If a random variable X satisfies poisson distribution with a mean value of 5 , then probability that…
If a random variable X satisfies poisson distribution with a mean value of 5 , then probability that $\mathrm{X} \lt 3$ is
$\frac{37}{2} e^5$
$6 e^5$
$6 e^{-5}$
$\frac{37}{2} e^{-5}$
Solution
Given $x$ satisfies poisson with mean value is 5
So, $\mathrm{P}(x \lt 3)=\mathrm{P}(x=0)+\mathrm{P}(x=1)+\mathrm{P}(x=2)$
$=e^{-5} \frac{5^0}{0!}+e^{-5} \frac{5^1}{1!}+e^{-5} \frac{5^2}{2!}$
$=e^{-5}\left(1+5+\frac{25}{2}\right)=\frac{37}{2} e^{-5}$