If sum_{i=1}^{10} (x_i - 15) = 12 and sum_{i= | vedclass

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(B) Let $y_i = x_i - 15$. Then the given values are $\sum_{i=1}^{10} y_i = 12$ and $\sum_{i=1}^{10} y_i^2 = 18$.Standard deviation is invariant under

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$\frac{3}{5}$

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(B) Let $y_i = x_i - 15$. Then the given values are $\sum_{i=1}^{10} y_i = 12$ and $\sum_{i=1}^{10} y_i^2 = 18$.Standard deviation is invariant under change of origin,so the standard deviation of $x_i$ is the same as the standard deviation of $y_i$.The formula for variance $\sigma^2$ is $\frac{1}{n} \sum y_i^2 - (\frac{1}{n} \sum y_i)^2$.Here $n = 10$,$\sum y_i = 12$,and $\sum y_i^2 = 18$.$\sigma^2 = \frac{18}{10} - (\frac{12}{10})^2 = 1.8 - (1.2)^2 = 1.8 - 1.44 = 0.36$.Standard deviation $\sigma = \sqrt{0.36} = 0.6 = \frac{6}{10} = \frac{3}{5}$.

If $\sum_{i=1}^{10} (x_i - 15) = 12$ and $\sum_{i=1}^{10} (x_i - 15)^2 = 18$,find the standard deviation of the observations $x_1, x_2, \dots, x_{10}$.

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