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  4. The variance of observations 112, 116, 120, 125, 132 is :-

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Q. The variance of observations $112, 116, 120, 125, 132$ is :-

Statistics

Solution:

Given observations $112,116,120,125,132$
$\bar{x}=\frac{\Sigma x_1}{n}=\frac{605}{5}=121 $
$\Sigma\left(x_1-\bar{x}\right)^2=81+25+1+16+121=244 $
$\therefore \sigma^2=\frac{\Sigma\left(x_1-\bar{x}\right)^2}{n}=\frac{244}{5}=48.8$