Question

Tutorial Exercise Based on long experience, an airline found that about 7% of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 270 ticket reservations for an airplane with only 255 seats. (a) What is the probability that a person holding a reservation will show up for the flight? (b) Let n 270 represent the number of ticket reservations. Let r represent the number of people with reservations who show up for the flight. What expression represents the probability that a seat will be available for everyone who shows up holding a reservation? (c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation? Step 1 (a) What is the probability that a person holding a reservation will show up for the flight? Let A be the event a person holding a reservation does not show up for this flight, and B be the event a person does show up for this flight. Because these are the only possible events, it must be the case that P(A) +P(B) Submit Skip (you cannot.come back)

Answer 1

Solution:

a) P[person shows up] = 1 - P[person does not show up] = 1 - .07= .93

b.)

The probability that a seat will be available for everyone who shows up holding a reservation is

P(r < 255)

c.)

mean(x) = np = 270 * 0.93 = 251.1 0 = Vnp(1- p) = V270 % 0.93 * 0.07 = 4.1925

P(X < 255) = P(X<255.5) (using contuinity correction) X-4 255.5 - 251.1 4.1925 = P(Z < 1.05) = P < = 0.8531