Suppose that the probability that a passenger will miss a flight is 0.09580. Air

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"?

Explanation / Answer

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)

=61C60*0.904260*0.09581+0.904261

=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)

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

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

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