The variance of a Poisson variate X is 2. The | vedclass

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

Question

(C) For a Poisson distribution,the mean and variance are both equal to $\lambda$. Given variance $\lambda = 2$. The probability mass function is $P(X=

Vedclass · Accepted Answer

$\frac{e^2-5}{e^2}$

Vedclass · Answer

(C) For a Poisson distribution,the mean and variance are both equal to $\lambda$. Given variance $\lambda = 2$. The probability mass function is $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$. We need to find $P(X \geq 3) = 1 - \{P(X=0) + P(X=1) + P(X=2)\}$. $P(X=0) = \frac{2^0 e^{-2}}{0!} = e^{-2}$. $P(X=1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2}$. $P(X=2) = \frac{2^2 e^{-2}}{2!} = \frac{4 e^{-2}}{2} = 2e^{-2}$. Summing these,$P(X Therefore,$P(X \geq 3) = 1 - \frac{5}{e^2} = \frac{e^2-5}{e^2}$.

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) = $

Similar Questions

$A$ $1$-rupee coin,a $2$-rupee coin,a $5$-rupee coin,and a $10$-rupee coin are tossed simultaneously. The expected value of the sum of the values of the coins that show heads up 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 $P(X > 2)$ is equal to:

Find the mean number of heads in three tosses of a fair coin.

If $X$ is a Poisson variate with mean $2$,then $P\left(X>\frac{3}{2}\right)=$

For a Poisson distribution,if mean $= l$,variance $= m$ and $l + m = 8$,then $e^4[1 - P(X > 2)] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module