A random variable X has a Poisson distributio | vedclass

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(A) For a Poisson distribution,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda = 2$.We need to

Vedclass · Accepted Answer

$1 - \frac{3}{e^2}$

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(A) For a Poisson distribution,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$,where $\lambda = 2$.We need to find $P(X > 1.5)$. Since $X$ takes only non-negative integer values,$P(X > 1.5) = P(X \ge 2)$.This can be calculated as $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$.For $k = 0$: $P(X = 0) = \frac{e^{-2} 2^0}{0!} = \frac{e^{-2} \cdot 1}{1} = \frac{1}{e^2}$.For $k = 1$: $P(X = 1) = \frac{e^{-2} 2^1}{1!} = \frac{e^{-2} \cdot 2}{1} = \frac{2}{e^2}$.Therefore,$P(X > 1.5) = 1 - (\frac{1}{e^2} + \frac{2}{e^2}) = 1 - \frac{3}{e^2}$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

$A$ random variable $X$ has a Poisson distribution with mean $\lambda = 2$. Then $P(X > 1.5)$ is equal to:

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