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

Option 1)
$2\sqrt{6}$

Option 2)
$2\sqrt{\frac{10}{3}}$

Option 3)
$4\sqrt{\frac{5}{3}}$

Option 4)
$\sqrt{6}$

Answers (1)

best_answer

$S.D=\sqrt{\frac{\Sigma xi^{2}}{n}-\left ( \frac{\Sigma xi}{n} \right )^{2}}=\sqrt{5}$

$\Rightarrow \frac{\Sigma xi^{2}}{n}-\left ( \frac{\Sigma xi}{n} \right )^{2}=5$

$\Rightarrow\frac{(-1^{2}+0^{2}+1^{2}+k^{2})}{4}-\left ( \frac{-1+0+1+k}{4} \right )^{2}=5$

$\Rightarrow\frac{2+k^{2}}{4}-\frac{k^{2}}{16}=5$

$8+4k^{2}-k^{2}=80$

$3k^{2}=72$

$k=2\sqrt{6}$

Option 1)

$2\sqrt{6}$

Option 2)

$2\sqrt{\frac{10}{3}}$

Option 3)

$4\sqrt{\frac{5}{3}}$

Option 4)

$\sqrt{6}$

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