The mean of four observations is 3 . If the sum of the squares of these observations is 48 , then their standard deviation is

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Aicharya · Accepted Answer

Let four observations are be $x_1, x_2, x_3$ and $x_4$ Given, mean and $ \begin{aligned} (\bar{x}) & =3 \ \Sigma x_i^2 & =48 \end{aligned} $ $\begin{aligned} 	herefore \quad \mathrm{SD} & =\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \ =\sqrt{\frac{48}{4}-(3)^2}=\sqrt{12-9} & =\sqrt{3}\end{aligned}$

The mean of four observations is 3 . If the sum of the squares of these observations is 48 , then their…

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