# Use the normal model to approximate the binomial to determine the probability of at least 191 passengers showing up

Because many passengers who make reservations do not show up, airlines often overbook flights (sell more than there are seats). A certain airplane holds 190 passengers. If the airline believes the rate of passenger no-shows is 9% and sells 208 tickets, is it likely they will not have enough seats and someone will get bumped?

$X$ - numbers of "shows". So, $X \in \mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) \geq 10$ we can make the following approximation, where the approximation was made in the last step  $$P_{\mathrm{Bin}}(X \geq 191) = 1- P_{\mathrm{Bin}}(X \leq 190) = 1- P_{\mathrm{Bin}}(X < 191) \approx 1- \mathrm{N}(\frac{190.5-np}{\sqrt{np(1-p)}} ).$$