both the sides, we obtain
$4^{2}=\frac{\sum x_{i}^{2}}{100}-(50)^{2}$
$\Rightarrow 16=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{100}-2500$
$\Rightarrow 16+2500=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{100}$
On rearranging we have
$\Rightarrow \frac{\sum \mathrm{x}_{i}^{2}}{100}=2516$
$\Rightarrow \sum \mathrm{x}_{\mathrm{i}}^{2}=2516 \times 100$
$\Rightarrow \sum \mathrm{x}_{\mathrm{i}}^{2}=251600$
As a result, the sum of the squares of all the 100 items is 251600 .