AP Statistics X-Bar Questions

Question

Based on long experience, an airline found that about 2% 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 267 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 = 267 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?

P(r ≤ 267)P(r ≥ 267)    P(r ≥ 255)P(r ≤ 255)

(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? (Round your answer to four decimal places.)

Answer 1

This is a binomial distribution with sample size 267 and success chance 0.98.

p=0.98

q=0.02

n=267

a) 98%. showing up is the logical complement of not showing up.

b) P(r≤255) represents the probability that everyone who shows has a seat.

c) Determine the mean and standard deviation of the binomial distribution.

mu = n*p = (267)*(0.98) = 261.66

sigma = sqrt(n*p*q) = sqrt(267*0.98*0.02) = 2.2876188

Make the continuity correction (I'm assuming you're expected to do this) and calculate the z-score.

z = (255.5-261.66)/2.2876188 = -2.69275

Determine the normal probability.

P(Z<-2.69) = 0.003572601 ≈ 0.0036

P(Z<-2.69275) = 0.003543269 ≈ 0.0035

So, depending on how you round you should get one of those answers. It seems unlikely that everyone who shows will have a seat!