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 ≤ 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

a) probability a person shows up =1-0.02 =0.98

b)

P(X<=255)

c)

here mean of distribution=μ=np=		261.66
and standard deviation σ=sqrt(np(1-p))=		2.29
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability =P(X<255.5)=(Z<(255.5-261.66)/2.288)=P(Z<-2.69)=0.0036

(please try 0.0018 if no continuity correction is required)