Mean and standard deviation of 100 items are 50 and 4,

Given mean and standard deviation of 100 items are 50 and 4, respectively Now we have to find the sum of all items and the sum of the squares of the items As per given criteria,

Number of items, $\mathrm{n}=100$

Mean of the given items, $\bar{x}=50$

But we know,

$\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$

Substituting the corresponding values, we get

$50=\frac{\sum x_{i}}{100}$

$\Rightarrow \sum x_{i}=50 \times 100=5000$

Hence the sum of all the 100 items $=5000$

Also given the standard deviation of the 100 items is 4

i.e., $\sigma=4$

But we know

$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$

Substituting the corresponding values, we get

$4=\sqrt{\frac{\sum x_{i}^{2}}{100}-\left(\frac{5000}{100}\right)^{2}}$

Now taking square on both sides, we get

$4^{2}=\frac{\sum x_{i}^{2}}{100}-(50)^{2}$

$\Rightarrow 16=\frac{\sum x_{i}^{2}}{100}-2500$

$\Rightarrow 16+2500=\frac{\sum x_{i}^{2}}{100}$

On rearranging we get

$\Rightarrow \frac{\sum x_{i}^{2}}{100}=2516$'

$\Rightarrow \sum x_{i}^{2}=2516 \times 100$

$\Rightarrow \sum x_{i}^{2}=251600$

And the sum of the squares of all the 100 items is 251600 .