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

If the variance of the numbers $-1,0,1, \mathrm{k}$ is 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. $\sqrt{6}$

Solution

$\begin{array}{ll} & \text { Variance }=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_i^2-\bar{x}^2 \\ & \text { Here, } \mathrm{n}=4 \text { and variance }=5 \\ \therefore \quad & 5=\frac{1}{4}\left[(-1)^2+(0)^2+(1)^2+\mathrm{k}^2\right]-\left(\frac{-1+0+1+\mathrm{k}}{4}\right)^2 \\ \therefore \quad & 5=\frac{2+\mathrm{k}^2}{4}-\frac{\mathrm{k}^2}{16} \\ \therefore \quad & 80=8+4 \mathrm{k}^2-\mathrm{k}^2 \\ \therefore \quad & 3 \mathrm{k}^2=72 \\ \therefore \quad & \mathrm{k}^2=24 \\ \therefore \quad & \mathrm{k}=2 \sqrt{6} \quad \quad \ldots[\because \mathrm{k}>0]\end{array}$

Asked in: MHT CET 2023 (12 May Shift 1)

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