The variance of the observations 112, 116, 12 | vedclass

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Question

(B) The given observations are $112, 116, 120, 125, 132$.The mean $\bar{x} = \frac{\Sigma x_i}{n} = \frac{112 + 116 + 120 + 125 + 132}{5} = \frac{605}

Vedclass · Accepted Answer

$48.8$

Vedclass · Answer

(B) The given observations are $112, 116, 120, 125, 132$.The mean $\bar{x} = \frac{\Sigma x_i}{n} = \frac{112 + 116 + 120 + 125 + 132}{5} = \frac{605}{5} = 121$.The variance $\sigma^2 = \frac{\Sigma (x_i - \bar{x})^2}{n}$.Calculating $(x_i - \bar{x})^2$ for each observation:$(112 - 121)^2 = (-9)^2 = 81$$(116 - 121)^2 = (-5)^2 = 25$$(120 - 121)^2 = (-1)^2 = 1$$(125 - 121)^2 = (4)^2 = 16$$(132 - 121)^2 = (11)^2 = 121$Sum of squares $\Sigma (x_i - \bar{x})^2 = 81 + 25 + 1 + 16 + 121 = 244$.Therefore,$\sigma^2 = \frac{244}{5} = 48.8$.

The variance of the observations $112, 116, 120, 125, 132$ is $......$

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