If a random variable X follows a Poisson dist | vedclass

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(D) For a Poisson distribution with mean $\lambda = 5$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.We nee

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$\frac{37}{2} e^{-5}$

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(D) For a Poisson distribution with mean $\lambda = 5$,the probability mass function is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.We need to find $P(X Substituting the values:$P(X = 0) = \frac{e^{-5} 5^0}{0!} = e^{-5}$$P(X = 1) = \frac{e^{-5} 5^1}{1!} = 5e^{-5}$$P(X = 2) = \frac{e^{-5} 5^2}{2!} = \frac{25}{2}e^{-5}$Summing these probabilities:$P(X < 3) = e^{-5} \left( 1 + 5 + \frac{25}{2} \right) = e^{-5} \left( 6 + 12.5 \right) = 18.5 e^{-5} = \frac{37}{2} e^{-5}$.

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

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