Question

Because many passengers who make reservations do not show​ up, airlines often overbook flights​ (sell more tickets than there are​ seats). A certain airplane holds 199 passengers. If the airline believes the rate of passenger​ no-shows is 8​% and sells 215 ​tickets, is it likely they will not have enough seats and someone will get​ bumped?

A)Use the normal model to approximate the binomial to determine the probability of at least 200 passengers showing up.

​b) Should the airline change the number of tickets they sell for this​ flight? Explain.

A. The proportion is very high​, so it is likely that they should sell less. ​However, the decision also depends on the relative costs of not selling seats and bumping passengers.

B. Since the proportion is so low​, they should not change the number of tickets they sell.

C. The proportion is fairly low​, so it is likely that they should not change the number of tickets they sell. ​However, the decision also depends on the relative costs of not selling seats and bumping passengers.

D. Since the proportion is so high​, they should change the number of tickets they sell.

Answer 1

a)

n=	215	p=	0.9200
here mean of distribution=μ=np=		197.8
and standard deviation σ=sqrt(np(1-p))=		3.9779
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability of at least 200 passengers showing up:

probability =

P(X>199.5)

=

P(Z>0.43)=

1-P(Z<0.43)=

1-0.6664=

0.3336

A. The proportion is very high​, so it is likely that they should sell less. ​However, the decision also depends on the relative costs of not selling seats and bumping passengers.