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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
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Verified Answer
The correct answer is:
$2 \sqrt{6}$
Given, numbers are $-1,0,1, k$.
Standard deviation, $\sigma=\sqrt{5}$
$$
\sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2
$$
$$
\begin{aligned}
& \Rightarrow \quad 5=\frac{1+0+1+k^2}{4}-\left(\frac{-1+0+1+k}{4}\right)^2 \\
& \Rightarrow \quad 5=\frac{2+k^2}{4}-\frac{k^2}{16} \Rightarrow 80=8+4 k^2-k^2 \\
& \Rightarrow \quad 72=3 k^2 \Rightarrow k^2=24 \Rightarrow k=2 \sqrt{6} .
\end{aligned}
$$
Standard deviation, $\sigma=\sqrt{5}$
$$
\sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2
$$
$$
\begin{aligned}
& \Rightarrow \quad 5=\frac{1+0+1+k^2}{4}-\left(\frac{-1+0+1+k}{4}\right)^2 \\
& \Rightarrow \quad 5=\frac{2+k^2}{4}-\frac{k^2}{16} \Rightarrow 80=8+4 k^2-k^2 \\
& \Rightarrow \quad 72=3 k^2 \Rightarrow k^2=24 \Rightarrow k=2 \sqrt{6} .
\end{aligned}
$$
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