Question:
If the standard deviation of the numbers $-1,0,1, \mathrm{k}$ is $\sqrt{5}$ where $\mathrm{k}>0$, then $\mathrm{k}$ is equal to
Correct Option: , 2
Solution:
$S . D=\sqrt{\frac{\sum(x-\bar{x})^{2}}{n}}$
$\bar{x}=\frac{\sum x}{4}=\frac{-1+0+1+k}{4}=\frac{k}{4}$
Now $\sqrt{5}=\sqrt{\frac{\left(-1-\frac{k}{4}\right)^{2}+\left(0-\frac{k}{4}\right)^{2}+\left(1-\frac{k}{4}\right)^{2}+\left(k-\frac{k}{4}\right)^{2}}{4}}$
$\Rightarrow 5 \times 4=2\left(1+\frac{\mathrm{k}}{16}\right)^{2}+\frac{5 \mathrm{k}^{2}}{8}$
$\Rightarrow 18=\frac{3 \mathrm{k}^{2}}{4}$
$\Rightarrow \mathrm{k}^{2}=24$
$\Rightarrow \mathrm{k}=2 \sqrt{6}$