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5.24 The U.S. Department of Transportation maintains statis- tics for mishandled bags per 1,000 airline passengers. In October 2013, Delta mishandled 1.55 bags per ...

Question

5.24 The U.S. Department of Transportation maintains statis- tics for mishandled bags per 1,000 airline passengers. In October 2013, Delta mishandled 1.55 bags per 1,000 passengers _ What is the probability that in the next 1,000 passengers, Delta will have a. no mishandled bags? b: at least one mishandled bag? C. at least two mishandled bags? Anind

5.24 The U.S. Department of Transportation maintains statis- tics for mishandled bags per 1,000 airline passengers. In October 2013, Delta mishandled 1.55 bags per 1,000 passengers _ What is the probability that in the next 1,000 passengers, Delta will have a. no mishandled bags? b: at least one mishandled bag? C. at least two mishandled bags? Anind



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USA Today reported that for all airlines, the number of lost bags was May: 6.02 per 1000 passengers December: 12.78 per 1000 passengers Note: A passenger could lose more than one bag. (a) Let $r=$ number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable $r$ What is $\lambda$ to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round $\lambda$ to the nearest whole number.)

Have occur at a rate of 0.05 Fatalities per 100 miles. So that's Set London Equal to 0.05. And our post on perimeter is consider your funds. Yeah. So in part a I think a good probability X is equal to zero. Just simply 0.05. Was your factorial to the paris room Times. You need to manage your friends here at five Which is equal to 0.954. Now the probability that X is greater equal to one is one minus the probability X less than one. Which is equal to 1 -1 submission. Yeah. Oh sorry. One minus the probability that X physical too soon. Which is your .04877. And finding the problem, the Xk one Is 1- the probability. Excellent Ankle to one. Which is one. That's your point nine. Yeah. 879. Which will give us It's your coins your 01- one. Mhm.

Okay, This problem is about lost luggage on airlines. And so we're gonna look at some information about some definitions of populations of statistics and variables. Um, so what we know for this we want to know that the population is so you can see what's given in your book. But the population is what it represents overall. So, really, the population in this case is the U. S. Commercial airline industry passengers from the U. S. Commercial airline industry. So that's the population variables is what we're measuring from the passengers better writing in the U. S. Commercial airlines. So it's a smaller subset. So in this case, the variable in this case have their street, even variables. So this the 1st 1 is in, which is the number of report ah reported luggage stolen specifically, and the second variable that's measured is the number of ah and censor number. The number of passengers in the third variable under consideration is the, uh, number of reports per 1000. Meaning how many people lose their luggage out of every 1000 people that right now take flights on the airlines. So there's three variables that case okay for a second part are these numbers? Reported data are statistics. Okay, so these numbers reported and specifically referring to the table. The table talks about, uh, 3.26 and they refer to 3.37 And what that means specifically as those are the number reported lost luggage Stolper those millions of passengers. So because it just measures of the variable. So they're actually just data, okay? Because they're just they're just one of the variable reports. We're not doing any map of them dividing by 1000 but it's still just a variables. Okay, So ah, those were just the data because in the value of the variables aren't see looking at the information we have is a value 5.36 So 5.36 That is a statistic. So if you look at it because we're actually taking all the variables and there's multiple airlines being reported here because it's the average witnessed a calculation with a bunch of different reported data. So it's the average from a bunch of different data from airlines. So it's a statistic party. Looking at a party in the industry average is that is the industry average the mean of the airline rates, Uh, reports per 1000. Is the industry average in the mean of airline rates of reports per 1000 if not explain in detail how the 20 airline values are related to the industry average. Okay, so what this is is the number we have. The values that we have information is what is called a sample statistic, meaning really sampling 12 airlines. And there's more, you know, airlines that fly or more information, and you don't example at racing on thing. So So it's called a sample statistic. So it's not the exact data for the whole entire airline industry. It just represents what that is. Okay, so it's a sample statistic of the airline industry. It's not a population statistics. To get that, we have to measure all of them chances. The industry average so and just to be clear, and that's going on right? No

In this problem, we assume that passengers have a 5% chance of not showing up further flights with an airline and so on. Airline has a plane that they book 275 tickets for, even though they only have space for 265 passengers, and were asked to calculate the probability that the airline will not have enough seats on that flight. So they have sold 275 tickets to passengers. And let's say a passenger not showing up is a success. So the probability of success is 5%. And let's further assume that each passenger is considered a Bernoulli trial in which there is a probability of success or failure. That's 0.5 and that that probability is the same for each passenger and that the passengers are independent from each other. That may not be entirely true. These assumptions, you know, for example, a family traveling together. If one of the family members doesn't make the flight, the probability that the other members don't make it is higher. But let's assume that we can say these air Bernoulli trials, and let's say X is the number of passengers who do not show up. So those are our successes. So we can say that X is a binomial random variable based on 275 trials with probability of success of 0.5 Now we also know that there is room on the plane for 265 passengers, which means that we need and minus 265 successes. So we need to have at least 10 successes in order for everybody to have a seat on the plane. That is, we need at least 10 people not to show up so that we have, at most 265 people for the 265 seats. So the probability that there will not be enough seats is the probability that we have, at most nine successes. Now this binomial could be computed with a computer. It's more practical to use the normal approximation for this binomial. We need to show that end Times P and end times Q are at least 10 in order to use the normal approximation. So the binomial has an expected number of successes, which is equal to N. P. which equals 13 0.75 So we can say that in times P equals 13.75 which is bigger than 10 and in times cute is 261.25 which is also bigger than 10. So that means that we can use the normal approximation. We also need to calculate the standard deviation on the expected number of successes and that is given by the square root of end times. P Times Q. And this comes out to 3.61 So now we're saying that X is approximately normally distributed, with the main of 13.75 and a standard deviation of 3.61 So now we can look at the probability that X is less nor equal to nine, using the normal approximation. This is equal to the probability that said is less than nine, minus the mean, which is 13.75 divided by the standard deviation of 3.61 and looking this up on a standard normal table. So it's minus 1.3 to, so it's 0.934 so there is slightly more than 9% chance that flights on this plane will not have enough seating for all the people who show up. And just in case you're interested, if you software to calculate the actual probability using the binomial random variable, now you get a probability of zero point 116

Now necessarily exercise to 0.44 in this exercise. The questions is that this tax is allocated to the airport at random. So in exercise, two points 43 we know that there are 504 different ways of allocating these taxes. And, uh hence their probability equally likely, which means for each way of the petitioning it's probability is one over 504 for part a question. See, that one taxi is indeed our re pyre. And what is the probability that this taxi is allocated to air policy? Since this taxi is already allocated and now they're eight taxis left, and for these taxis, three of them go through airport eight and four of them go through Airport B. And according to this formula, we want to know how many ways of petitioning there are. And we do this formula. It's just, uh, this 823 and five. And the answer is 56 enhance. The probability is just 56 over 504 and the answer is one over or not. For Barbie, there are three taxis which are in need of repair. So what is the probability that each airport receives one of these taxis that needs repair. So, uh, first, these three taxis are allocated, and now there are only six taxes left, and these taxes should be allocated to two airports. Two of them go to airport A and four of them go to Airport B. So there are six choose two and four different ways of allocation. And also for these three taxis that needs repair, there are also three times, two times one different ways to allocate these three taxes. For example, for airport A, there are three days to allocate a taxi. You can choose one taxi from this three and allocated to Airport A and then for airport, be There are only two taxis left, and then for air policy. There's only the last one left, and hence you should also multiple this number. And now the total number of partitioning these taxis is 90. And hence the probability is just 90 over 504 and the answer is 5/28


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