MD Airways fleet is comprised of Airbus A330 with a capacity of 293 passengers. With overbookings...

MD Airways fleet is comprised of Airbus A330 with a capacity of 293 passengers. With overbookings risks in mind, MD Airways decides to sell more than 293 tickets. If they wish to keep the probability of having more than 293 passengers show up to get on the flight to less than 0.08, how many tickets should they sell? Define RVs, distributions, and show your calculation in detail.

Expert Solution

Supoose the Airlines sell n tickets. Let X be a random variable indicating the number of passengers who show up. Since the passengers show up randomly, we can assume the probability of every passenger show up is 0.5. Hence we can assume X to have Binomial(n,0.5) distribution where n>293.

Then we define the probability p(n) as

Thus we need to find n such that p(n)<0.08. Now from compution we find that p(551)=0.07371221 and p(552)=0.08004345 which means that the required value of n is 551.

Hence the Airline can sell upto 551 ticket to keep the probability less than 0.08.

orchestra answered 2 years ago

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