Question

Because many passengers who make reservations do not show up, airlines often over book flights ( sell more tickets than there are seats) A certain airplane hold 287 passengers. If the airline believes the rate of passenger no-shows is 5% and sells 302 tickets, it is likely they will not have enough and someone will get bumped? answer A &B

A. Use the normal model to approximate the binomial to determine the probability of atleast 288 passengers show up.

B. Should the airline change the number of tickets they sell? explain

Answer #1

A)

n= | 302 | p=probability of showing =1-0.05= | 0.9500 |

here mean of distribution=μ=np= | 286.90 | |

and standard deviation σ=sqrt(np(1-p))= | 3.79 | |

for normal distribution z score =(X-μ)/σx |

therefore from normal approximation of binomial distribution and continuity correction: |

probability of atleast 288 passengers show up:

probability
=P(X>287.5)=P(Z>(287.5-286.9)/3.787)=P(Z>0.16)=1-P(Z<0.16)=1-0.5636=0.4364 |

B)

yes since the probability of having 288 or more passenger is not unusual ( probability >0.05)

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