MHT CET · Maths · Statistics
Variance of first \(2 n\) natural numbers is
- A \(\frac{4 n^2+1}{12}\)
- B \(\frac{(2 \mathrm{n}-1)^2}{12}\)
- C \(\frac{\mathrm{n}^2}{3}-1\)
- D \(\frac{4 n^2-1}{12}\)
Answer & Solution
Correct Answer
(D) \(\frac{4 n^2-1}{12}\)
Step-by-step Solution
Detailed explanation
\(\sigma^2=\frac{1}{2 n}\left[1^2+2^2+3^2+\ldots\right. \left.+(2 n)^2\right] \) \( -\left(\frac{1+2+3+\ldots+2 n}{2 n}\right)^2\)
\(=\frac{1}{2 n}\left[\frac{2 n(2 n+1)(4 n+1)}{6}\right]-\left[\frac{1}{2 n} \times \frac{2 n(2 n+1)}{2}\right]^2 \)
\( =\frac{(2 n+1)(4 n+1)}{6}-\left(\frac{2 n+1}{2}\right)^2 \)
\( =\frac{2 n+1}{2}\left(\frac{4 n+1}{3}-\frac{2 n+1}{2}\right) \)
\( =\frac{2 n+1}{2}\left(\frac{2 n-1}{6}\right) \)
\( =\frac{4 n^2-1}{12}\)
\(=\frac{1}{2 n}\left[\frac{2 n(2 n+1)(4 n+1)}{6}\right]-\left[\frac{1}{2 n} \times \frac{2 n(2 n+1)}{2}\right]^2 \)
\( =\frac{(2 n+1)(4 n+1)}{6}-\left(\frac{2 n+1}{2}\right)^2 \)
\( =\frac{2 n+1}{2}\left(\frac{4 n+1}{3}-\frac{2 n+1}{2}\right) \)
\( =\frac{2 n+1}{2}\left(\frac{2 n-1}{6}\right) \)
\( =\frac{4 n^2-1}{12}\)
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