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 $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$: $\sum x_i^2 = 4 	imes \frac{50 	imes 51 	imes 101}{6} = 2 	imes 50 	imes 17 	imes 101 = 171700$. $\sigma^2 = \frac{171700}{50} - (51)^2 = 3434 - 2601 = 833$.

$X$	$5$	$6$	$7$	$8$	$10$
Frequency	$3$	$7$	$4$	$2$	$4$

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

Similar Questions

For the following frequency distribution,find the variance:
$X$ $5$ $6$ $7$ $8$ $10$
Frequency $3$ $7$ $4$ $2$ $4$

The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^2$ respectively. If the variance of all the $30$ numbers in the two sets is $13$,then $\sigma^2$ is equal to $.........$.

If $x_1, x_2, ..., x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 100$,then which of the following is a possible value of $n$?

If $\sum_{i=1}^{n}(x_{i}-a)=n$ and $\sum_{i=1}^{n}(x_{i}-a)^{2}=na$,where $n, a > 1$,then the standard deviation of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is

The standard deviation of $a, a+d, a+2 d, \ldots, a+2 n d$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module