If the mean of 100 observations is 50 and their standard deviation is 5 , then the sum of squares of all observations is

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Aicharya · Accepted Answer

Given; $\bar{x}=50$ $n=100$ and $S . D(\sigma)=5$ As, $\bar{x}=\frac{\Sigma x_i}{n} \Rightarrow \Sigma x_i=n . \bar{x}=5000$. Also, S.D. is given by, $\begin{aligned} \sigma & =\sqrt{\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2} \ \Rightarrow \quad \sigma^2 & =\frac{\Sigma x_i^2}{n}-(\bar{x})^2 \Rightarrow 25=\frac{\Sigma x_i^2}{100}-(50)^2 \ \Rightarrow \quad \Sigma x_i^2 & =252500 \end{aligned}$

If the mean of 100 observations is 50 and their standard deviation is 5 , then the sum of squares of all…

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