Question

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

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

Answer 1

Answer: (D)

$\sigma^{2}=5, \overline{ x }=\frac{ k }{4}$
$\frac{1}{4}\left(1+0+1+ k ^{2}\right)-\frac{ k ^{2}}{16}=5$
$\frac{ k ^{2}+2}{4} \frac{- k ^{2}}{16}=5$
$\frac{4 k ^{2}+8- k ^{2}}{16}=5 $
$\Rightarrow 3 k ^{2}+8=80$
$3 k ^{2}=72$
$k ^{2}=24$
$k =\pm \sqrt{24}=2 \sqrt{6}$