Suppose that 15% of people dont show up for a flight, and suppose that their decisions are independent. how many tickets can you sell for a plane with 144 seats and be 99% sure that not too many people will show up. The book says to do this by using the normal distribution function and that the answer is selling 157 tickets. Solution The people not showing for the flight can be treated as from Binomial dist. The binomial distribution B(n,p) is approximately normal N(np,np(1 p)) for large n and for p not too close to zero or one. let = n be as n p=0.15 soBi(n,0.15n) follows Normal dist. u=n ^2 = 0.15n We have to calculate P(X<144) with 99% accuracy P(X<144) = P(Z.