Question

Based on long experience, an airline found that about 4% 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. (Round your answers to four decimal places.)

(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?

P(r ≥ 270)P(r ≥ 255)    P(r ≤ 270)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?

Answer 1

Solution:-

a) The probability that a person holding a reservation will show up for the flight is 0.96.

P(Do not show) = 0.04

P(Show) = 1 - 0.04

P(Show) = 0.96

b) The expression represents the probability that a seat will be available for everyone who shows up holding a reservation is P(r ≤ 255).

The number of people showing up should be equal to or less than 255, so that seat will be available for everyone who shows up holding a reservation.

c) The probability that a seat will be available for every person who shows up holding a reservation is 0.125.

x = 255

After applying continuity correction, we will be finding out P(x < 255.5)

By applying normal distribution:-

z = - 1.149

P(x < 255.5) = P(z < - 1.46)

P(x < 255.5) = 0.125