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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Statistics

JEE Main 2009 · Mathematics · Statistics

Statement-1 : The variance of first n even natural numbers is...

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

Options

  1. AStatement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. BStatement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. CStatement-1 is true, Statement-2 is false
  4. DStatement-1 is false, Statement-2 is true

Correct answer

D. Statement-1 is false, Statement-2 is true

Step-by-step solution

Statement-2 is true Statement-1: Sum of nn even natural numbers =n(n+1)=n(n+1) Mean (xˉ)=n(n+1)n=n+1 Variance =[1n∑(x1)2]−(xˉ)2=1n[22+42+….+(2n)2]−(n+1)2=1n22[12+22+….+n2]−(n+1)2=4nn(n+1)(2n+1)6−(n+1)2=(n+1)[2(2n+1)−3(n+1)]3=(n+1)[4n+2−3n−3]3=(n+1)(n−1)3=n2−13\begin{aligned} & \text{ Mean }(\bar{x})=\frac{n(n+1)}{n}=n+1 \\ & \text{ Variance }=\left[\frac{1}{n} \sum\left(x_1\right)^2\right]-(\bar{x})^2=\frac{1}{n}\left[2^2+4^2+\ldots .+(2 n)^2\right]-(n+1)^2 \\ & =\frac{1}{n} 2^2\left[1^2+2^2+\ldots .+n^2\right]-(n+1)^2=\frac{4}{n} \frac{n(n+1)(2 n+1)}{6}-(n+1)^2 \\ & =\frac{(n+1)[2(2 n+1)-3(n+1)]}{3}=\frac{(n+1)[4 n+2-3 n-3]}{3}=\frac{(n+1)(n-1)}{3}=\frac{n^2-1}{3} \end{aligned} \thereforeStatement1isfalse.Statement 1 is false.
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