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Question

Mathematics Question on Variance and Standard Deviation

Standard deviation of first nn odd natural numbers is

A

n\sqrt{n}

B

(n+2)(n+1)3\sqrt{ \frac{(n+2)(n+1)}{3}}

C

n2−13\sqrt{ \frac{n^2 - 1 }{3}}

D

nn

Answer

n2−13\sqrt{ \frac{n^2 - 1 }{3}}

Explanation

Solution

Standard deviation, σ=Σxi2N−(Xˉ)2\sigma=\sqrt{\frac{\Sigma x_{i}^{2}}{N}-(\bar{X})^{2}}
∴Xˉ=ΣXiN\therefore \bar{X}=\frac{\Sigma X_{i}}{N}
=1+3+5+…(2n−1)n=\frac{1+3+5+\ldots(2 n-1)}{n}
=n2[1+2n−1]n=\frac{\frac{n}{2}[1+2 n-1]}{n}
=n2n=n=\frac{n^{2}}{n}=n
Again, Σxi2=12+32+52+…(2n−1)2\Sigma x_{i}^{2}=1^{2}+3^{2}+5^{2}+\ldots(2 n-1)^{2}
=Σ(2n−1)2=\Sigma(2 n-1)^{2}
=Σ(4n2−4n+1)=\Sigma\left(4 n^{2}-4 n+1\right)
=4Σn2−4Σn+Σ1=4 \Sigma n^{2}-4 \Sigma n+\Sigma 1
=4n(n+1)(2n+1)6−4n(n+1)2+n=\frac{4 n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n
=n[23(n+1)2n+1)−2(n+1)+1]\left.=n\left[\frac{2}{3}(n+1) 2 n+1\right)-2(n+1)+1\right]
=n3[2(2n2+3n+1)−6(n+1)+3]=\frac{n}{3}\left[2\left(2 n^{2}+3 n+1\right)-6(n+1)+3\right]
=n3[4n2+6n+2−6n−6+3]=\frac{n}{3}\left[4 n^{2}+6 n+2-6 n-6+3\right]
=n3[4n2−1]=\frac{n}{3}\left[4 n^{2}-1\right]
∴σ=n(4n2−1)3n−n2\therefore \sigma=\sqrt{\frac{n\left(4 n^{2}-1\right)}{3 n}-n^{2}}
=4n2−13−n2=\sqrt{\frac{4 n^{2}-1}{3}-n^{2}}
=4n2−1−3n23=\sqrt{\frac{4 n^{2}-1-3 n^{2}}{3}}
=n2−13=\sqrt{\frac{n^{2}-1}{3}}