Question

If the standard deviation of the numbers $-1,0,1,k$ is $\sqrt{5}$, then value of $k^2$ is

Answer 1

Mean
$=\overline{x}=\frac{\sum x}{n}=\frac{-1+0+1+k}{4}=\frac{k}{4}$

Standard deviation
$\sqrt{\frac{\sum x^2}{n}-(\overline{x}{)}^2}=\sqrt{5}\Rightarrow \sqrt{\frac{k^2+2}{4}-\frac{k^2}{16}}=\sqrt{5}\Rightarrow \frac{3k^2+8}{16}=5\therefore k^2=24$