Question

Suppose that the probability that a passenger will miss a flight is 0.09580. Airlines do not like flights with empty? seats, but it is also not desirable to have overbooked flights because passengers must be? "bumped" from the flight. Suppose that an airplane has a seating capacity of 59 passengers.

**?(a)** If 61 tickets are? sold, what is the
probability that 60 or 61 passengers show up for the flight
resulting in an overbooked? flight?

**?(b)** Suppose that 65 tickets are sold. What is
the probability that a passenger will have to be? "bumped"?

**?(c)** For a plane with seating capacity of 56
?passengers, how many tickets may be sold to keep the probability
of a passenger being? "bumped" below 55?%? *Please
explain all steps for this section of the problem. If using a
binomial calculator, for example, what would go in "x" "n" and
"p"?*

Answer #1

Ans:

Probability that a passenger shows up for flight,p=1-0.0958=0.9042

a)n=61,p=0.9042

P(x=60 or 61)=P(x=60)+P(x=61)

=61C_{60}*0.9042^{60}*0.0958^{1}+0.9042^{61}

=0.0139+0.0021

=**0.016**

b)n=65,p=0.9042

P(overbooked or bumped)=P(x>59)=1-P(x<=59)

=1-binomcdf(65,0.9042,59)

=1-0.5996

**=0.4004**

c)P(overbooked or bumped)=1-binomcdf(n,0.9042,56)

n | 1-binomcdf(n,0.9042,56) |

57 | 0.003 |

58 | 0.021 |

59 | 0.070 |

60 | 0.161 |

61 | 0.293 |

62 |
0.448 |

63 |
0.600 |

64 | 0.732 |

So,**62 tickets** may be sold to keep the
probability of a passenger being? "bumped" below 55?%.

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