The variance of the first 50 even natural num | vedclass

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

Question

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$. The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2

Vedclass · Accepted Answer

$833$

Vedclass · Answer

(A) The first $50$ even natural numbers are $2, 4, 6, \dots, 100$. The mean $\bar{x} = \frac{2+4+\dots+100}{50} = \frac{2(1+2+\dots+50)}{50} = \frac{2 	imes \frac{50 	imes 51}{2}}{50} = 51$. The variance is given by $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. $\sum x_i^2 = 2^2 + 4^2 + \dots + 100^2 = 4(1^2 + 2^2 + \dots + 50^2)$. Using the sum of squares formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$: $\sum x_i^2 = 4 	imes \frac{50 	imes 51 	imes 101}{6} = 2 	imes 50 	imes 17 	imes 101 = 171700$. $\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Group $A$	$9$	$17$	$32$	$33$	$40$	$10$	$9$
Group $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

The variance of the first $50$ even natural numbers is

Similar Questions

The variance of the numbers $8, 21, 34, 47, \ldots, 320$ is . . . . . . .

The mean of $100$ observations is $50$ and their standard deviation is $5$. Then,the sum of the squares of all the observations is:

The variance of the first ten multiples of $3$ is

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

From the data given below,state which group is more variable,$A$ or $B$?

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Group $A$ $9$ $17$ $32$ $33$ $40$ $10$ $9$

Group $B$ $10$ $20$ $30$ $25$ $43$ $15$ $7$

Vedclass Test Series

Exam Paper Generator

Online Exam Module