Find the variance of the first n natural numb | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The variance $\sigma^2$ is given by the formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$. For the first $n$ natura

Vedclass · Accepted Answer

$\frac{n^2 - 1}{12}$

Vedclass · Answer

(C) The variance $\sigma^2$ is given by the formula: $\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2$. For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$. Substituting these values: $\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n(n+1)}{2n} \right)^2$. $\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$. Taking the common denominator $12$: $\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$. $\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12} = \frac{(n+1)(4n+2 - 3n - 3)}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

	Physics	Chemistry	Mathematics	Biology
Mean	$20$	$25$	$23$	$27$
$S$.$D$.	$3$	$2$	$4$	$5$

Find the variance of the first $n$ natural numbers.

Similar Questions

The standard deviation and mean of five observations are $0$ and $9$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10$,then their standard deviation is

If each of the observations $x_{1}, x_{2}, \ldots, x_{n}$ is increased by $a$,where $a$ is a positive or negative number,show that the variance remains unchanged.

For given data $N=60, \sum X^2=18000$ and $\sum X=960$,then variance of data is

What is the standard deviation of the numbers $31, 32, 33, \dots, 47$?

The following table shows the information about marks obtained in Physics,Chemistry,Mathematics,and Biology by $100$ students in a class. Which subject shows the highest variability in marks?
Physics Chemistry Mathematics Biology
Mean $20$ $25$ $23$ $27$
$S$.$D$. $3$ $2$ $4$ $5$

Vedclass Test Series

Exam Paper Generator

Online Exam Module