Question

5. The no-show rate for airline tickets on a particular airline is about 4%. That is, about 4% of people who buy a seat on an airplane do not show up for their flight. To make more money, the airlines will often overbook flights, which means they will sell more tickets than they have seats, counting on some people to not show up. If more people show up for the flight than the plane can hold, this flight is considered "overbooked”, and the airline will often have to ask some people to take a later flight, pay for meals and hotels, and generally make people upset. Airlines don't like doing this. On a particular flight, there are 300 seats on the plane, and the airline sells 314 tickets for that flight. What is the probability that the flight will be overbooked? In other words: What is the probability that there will be in this group of 314 tickets, fewer than 14 no-shows? Possibly helpful: z = P SE = p(1.-p)

Answer 1

The probability of no-show, p = 0.04

Since the sample size is large enough, the normal approximation to binomial is used, where the mean and standard deviation of the normal distribution is,

u=p=0.04

px (1-P) SE=1 0.04 x (1 -0.04) 2=0.0111 314

p = = 0.046

The probability that less than 14 of no show in the sample of 314 is obtained by calculating the z score,

P(p<P)=P(z<)

1 P(p < 0.0446) =P < 0.0446 -0.04 0.0111

P(p < 0.0446) = P(Z < 0.04147)

from the z distribution table,

P(p < 0.0446) = 0.6608