If sum_{i=1}^9(x_i-5)=9 and sum_{i=1}^9(x_i-5 | vedclass

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(A) Let $y_i = x_i - 5$. Then the given sums are $\sum_{i=1}^9 y_i = 9$ and $\sum_{i=1}^9 y_i^2 = 45$.The variance of the observations $x_i$ is the sa

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(A) Let $y_i = x_i - 5$. Then the given sums are $\sum_{i=1}^9 y_i = 9$ and $\sum_{i=1}^9 y_i^2 = 45$.The variance of the observations $x_i$ is the same as the variance of $y_i$ because shifting the data by a constant does not change the variance.The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum y_i^2 - (\bar{y})^2$.Here,$n = 9$,$\sum y_i = 9$,and $\sum y_i^2 = 45$.First,calculate the mean of $y$: $\bar{y} = \frac{1}{9} \sum_{i=1}^9 y_i = \frac{9}{9} = 1$.Now,calculate the variance: $\sigma^2 = \frac{45}{9} - (1)^2 = 5 - 1 = 4$.The standard deviation $\sigma$ is the square root of the variance: $\sigma = \sqrt{4} = 2$.

If $\sum_{i=1}^9(x_i-5)=9$ and $\sum_{i=1}^9(x_i-5)^2=45$,then the standard deviation of the nine observations $x_1, x_2, \ldots, x_9$ is

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