If both the mean and the standard deviation of 50 observations x_1,x_2,,x_50 are equal to …

Question

If both the mean and the standard deviation of 50 observations $x_1,x_2,\ldots,x_{50}$ are equal to 16, then the mean of $(x_1-4)^2,(x_2-4)^2,\ldots,(x_{50}-4)^2$ is:

Accepted Answer

$\sum x_i = 50(16)=800$; $16^2 = \dfrac{\sum x_i^2}{50}-16^2 \implies \sum x_i^2 = 50(2\times256)=25600$

$$\text{Mean of }(x_i-4)^2 = \frac{\sum x_i^2-8\sum x_i+50(16)}{50} = \frac{25600-6400+800}{50} = \frac{20000}{50}=400$$

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