Question

The number of customers arriving per hour at a certain automobile service facility is assumed to follow a Poisson distribution with mean 𝜆 = 7. Compute the probability that more than 10 customers will arrive in a 2-hour period.

Answer 1

$X_2\sim Poiss\left( 2\lambda \right) =Poiss\left( 14 \right) \\P\left( X_2>10 \right) =1-P\left( X_2\leqslant 10 \right) =\\=1-\sum_{n=0}^{10}{\frac{\lambda ^ne^{-\lambda}}{n!}}=\\=1-e^{-14}\left( 1+\frac{14}{1!}+\frac{14^2}{2!}+\frac{14^3}{3!}+\frac{14^4}{4!}+\frac{14^5}{5!}+\frac{14^6}{6!}+\frac{14^7}{7!}+\frac{14^8}{8!}+\frac{14^9}{9!}+\frac{14^{10}}{10!} \right) =0.824319$