If a Poisson variate X satisfies the relation | vedclass

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

Question

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=3) = P(X=5)$, we have

Vedclass · Accepted Answer

$\frac{50}{3 e^{\sqrt{20}}}$

Vedclass · Answer

(A) The probability mass function of a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.Given $P(X=3) = P(X=5)$, we have $\frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!}$.Canceling $e^{-\lambda}$ and $\lambda^3$ from both sides, we get $\frac{1}{6} = \frac{\lambda^2}{120}$.Thus, $\lambda^2 = \frac{120}{6} = 20$, which implies $\lambda = \sqrt{20}$.Now, we need to find $P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!}$.Substituting $\lambda = \sqrt{20}$ and $\lambda^2 = 20$, we get $P(X=4) = \frac{e^{-\sqrt{20}} (20)^2}{24} = \frac{400}{24 e^{\sqrt{20}}} = \frac{50}{3 e^{\sqrt{20}}}$.

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$

If a Poisson variate $X$ satisfies the relation $P(X=3)=P(X=5)$, then $P(X=4)=$

Similar Questions

In a book consisting of $600$ pages, there are $60$ typographical errors. The probability that a randomly chosen page will contain at most two errors is

$A$ die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.

$X$ $0$ $1$ $2$

$P(X)$ $0.4$ $0.4$ $0.2$

$A$ die is rolled. If $X$ denotes the number of positive divisors of the outcome,then the range of the random variable $X$ is

Two cards are drawn simultaneously from a well-shuffled pack of $52$ cards. If $X$ is the random variable representing the number of queens obtained,then the value of $2 E(X) + 3 E(X^2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module