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 |

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