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Problem Fanl Binpbing; who working (or American Airlines; dlecided t0 sell 152 (ickets (ar & certaln Ilight ta Chlna on an airplane that has only 150 seals; She...

Question

Problem Fanl Binpbing; who working (or American Airlines; dlecided t0 sell 152 (ickets (ar & certaln Ilight ta Chlna on an airplane that has only 150 seals; She estirated (hal, On1 (he average percentol purchaser; of airline tIckets do not appear far Ihe departure oftheir (light;(1) Among (he 152 passengers who bought the ticket lind the average number of passengers who will not appear for Ihe departure af the fentSelect one:2,25 passengers 2.5 passengers 3.8 passengers None of the otner Io

Problem Fanl Binpbing; who working (or American Airlines; dlecided t0 sell 152 (ickets (ar & certaln Ilight ta Chlna on an airplane that has only 150 seals; She estirated (hal, On1 (he average percentol purchaser; of airline tIckets do not appear far Ihe departure oftheir (light; (1) Among (he 152 passengers who bought the ticket lind the average number of passengers who will not appear for Ihe departure af the fent Select one: 2,25 passengers 2.5 passengers 3.8 passengers None of the otner Iour 4.23 passengers {21 Find thie probability thaLexctly 50 passengers will show up for the fEnt Select one 65.4223 76.3296 22.1756 84.5890 None 0i the other (cur (3) What is the probability that everyone who appears for the departure ofIhis Might wil have & seat? Selectone: 65.41200 76.3256 80.9795 84.589



Answers

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable $Y$ as the number of ticketed passengers who actually show up for the flight. The probability mass function of $Y$ appears in the accompanying table.
(a) What is the probability that the flight will accommodate all ticketed passengers who show up?
(b) What is the probability that not all ticketed passengers who show up can be accommodated?
(c) If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Problem. 15 which in a given probability of person not showing up for a flight, it's 6%. So the probability of the showing up it then 100 degrees vice Ah, 100 minus six, which is 49% uh which in B is 267 So the property of our is sooner than 2 55 for which in C on is ableto 267 b is ableto open 94 So, uh, the that is equal to x minus in be over square root off nb one minus me which is to 55.5 minus 267 times 4.94 over square root off 267 I'll find 941 minus opened 94 which is 1.56 eso the probability that X is smaller than 2045 Probability that exist 2 to 55.5 property that that is smaller than 1.6, which is open 9452

So we know that there is a probability of 6% of no show and we know that they are making 267 Reservations. However, there are only 255 seats. I have had a plane that I was overbooked on and didn't get to fly and wasn't making me happy. What's the probability that a person will show up? That is 94%. And then part B. The probability statement that we want to show that everybody will every seat will be used Or every person will have a seat is if this is how many people show up, if more than 255 people show up, then everybody won't have a seat. So this distribution, we want to find that probability in part C. And before we do that, let's find what the mean. Is the mean. Is that n times P. And that end is that to 67 To 67 times the .94. That's the problem. That's the expected number. That will show up. The expected number is 250.98. The standard deviation for that approximate normal distribution. And by the way end times 1 -2 will be greater than or equal to five. So this distribution is approximately normal and we know we need to take n times p times one minus P. So I need to multiply this value times that point or six and then we need to square root that answer to the 0.5 power. And we get the standard deviation is 3.880 I'll say six, but I'm just going to store that value is X. So we want to find the likelihood of our being less than or equal to 2 55. And let's use that continuity correction. And this would bump up to 2 55.5 when we convert it to a Z value, so that to 55.5 minus that mean we just found divided by that standard deviation, let's see what that Z value is, And we've got that to 55.5 minus that mean of 250.98 and then divided by that standard deviation. That I started my calculators like at 1.16 And we want to find the area below 1.16 And when we do, we find that that is .8770. So it looks like we are going to have about 87 or close to 88% chance that everybody will have a seat.

So we're told that tickets for a particular flight are $215, each. The plane seats 120 passengers, but the airline will overbook will purposely sell more than 120 tickets because not every paid passenger will end up showing up. So we're going to have D. Denotes the number of tickets the Arab line is going to sell and will assume that the number of passengers that actually show up for the flight X, follows a distribution. Um So ultimately we want to write the program to simulate the scenario um and we know that they wish to determine the optimal value. So running our program, we record the average profit from each run and it appears that the target value T. Is going to equal 142 tickets. So that would be our final result. Once we develop that program.

Question. We're using an airline example and we're having that the probability of a no show is 16%. And we're having what happens on the next 42 reservations. And again, assuming Lee that these reservations are each independent begin. And we know that that's not really a justifiable situation because you have people that go together in pairs or families that travel together or business people that are traveling together. And if we find N times P, we find that end times B. Which would be the mean of our distribution is equal to 6.72 And I happen to store that in my calculator as alpha A. That would just be helpful for me later on. And the standard deviation of that distribution is a square root of N times p times one minus P. And I did that calculation and I got 2.376 And I happened to store that in my calculator as X. Just so I don't have to type all that. And again, so when I write them down, I'm going to refer to him with those letters. Now we know that N. Times P and end times one minus P. And the one minus P would be that 10.8 for their both greater than a wrinkle to five. So this distribution is approximately normal, centered at 6.72 With this standard deviation, I could draw the picture of the distribution, but we have quite a few things we want to find And we want to know what's the probability if 42 reservations are made, what's the probability that there are exactly five, no shows. And so to do that normal uh that continuity correction and find this with a binomial excusing with the normal probability, we need to go half a unit below four point 5.5 a unit higher 5.5 and find that area between those two numbers. To use the normal approximation. Now we need to convert these two Z values and I'm going to write this one down and then I won't won't write down all the rest. We have 4.5 minus are mean provided by our standard deviation and then that will be a Z value and then we'll take that 5.5 minus are mean and then divided by our standard deviation and let me quick do that on my calculator here. So I have that 4.5 minus four mean, which I stored as alpha A divided by the standard deviation. And that Z value rounds to a Z value of negative 40.93 Mhm. And then the other z value which again I can just arrow back up and change that 4.5-5.5 I get negative .51. Now, I'm going to end up using my normal CDF with these two values, you can also use your table very easily and look up But negative .93 would be my low number and negative .51 would be my upper number and leave my mean and standard deviation 01 respectively. And when I do that I get .1288. You can look these values up in your table as well, defend that area between next we want to find from uh nine people out of the 42 all the way up to 12 people and it says inclusive. So we want to include those values now For using the continuity correction. This needs to bump down to 8.5 and this one needs to bump up to 12.5 and then I'm going to do that same calculation right here, just substituting 8.5 and 12.5 in of those values. So let's see what we get for the Z values. So let me just arrow back up And we will change that value to an 8.5 And that z value becomes .75 .75. Is that CLUCR 3/4 of a standard deviation higher than the mean. And this 12.5. Let me do a little insert there to get that value in 12.5 And that's the value becomes 2.43. Now, once again, you can look those values up in your table. I'm going to use my normal CBF to find those values and I'm going to use this .75 as the lower and there's 2.43 as the upper And then find that area between. And that comes out to be .2191. And likewise you can look those values up in your table. Part C. Now we want to find the probability that at least one no show. Well in order to calculate that as a Z. value again we are going to use the continuity correction and that means I have to go down .5. so we're going to use 0.5 and use that value in place of our 4.5. So I can go up in do that little change and we want zero wait five. That's the value. I'll leave it in blue. Nice to see a color different black all the time Is negative 2.62. And then you can look that up in the table. But remember you are finding a value that here is negative 2.62 and you're finding this. So if you're gonna look it up in the table, it's actually easier to look up positive 2.62 and find the area below And positive 2.6. To looking it up on the table, I get .99 56 And now we're almost done. We have one more to do. Let's go green. And we want to find the probability that at most two now for our continuity correction, we need to bump this up. So this is what we'll use in place of uh this value right here and we'll convert that to a Z. Value. So let's find out what that Z value is. So I just have to arrow up, Change that to a 2.5. And we get that Z value Is negative 1.7 and that would round to eight. So I'm just going to click look that up in my help and my table and not using my software. And when I do .8 that comes out to be .0375. Mhm. And I think we're all done. Yeah.


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