The variance of the variates 112, 116, 120, 1 | vedclass

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Question

(C) The given observations are $112, 116, 120, 125, 132$. First,calculate the Arithmetic Mean $(\bar{x})$: $\bar{x} = \frac{112 + 116 + 120 + 125 + 13

Vedclass · Accepted Answer

$48.8$

Vedclass · Answer

(C) The given observations are $112, 116, 120, 125, 132$. First,calculate the Arithmetic Mean $(\bar{x})$: $\bar{x} = \frac{112 + 116 + 120 + 125 + 132}{5} = \frac{605}{5} = 121$. Next,calculate the sum of the squares of the deviations from the mean: $\sum(x_i - \bar{x})^2 = (112 - 121)^2 + (116 - 121)^2 + (120 - 121)^2 + (125 - 121)^2 + (132 - 121)^2$ $= (-9)^2 + (-5)^2 + (-1)^2 + (4)^2 + (11)^2$ $= 81 + 25 + 1 + 16 + 121 = 244$. Finally,the variance $(\sigma^2)$ is given by: $\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{244}{5} = 48.8$.

The variance of the variates $112, 116, 120, 125, 132$ about their $A.M.$ is

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