For the set of 50 observation, the sum of their squares is 3050 , their arithmetic mean is 6 . Hence the standard deviation of these observations is

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Aicharya · Accepted Answer

We have $\mathrm{n}=50, \Sigma \mathrm{x}_{\mathrm{i}}^2=3050, \overline{\mathrm{x}}=6$ $\begin{aligned} & 	ext { S.D. }=\sqrt{\frac{1}{n} \Sigma x_i^2-(\bar{x})^2} \ & =\sqrt{\frac{3050}{50}-(6)^2}=\sqrt{61-36}=\sqrt{25}=5 \end{aligned}$

For the set of 50 observation, the sum of their squares is 3050 , their arithmetic mean is 6 . Hence the…

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