Solveeit Logo
Login

Question

Mathematics Question on Variance and Standard Deviation

: The variance of first nn even natural numbers is n2−14\frac{n^{2}-1}{4} : The sum of first nn natural numbers is n(n+1)2\frac{n\left(n+1\right)}{2} and the sum of squares of first nn natural numbers is n(n+1)(2n+1)6\frac{n\left(n+1\right)\left(2n+1\right)}{6}

A

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C

Statement-1 is true, Statement-2 is false

D

Statement-1 is false, Statement-2 is true

Answer

Statement-1 is false, Statement-2 is true

Explanation

Solution

Statement-2 is true Sum of n even natural numbers =n(n+1)= n (n + 1) Mean (xˉ)=n(n+1)n=n+1\left(\bar{x}\right)=\frac{n\left(n+1\right)}{n}=n+1 Variance =[1n∑(xi)2]−(xˉ)2=1n[22+42+.....+(2n)2]−(n+1)2=\left[\frac{1}{n}\sum\left(x_{i}\right)^{2}\right]-\left(\bar{x}\right)^{2}=\frac{1}{n}\left[2^{2}+4^{2}+.....+\left(2n\right)^{2}\right]-\left(n+1\right)^{2} =1n22[12+22+.....+n2]−(n+1)2=4nn(n+1)(2n+1)6−(n+1)2=\frac{1}{n}2^{2}\left[1^{2}+2^{2}+.....+n^{2}\right]-\left(n+1\right)^{2}=\frac{4}{n} \frac{n\left(n+1\right)\left(2n+1\right)}{6}-\left(n+1\right)^{2} =(n+1)[2(2n+1)−3(n+1)]3=(n+1)[4n+2−3n−3]3=(n+1)(n−1)3=n2−13=\frac{\left(n+1\right)\left[2\left(2n+1\right)-3\left(n+1\right)\right]}{3}= \frac{\left(n+1\right)\left[4n+2-3n-3\right]}{3}=\frac{\left(n+1\right)\left(n-1\right)}{3}=\frac{n^{2}-1}{3} ?? Statement 1 is false.