If the standard deviation of the numbers -1, | vedclass

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(D) Given,the numbers are $-1, 0, 1, k$. Standard deviation,$\sigma = \sqrt{5}$. We know that the variance $\sigma^2$ is given by $\sigma^2 = \frac{\s

Vedclass · Accepted Answer

$2 \sqrt{6}$

Vedclass · Answer

(D) Given,the numbers are $-1, 0, 1, k$. Standard deviation,$\sigma = \sqrt{5}$. We know that the variance $\sigma^2$ is given by $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$. Here,$n = 4$. $\sigma^2 = 5 = \frac{(-1)^2 + 0^2 + 1^2 + k^2}{4} - \left(\frac{-1 + 0 + 1 + k}{4}\right)^2$. $5 = \frac{2 + k^2}{4} - \left(\frac{k}{4}\right)^2$. $5 = \frac{2 + k^2}{4} - \frac{k^2}{16}$. Multiply the entire equation by $16$: $80 = 4(2 + k^2) - k^2$. $80 = 8 + 4k^2 - k^2$. $72 = 3k^2$. $k^2 = 24$. Since $k > 0$,$k = \sqrt{24} = 2 \sqrt{6}$.

Class Interval	$0$-$5$	$5$-$10$	$10$-$15$	$15$-$20$	$20$-$25$
Frequency	$4$	$1$	$10$	$3$	$2$

Class	$10-20, 20-30, 30-40$
Frequency	$2, x, 2$

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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