The random variable X takes the values 1, 2, | vedclass

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

Question

(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.The variance of $X$ is given by $	ext{Var}(X) = E(X^2) - [E(X)]^2$.First,calculate $E(X^2) = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.Now,$	ext{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left( \frac{m+1}{2} ight)^2 = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4}$.Taking $\frac{m+1}{2}$ as a common factor: $	ext{Var}(X) = \frac{m+1}{2} \left[ \frac{2m+1}{3} - \frac{m+1}{2} ight] = \frac{m+1}{2} \left[ \frac{4m+2 - 3m - 3}{6} ight] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

The random variable $X$ takes the values $1, 2, 3, \ldots, m$. If $P(X=n) = \frac{1}{m}$ for each $n$,then the variance of $X$ is

Similar Questions

$A$ random variable $X$ has the following probability distribution:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X=x)$ $k$ $2k$ $3k$ $4k$ $4k$ $3k$ $2k$ $k$ $k$

Then $P(3 < X \leq 6) = $

$A$ coin is biased so that the head is $3$ times as likely to occur as tail. If the coin is tossed twice,find the probability distribution of number of tails.

$A$ box contains $6$ pens,$2$ of which are defective. Two pens are taken randomly from the box. If random variable $x$ represents the number of defective pens obtained,then the standard deviation of $x$ is:

$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:

If $X$ follows a Poisson distribution with variance $2$,then $P(X \geq 3) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module