Question

Overbooking flights American Airlines Flight 171 from New York's JFK to LAX with 189 seats available for passengers. American Airlines can overbook by accepting more reservations than there are seats available. Assume that there is a 0.0995 probability that a passenger with a reservation will not show up for the flight. Also, assume that American Airlines accepts 205 reservations the 189 seats that are available.

1. Find the probability that when 205 reservations are accepted, there are more passengers showing up than there are seats available Is the probability of overbooking small enough so that it does not happen very often, or does it seem too high that changes must be made to make it lower?

2. Use trial an error to find the maximum number of reservations that could be accepted so that the probability of having more passengers than sets is 0.05 or less.

I need help with number 2.

Answer #1

(2)

Let X is a random variable shows the number of people show up of n. Here X has binomial distribution with parameter n and p= 1 -0.0995 = 0.9005.

Here we need to find n such that

Following table shows the values of above expression for various n:

X | P(X>=190) |

195 | 6.30446E-05 |

196 | 0.000217213 |

197 | 0.000646724 |

198 | 0.001699109 |

199 | 0.004002779 |

200 | 0.008564161 |

201 | 0.016816116 |

202 | 0.03056903 |

203 | 0.051832093 |

204 | 0.082509377 |

205 | 0.124021878 |

206 | 0.176943831 |

207 | 0.24075214 |

208 | 0.313764798 |

So required value of n is 202 because for n=203 this probability is greater than 0.05.

Excel function used to find the probabilities: "=1-BINOMDIST(189,n,0.9005,TRUE)"

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