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

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Aicharya · Accepted Answer

$\begin{aligned} & \bar{x}=50, 	ext { Standard deviation }=5 \ & \mathrm{n}=100 \ & 	ext { Mean } \bar{x}=50 \ & \Rightarrow \frac{\sum x_{\mathrm{i}}}{100}=50 \ & \Rightarrow \sum x_{\mathrm{i}}=5000 \ & 	ext { Now, standard deviation }=\sqrt{\frac{\sum x_i^2}{\mathrm{n}}-\left(\frac{\sum x_{\mathrm{i}}}{\mathrm{n}}ight)^2} \ & \Rightarrow 5=\sqrt{\frac{\sum x_{\mathrm{i}}^2}{100}-\left(\frac{5000}{100}ight)^2} \ & \Rightarrow 25=\frac{\sum x_i^2}{100}-(50)^2 \ & \Rightarrow \frac{\sum x_i^2}{100}=25+2500 \ & \Rightarrow \sum x_i^2=252500 \end{aligned}$

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

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