If sumlimits_{i = 1}^{18} {(x_i - 8) = 9} and | vedclass

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(C) Let $y_i = x_i - 8$. The variance of $y_i$ is given by:$Var(y) = \frac{1}{n} \sum y_i^2 - \left(\frac{1}{n} \sum y_i\right)^2$$Var(y) = \frac{45}{

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$3/2$

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(C) Let $y_i = x_i - 8$. The variance of $y_i$ is given by:$Var(y) = \frac{1}{n} \sum y_i^2 - \left(\frac{1}{n} \sum y_i\right)^2$$Var(y) = \frac{45}{18} - \left(\frac{9}{18}\right)^2$$Var(y) = \frac{5}{2} - \left(\frac{1}{2}\right)^2 = \frac{5}{2} - \frac{1}{4} = \frac{10-1}{4} = \frac{9}{4}$Since the standard deviation is invariant under the change of origin,the standard deviation of $x_i$ is the same as the standard deviation of $y_i$.$S.D. = \sqrt{Var(y)} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

If $\sum\limits_{i = 1}^{18} {(x_i - 8) = 9}$ and $\sum\limits_{i = 1}^{18} {(x_i - 8)^2 = 45}$,then the standard deviation of $x_1, x_2, \dots, x_{18}$ is:

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