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2.38 Baggage fees: An airline charges the following baggage fees: S25 for the first bag and S40 for the second: Suppose 47%/ of passengers have no checked luggage, ...

Question

2.38 Baggage fees: An airline charges the following baggage fees: S25 for the first bag and S40 for the second: Suppose 47%/ of passengers have no checked luggage, 32% have only one piece of checked luggage and 21% have two pieces_ We suppose a negligible portion of people check more than two bagsThe average baggage-related revenue per passenger is: $ 16.4 (please round to the nearest cent) b) The standard deviation of baggage-related revenue is: $ 16.34 (please round to the nearest cent) About

2.38 Baggage fees: An airline charges the following baggage fees: S25 for the first bag and S40 for the second: Suppose 47%/ of passengers have no checked luggage, 32% have only one piece of checked luggage and 21% have two pieces_ We suppose a negligible portion of people check more than two bags The average baggage-related revenue per passenger is: $ 16.4 (please round to the nearest cent) b) The standard deviation of baggage-related revenue is: $ 16.34 (please round to the nearest cent) About how much revenue should the airline expect for a flight of 120 passengers? $ 1968 (please round to the nearest dollar) Get Help:



Answers

The mean weight of luggage checked by a randomly selected tourist-class passenger flying
between two cities on a certain airline is $40 \mathrm{lb},$ and the standard deviation is 10 $\mathrm{lb.}$ The mean and standard deviation for a business-class passenger are 30 lb and 6 lb, respectively.
(a) If there are 12 business-class passengers and 50 tourist-class passengers on a particular
flight, what are the expected value of total luggage weight and the standard deviation of
total luggage weight?
(b) If individual luggage weights are independent, normally distributed rvs, what is the probability that total luggage weight is at most 2500 $\mathrm{lb} ?$

So they were given a mean of 380 a sinner deviation of 20. We want to find out what percentage lies from 342 4 120 So we'll start by labeling occur. We have our mean at 380 at the first Energy Aviation's. We have 364 100 and her second standard deviations. We have 344 120. So we're looking for the percentage of lies within this area so we can add our percentages all with ends to enter deviations because it's normally distributed. So 34% lie within the state at FirstEnergy Aviation to the means on either side. We have 34% from the first energy aviation. The second we have 13.5% on either side. So to find the total percentage, we just addle of these up. So we'll have 13 point by plus 34 plus 34 plus 13.5, and that's equal to 95%

I'm going to draw a normal Kurt to get an idea of what's going on here and says that number pieces of checked luggage has a mean A 380 and a standard division of 20. Meaning I'm going to add 20 or subtract Winnie to create this. It's normal crew. Okay, so it says, what number of pieces of checked luggage is three standard deviations about the mean. So maybe I didn't have to do all this work because I just have to remember percentages. What percent of the data lies? Three Center deviations of move above the meeting. Let's see if you remember number. Oh, okay. Sorry. I'm just struggling a bit. What number? Checked pieces. So we had to do Was do for 40 minus 3 80 Cool. 60. So that's just saying 60 pieces of luggage is three standard deviations about the meeting

The F A A. Gave information to the airlines at the average weight of about the average weight of passengers and in the summer, and what they said is that the average weight of passengers in the summer have a mean weight of 190 but they didn't give any cedar deviation. But a standard deviation that would be reasonable to consider is £35. And I know that my distribution is not normal because it contains both male and female weights. But it's not too far away from normal. So I'm doing my population. It's gonna be average passenger weight in the summer, and we're giving that The mean is 100 £90. And then we estimated that a reasonable standard deviation was £35. So this distance right here is £35. So part a. Ask us why you can't calculate the probability of one selected passenger that weighs more than £200. And the reason why that we can't calculate the probability for one randomly selected passenger is because our population data is it normal and you can't use it to calculate probability. If it's not normal, then part B says find the probability of the total weight of passengers on a full flight is greater than or equal to £6000. Okay, so they give the information about a commuter flight where 30 passengers. So this is a computer commuter plane. There's 30 passengers, and I know that the mean of this is going to be the mean of the population. So that's gonna be 100 and 90 and I need to find the standard deviation. So if there's 30 passengers, I know that in is equal to 30. So when I go to find my standard deviation, it's the standard deviation of the population, which in this case is 35 over the square root of the sample size, and in this case it's 30 so that's gonna equal to 6.39 So my standard deviation is about 6.39 Wants to intervene ation distance to the one standard deviation mark is approximately £6.39. Now. If I took take a look at the question. My question says it's the total weight of passengers on the full flight, and I know that there are 30 passengers on the flight in the hole. It needs to be greater than £6000. So if I take £6000 and divided by the 30 passengers, I'm really looking for an average weight of the passengers to be 202 £100. So, in reality, I'm looking for the probability that marine dimly selected passenger is going to be greater than or equal to £200. So what I need to do IHS, there's 1 90 once injured. Aviation is around 91 96. So 200 it's gonna be passed one standard deviation. So right here is where I'm estimating 200. Then I need to get what the standard deviation is for 200. So I'm gonna use mine Z score. So if I used my Z score, I'm gonna have the number of looking for just 200 minus the mean, which is 1 90 divided by the standard deviation, which is 6.39 and I get a Z score of 1.56 You know, I'm looking for the probability than it's greater than 100. So greater than 100 is going greater than 200 is going to be this probability. But when I look up the Z score of 1.56 it's actually going to give me a probability to the left. So to get the probability that I'm looking for harmony, have to look for the probability four Z is greater than or equal to 1.56 You remember the area under the total curve ISS one. So you're going to have to look for one minus the probability that Z is going to be less than or equal to 1.56 So that's gonna be one minus. I'm gonna look up 1.56 in the table and I get 0.94062 and that's going to equal zero point, he wrote five 938 And that's the probability that one person, the sample it's greater than are equal to £200 for the In reality, I know that there is basically a 6% chance that the total weight of the past teachers on a full flight is greater than people to £6000

This problem is about about a boat accident where a boat sank in Lake George back in around 1960. And when that sank, they assumed that the average person wait on that boat. The assumed average was 140. Now, keep in mind when the passengers air on the boat, you're going to have male passengers. You're gonna have female passengers and you're going to have Children. Eso some of the people could have weighed more than 140 others could have weighed less. So in solving this problem, we're going to make an assumption that it was all males on the boat because they're usually ah, heavier population than the females or the Children. So throughout this problem, they gave us some information. They told us that men's weights are normally distributed. The average male weighs £189 and the standard deviation of the population of men would be £39. And there are two parts to this problem. The first part, let's look at part a in part, a. It's saying, given that the boat was rated for 50 passengers and they assumed that the mean weight was £140. Then the boat limit was £7000. So we're going to do a sample from the population of men and we're going to assume that there were 50 people on this boat and all of them were men. So we're talking about a sample size of 50 and we want to find the probability that the boat was overloaded. Well, if the boat is overloaded, that means the average passenger weighed more than 140. So in order to solve this, we're going to have to discuss the average of the sample means on the standard deviation of the sample means. And in order to calculate those we're going to use the central Limit theorem and the Central Limit Theorem says that the average of the sample means is going to be the equal to the average of the population, and in this scenario it was all men, so the average was £189. The standard deviation of the sample means will equal the standard deviation off the population divided by thes square root of the sample size. So in this case, it's going to be 39 divided by the square root of 50 because they told us that men's weights are normally distributed. We can draw our bell shaped curve, and we can put the average in the center at 189. And this problem is asking us to determine the probability that your average was greater than 140 now. In order to do this, we are going to have to use a C score, and we're going to have to use the Z score, associate it with sample means, and to refresh your memory, Z is going to equal X bar minus the average of the sample means over the standard deviation of the sample means so therefore, the Z score associated with 140 is going to be 140 minus 189. And in place of the standard deviation of the sample means we're going to use the expression 39 over the square root of 50 and we're going to calculate a C score out to be negative 8.88 I always like to put that up on the picture, so I'm gonna put negative 8.88 in correspondence with the 140. So when we're solving the problem where the probability is greater than 140 it's no different than saying, What's the probability that the Z score is greater than that negative 8.88 So you're going to need the standard normal table in the back of your textbook, which is table eight to. And if you look at the picture on that table, the picture shows that the area or the probability that we're trying to find extends into the left table or left tail and our picture is extending into the right tail. So we're going to have to rewrite our probability as one minus the probability that Z is less than negative 8.88 and again looking at that table, Um, when you're on the negative side of the table. If you wanted a new area or a probability, associate it with Z that is less than or equal to 3.50 it's going to be 0.0 001 So we're going to do one minus 10.1 and we're going to get a probability of 0.9999 So just summarizing part A. It was asking us what was the probability that the boat was overloaded. Being overloaded would mean that the average person wait more than £140 and that probability is 1400.9999 So after they discovered this, they changed the average weight. They made the assumption that you know what the average weight of the people on that boat was not 1 40 but they upped it to approximately 174 which takes us into part B in Part B. If they brought the average wait up to 174 and they downgraded the amount of passengers on the boat tow 14. So in part B, the boat was later rated to carry only 14 passengers. Um, they upped the estimate of the average person's weight, so it created a load limit of 2436. So now we're only gonna have 14 passengers were going to make the assumption there all male and again, we're using the mail because they're generally, um, heavier than females or the Children. And in Part B. We want to find the probability that those 14 passengers, how they mean wait greater than 174. So again, we are using a sample. This time, our sample size is only 14. We are going to need to calculate the average of the sample means and the standard deviation of the sample means again we're using our central limit theorem. The average of the sample means is going to equal the average of the population. Are population being the weights of men, which was 189. And our standard deviation of sample means according to the central limit, their, um will be equivalent to the standard deviation of the population divided by the square root of the sample size. Since our population is the weights of men are standard, deviation of men was £39. And this time we're going to divide by the square root of 14. Because our sample size is now only 14 passengers on the boat we're going to construct are bell shaped curve, gonna put our average in the center and we're trying to find the probability that were greater than on average weight of 1 74. We're going to use the same Z score formula. So we're going to use the Z score formula Z equals X bar minus mu sub X bar over Sigma sub X bar. So we want the Z score, associate it with 174. So we're gonna do C equals 1 74 minus 189 in place of the standard deviation of the sample means we're going to use the expression 39 divided by the square root of 14. And in doing so are Z score turns out to be negative 1.44 again, I like to always put that z score back on the picture. So we're gonna put negative 1.44 here. Keep in mind in part B, we were trying to find the probability that the average was greater than £174. So when we're going greater than £174 it's no different than saying What's the probability that Z is greater than negative 1.44 again, our tables were set up to go into the left tail. Our picture is going into the right tail, so we're going to have to rewrite our probability as one, minus the probability that sees less than negative 1.44 We're going to go to our standard normal table, which is tabled a to, and we're going to get the probability that see us less than negative 1.44 to be 0.749 And when we subtract that, we get an overall probability of 0.9 to 51 So in summary of part B, we readjusted theme passenger limit down to 14. We made a new assumption that the average passenger weighed 174 and we wanted to, um, know what the probability that the average weight of the 14 passengers on the boat was greater than 174 and that was 1740.9 to 51 Now, in part B, there is a second question, and that question says, Do the new ratings appear to be safe when the boat is loaded with 14 male passengers? And our answer to that would be because there's still a high probability, and that probability is 0.9 to 51 That's pretty high. The new ratings do not appear safe when the boat is loaded with 14 male passengers


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