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Lionheart3388
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Hi this is a problem I had in statistics and probability. It may actually be very simple, but it just seems like I did way too little work. Thanks for the help!
An airline charges each passenger $300 for a trip between City A and City B. The airplane holds exactly 100 passengers. Each passenger has 70% probability of showing up, independent of the other passengers. All tickets can be rebooked for a later flight at no extra charge, so the airline cannot penalize a no-show passenger. Clearly, the airline should book somewhat more than 100 passengers because if they do not overbook somewhat, they will make on average $300×70 = $21,000 per flight, while if the plane were completely full, they would make $30,000 per flight. However, if they book too many people and more than 100 show up, the airline has to pay $500 to each passenger who is denied passage on that flight. For example, if they book 300 people and all 300 show up, they actually lose money: $300×300 − $500×200 = −$10,000.
How many additional passengers (beyond 100) should they book in order to maximize their mean revenue? Although this problem can be solved with JMP, you are not required to use JMP for it. You may use any methods and/or software you wish. The number of passengers who show up follows a binomial probability model. However, the associated random variable is not just $300 times the number of passengers.
P(X=x)=nCx p^x q^(n-x)
mean=np
SD=root(npq)
q=1-p
I assumed that using a binomial distribution, having a mean of 100 passengers would result in maximum revenue, so 100=np=n*.70, resulting in n=143, but this seems too easy.
Homework Statement
An airline charges each passenger $300 for a trip between City A and City B. The airplane holds exactly 100 passengers. Each passenger has 70% probability of showing up, independent of the other passengers. All tickets can be rebooked for a later flight at no extra charge, so the airline cannot penalize a no-show passenger. Clearly, the airline should book somewhat more than 100 passengers because if they do not overbook somewhat, they will make on average $300×70 = $21,000 per flight, while if the plane were completely full, they would make $30,000 per flight. However, if they book too many people and more than 100 show up, the airline has to pay $500 to each passenger who is denied passage on that flight. For example, if they book 300 people and all 300 show up, they actually lose money: $300×300 − $500×200 = −$10,000.
How many additional passengers (beyond 100) should they book in order to maximize their mean revenue? Although this problem can be solved with JMP, you are not required to use JMP for it. You may use any methods and/or software you wish. The number of passengers who show up follows a binomial probability model. However, the associated random variable is not just $300 times the number of passengers.
Homework Equations
P(X=x)=nCx p^x q^(n-x)
mean=np
SD=root(npq)
q=1-p
The Attempt at a Solution
I assumed that using a binomial distribution, having a mean of 100 passengers would result in maximum revenue, so 100=np=n*.70, resulting in n=143, but this seems too easy.