Overbooking a Boeing 767-300 A Boeing 767-300 aircraft has 213...

Question

Overbooking a Boeing 767-300 A Boeing 767-300 aircraft has 213 seats. When someone buys a ticket for a flight, there is a 0.0995 probability that the person will not show up for the flight (based on data from an IBM research paper by Lawrence, Hong, and Cherrier). How many reservations could be accepted for a Boeing 767-300 for there to be at least a 0.95 probability that all reservation holders who show will be accommodated?

Accepted Answer

All right, hello everyone. So this question says a hotel conference hall has 180 seats available for attendees. When individuals register for a conference, there is a 0.082 chance that a registrant will not attend based on prior records. How many registrations can the hotel accept so that the probability of having enough seats for all attendees who do show up is at least 0.95? Option A says 193, B says 191, C says 188, and D says 189. So, first and foremost, let's define the variables that we need to find, or the variables that we have rather. It's important to recognize that this problem is an example of binomial probability. And What's noteworthy here is that not only is our sample size relatively large, our P is relatively small, which means that we can use the normal approximation of the binomial to solve it. Now, recall that the normal approximation is used when both NP and NQ are greater than 5. Which does apply in this case. So going back to our variables that we were talking about earlier, and corresponds to the number of registrations that are accepted. And each registrant is going to show up with the following probability. Right, so the probability that the registrant actually does show up is one subtracted by the probability that they don't, which was 0.082. And so this gives you 0.918. So in this case, what we're solving for is the probability. That the number of attendees is. Less than or equal to 180. And we specifically want this probability to be greater than or equal to 0.95. So now let's go back to the normal approximation. So here X. Corresponds to the binomial. Of the number of attendees and and the probability of their arrival at 0.918. So using the normal approximation, mu is equal to We want mu equal to and multiplied by 0.918 sigma. Which then equals the square root of N multiplied by 0.918. Multiplied by 1 subtracted by 0.918. And so this ultimately equals. The square root of N multiplied by 0.918, multiplied by 0.082. So at this point, recall what we're actually solving for, right? We're solving for the probability. Of X Being less than or equal to 180. Being greater than or equal to 0.95. So this gives you The probability Or this corresponds to the probability of z being. Less than or equal to. 180 Added to 0.5, subtracted by mu. Divided by sigma. And this expression should still be greater than or equal to 0.95. So now, subtracting the information that we have from earlier. We can say That 180.5, subtracted by 0.918 N. Divided by The square root of N multiplied by 0.918. And multiplied by 0.082. Is greater than or equal to 1.645. So now we can move on to solve the inequality numerically. So it's at this point that you would solve or you would apply each of the four different entry choices to this formula. So specifically, you would solve for mu plugging in each value for N and then find the corresponding Z score. Now, the Z score that you get from each value should be greater than 1.645. When you carry out the calculation for. N equals 193, and N equals 191, you'll see that the corresponding Z score is actually less than 1.645, which in this context is not acceptable, because that would mean that That the probability of 180 or fewer people showing up is less than 0.95. But look what happens when you apply N equals 188. And N equals 189, but let's do 188 1st. So as mentioned before, for each value of N, what you are to do is. Solve from me and then find the corresponding Z score. So for So for mu, right, recall that that's N, multiplied by 0.918. So, 188, multiplied by 0.918, gives you mu. Of approximately 3.762. So now the corresponding Z score is equal to 180.5. Divided by, excuse me, subtracted by. 172. Whoa, my apologies. Let me go back here, because 188, multiplied by 0.918 is actually 172.584. And so sigma, right, the standard deviation. A standard deviation is equal to the square root. Of 188. Multiplied by 0.918, and divided by 0.082. And so this equals 3.762. That's my bad. Now, however, we can find our Z score. And Z is equal to 180.5, subtracted by 172.584. Divided by 3.762. And this equals approximately 2.104. Which is indeed greater than 1.645, which means that it is valid for this context. But for the sake of completeness, let's check N equals 189, because recall that we're looking for the maximum number of registrations. So when you apply these calculations for N equals 189, mu. is equal to 173.502. And sigma is equal to approximately 3.772. So the corresponding Z score is equal to 180.5. Subtracted by 173.502. And divided by 3.772. Which gives you approximately 1.855. So let's bring everything back, right? 188 and 189 both have Z scores that are greater than or equal to 1.645. But because we want the maximum number of registrations possible, the hotel can accept up to 189 registrations. Which means that our correct answer. Is option D in the multiple choice. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Probability of No-Show

Binomial Distribution

Cumulative Probability

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