Based on long experience, an airline found that about 9% of the people making re

Question

Based on long experience, an airline found that about 9% 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 greaterthanorequalto 255) P(r greaterthanorequalto 270) P(r lessthanorequalto 255) P(r lessthanorequalto 270) (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?

Explanation / Answer

Question a)

100% - 9% = 91%

0.9100

Answer: 0.9100

Question b)

P ( r <= 255)

Question c)

Mean = np = 270 *0.91 = 245.7

Standard deviation = sqrt (npq) = sqrt (270*0.91*(1-0.91)) = 4.7024

Here we need 0.5 as correction factor

z = (255.5 – 245.7) / 4.7024

= 2.08

P (z<2.08)

By using Normal Distribution Table we get,

P (z<2.08) = 0.9812

Answer: 0.9812

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