MHT CET · Maths · Statistics
If the variance of the numbers \(-1,0,1, \mathrm{k}\) is 5 , where \(\mathrm{k}>0\), then \(\mathrm{k}\) is equal to
- A \(2 \sqrt{\frac{10}{3}}\)
- B \(2 \sqrt{6}\)
- C \(4 \sqrt{\frac{5}{3}}\)
- D \(\sqrt{6}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Variance \(=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_i^2-\bar{x}^2 \)
Here, \(\mathrm{n}=4 \text { and variance }=5 \)
\( \therefore 5=\frac{1}{4}\left[(-1)^2+(0)^2+(1)^2+\mathrm{k}^2\right]-\left(\frac{-1+0+1+\mathrm{k}}{4}\right)^2 \)
\( \therefore 5=\frac{2+\mathrm{k}^2}{4}-\frac{\mathrm{k}^2}{16} \)
\( \therefore 80=8+4 \mathrm{k}^2-\mathrm{k}^2 \)
\( \therefore 3 \mathrm{k}^2=72 \)
\( \therefore \mathrm{k}^2=24 \)
\( \therefore \mathrm{k}=2 \sqrt{6} \ldots[\because \mathrm{k}>0]\)
Here, \(\mathrm{n}=4 \text { and variance }=5 \)
\( \therefore 5=\frac{1}{4}\left[(-1)^2+(0)^2+(1)^2+\mathrm{k}^2\right]-\left(\frac{-1+0+1+\mathrm{k}}{4}\right)^2 \)
\( \therefore 5=\frac{2+\mathrm{k}^2}{4}-\frac{\mathrm{k}^2}{16} \)
\( \therefore 80=8+4 \mathrm{k}^2-\mathrm{k}^2 \)
\( \therefore 3 \mathrm{k}^2=72 \)
\( \therefore \mathrm{k}^2=24 \)
\( \therefore \mathrm{k}=2 \sqrt{6} \ldots[\because \mathrm{k}>0]\)
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