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

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(D) Let $d_i = x_i - 8$. The standard deviation of $x_i$ is the same as the standard deviation of $d_i$,denoted as $\sigma_d$. $\sigma_d = \sqrt{\frac

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$3/2$

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(D) Let $d_i = x_i - 8$. The standard deviation of $x_i$ is the same as the standard deviation of $d_i$,denoted as $\sigma_d$. $\sigma_d = \sqrt{\frac{\sum d_i^2}{n} - \left( \frac{\sum d_i}{n} \right)^2}$ Given $n = 18$,$\sum d_i = 9$,and $\sum d_i^2 = 45$. $\sigma_d = \sqrt{\frac{45}{18} - \left( \frac{9}{18} \right)^2}$ $\sigma_d = \sqrt{\frac{5}{2} - \left( \frac{1}{2} \right)^2}$ $\sigma_d = \sqrt{\frac{5}{2} - \frac{1}{4}} = \sqrt{\frac{10 - 1}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$

If $\sum_{i=1}^{18} (x_i - 8) = 9$ and $\sum_{i=1}^{18} (x_i - 8)^2 = 45$,find the standard deviation of $x_1, x_2, \dots, x_{18}$.

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