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(C) The formula for standard deviation is $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$.Given numbers are $-1, 0, 1, k$. The mean $\bar{x} = \fra

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$2\sqrt{6}$

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(C) The formula for standard deviation is $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$.Given numbers are $-1, 0, 1, k$. The mean $\bar{x} = \frac{-1 + 0 + 1 + k}{4} = \frac{k}{4}$.The variance $\sigma^2 = 5$. Thus,$\frac{(-1 - \frac{k}{4})^2 + (0 - \frac{k}{4})^2 + (1 - \frac{k}{4})^2 + (k - \frac{k}{4})^2}{4} = 5$.Expanding the terms: $\frac{(1 + \frac{k}{2} + \frac{k^2}{16}) + \frac{k^2}{16} + (1 - \frac{k}{2} + \frac{k^2}{16}) + (\frac{3k}{4})^2}{4} = 5$.$\frac{2 + \frac{3k^2}{16} + \frac{9k^2}{16}}{4} = 5$.$\frac{2 + \frac{12k^2}{16}}{4} = 5$.$2 + \frac{3k^2}{4} = 20$.$\frac{3k^2}{4} = 18$.$k^2 = 24$.Since $k > 0$,$k = \sqrt{24} = 2\sqrt{6}$.

If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt{5}$ where $k > 0,$ then $k$ is equal to

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