If a random variable X has the probability di | vedclass

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

Question

(D) Given,the probability distribution of $X$ is defined by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$,and $P(X=4)=4C-1$.We know that the sum of all probabilitie

Vedclass · Accepted Answer

$\frac{128}{81}$

Vedclass · Answer

(D) Given,the probability distribution of $X$ is defined by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$,and $P(X=4)=4C-1$.We know that the sum of all probabilities in a distribution must be $1$,so $\Sigma P(X)=1$.$3C^3 + (5C-10C^2) + (4C-1) = 1$$3C^3 - 10C^2 + 9C - 2 = 0$Factoring the cubic equation: $(C-1)(3C^2-7C+2) = 0$$(C-1)(3C-1)(C-2) = 0$This gives $C = 1, \frac{1}{3}, 2$.Since probabilities must be between $0$ and $1$,we test the values. If $C=1$,$P(X=4)=4(1)-1=3 > 1$ (impossible). If $C=2$,$P(X=4)=4(2)-1=7 > 1$ (impossible). Thus,$C=\frac{1}{3}$.Substituting $C=\frac{1}{3}$ into the probabilities:$P(X=0) = 3(\frac{1}{3})^3 = \frac{3}{27} = \frac{1}{9}$$P(X=2) = 5(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{5}{3} - \frac{10}{9} = \frac{15-10}{9} = \frac{5}{9}$$P(X=4) = 4(\frac{1}{3}) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$The mean $E(X) = \Sigma X P(X) = (0 	imes \frac{1}{9}) + (2 	imes \frac{5}{9}) + (4 	imes \frac{1}{3}) = 0 + \frac{10}{9} + \frac{4}{3} = \frac{10+12}{9} = \frac{22}{9}$.The mean of squares $E(X^2) = \Sigma X^2 P(X) = (0^2 	imes \frac{1}{9}) + (2^2 	imes \frac{5}{9}) + (4^2 	imes \frac{1}{3}) = 0 + \frac{20}{9} + \frac{16}{3} = \frac{20+48}{9} = \frac{68}{9}$.Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{68}{9} - (\frac{22}{9})^2 = \frac{68}{9} - \frac{484}{81} = \frac{612-484}{81} = \frac{128}{81}$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$K$	$2K$	$3K$	$4K$	$5K$	$6K$

If a random variable $X$ has the probability distribution given by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$ and $P(X=4)=4C-1$,then the variance of that distribution is

Similar Questions

The probability distribution of a random variable $X$ is given below.

$X = x$ $0$ $1$ $2$ $3$

$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then:

The probability distribution of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $K$ $2K$ $3K$ $4K$ $5K$ $6K$

Find the value of $P(2 < X < 6)$.

If a random variable $X$ follows a Poisson distribution with a mean value of $5$,then the probability that $X < 3$ is:

Which of the following functions is not a probability density function $(p.d.f.)$ of a continuous random variable $X$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module