Question

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

• A. $2\sqrt{\frac{10}{3}}$
• B. $2\sqrt{6}$
• C. $4\sqrt{\frac{5}{3}}$
• D. $\sqrt{6}$

Answer 1

Answer: (B)

$2\sqrt{6}$

S.D$=\sqrt{\frac{\sum (x-x{)}^2}{n}}$
$\overline{x}=\frac{\sum }{4}=\frac{-1+0+1+k}{4}=\frac{k}{4}$
Now $\sqrt{5}=\sqrt{\frac{{(-1-\frac{k}{4})}^2+{(0-\frac{k}{4})}^2+{(1+\frac{k}{4})}^2+{(k-\frac{k}{4})}^2}{4}}$
$\Rightarrow 5\times 4=2{(1+\frac{k}{16})}^2+\frac{5k^2}{8}$
$\Rightarrow 18=\frac{3k^2}{4}$
$\Rightarrow k^2=24$
$\Rightarrow k=2\sqrt{6}$.