KCET · Maths · Statistics
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 \(\sqrt{6}\)
- C \(2 \sqrt{\frac{10}{3}}\)
- D \(2 \sqrt{6}\)
Answer & Solution
Correct Answer
(D) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Given, numbers are \(-1,0,1, k\).
Standard deviation, \(\sigma=\sqrt{5}\)
\[
\sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2
\]
\[
\begin{aligned}
& \Rightarrow \quad 5=\frac{1+0+1+k^2}{4}-\left(\frac{-1+0+1+k}{4}\right)^2 \\
& \Rightarrow \quad 5=\frac{2+k^2}{4}-\frac{k^2}{16} \Rightarrow 80=8+4 k^2-k^2 \\
& \Rightarrow \quad 72=3 k^2 \Rightarrow k^2=24 \Rightarrow k=2 \sqrt{6} .
\end{aligned}
\]
Standard deviation, \(\sigma=\sqrt{5}\)
\[
\sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2
\]
\[
\begin{aligned}
& \Rightarrow \quad 5=\frac{1+0+1+k^2}{4}-\left(\frac{-1+0+1+k}{4}\right)^2 \\
& \Rightarrow \quad 5=\frac{2+k^2}{4}-\frac{k^2}{16} \Rightarrow 80=8+4 k^2-k^2 \\
& \Rightarrow \quad 72=3 k^2 \Rightarrow k^2=24 \Rightarrow k=2 \sqrt{6} .
\end{aligned}
\]
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