Question

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

• A. 58.8
• B. 48.8
• C. 61.8
• D. None of these

Answer 1

Answer: (B)

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