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

Solution :
$$\eqalign{ & {\sigma ^2} = \frac{{\sum {{x^2}} }}{n} - {\left( {\frac{{\sum x }}{n}} \right)^2} \cr & \Rightarrow \,\,\frac{{{k^2} + 2}}{4} - {\left( {\frac{k}{4}} \right)^2} = 5 \cr & \Rightarrow \,\,4{k^2} + 8 - {k^2} = 80 \cr & \Rightarrow \,\,3{k^2} = 72 \cr & \Rightarrow \,\,k = 2\sqrt 6 \cr} $$