Solve this following


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


  1. $2 \sqrt{\frac{10}{3}}$

  2. $2 \sqrt{6}$

  3. $4 \sqrt{\frac{5}{3}}$

  4. $\sqrt{6}$

Correct Option: , 2


$S . D=\sqrt{\frac{\sum(x-\bar{x})^{2}}{n}}$

$\bar{x}=\frac{\sum x}{4}=\frac{-1+0+1+k}{4}=\frac{k}{4}$

Now $\sqrt{5}=\sqrt{\frac{\left(-1-\frac{k}{4}\right)^{2}+\left(0-\frac{k}{4}\right)^{2}+\left(1-\frac{k}{4}\right)^{2}+\left(k-\frac{k}{4}\right)^{2}}{4}}$

$\Rightarrow 5 \times 4=2\left(1+\frac{\mathrm{k}}{16}\right)^{2}+\frac{5 \mathrm{k}^{2}}{8}$

$\Rightarrow 18=\frac{3 \mathrm{k}^{2}}{4}$

$\Rightarrow \mathrm{k}^{2}=24$

$\Rightarrow \mathrm{k}=2 \sqrt{6}$


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