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Numerade Educator

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Problem 61 Hard Difficulty

Airline passengers get heavier In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) in 2003 told airlines to assume that passengers average 190 pounds in the summer, including clothes and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 30 passengers.
(a) Explain why you cannot calculate the probability that a randomly selected passenger weighs more than 200 pounds.
(b) Find the probability that the total weight of the passengers on a full flight exceeds 6000 pounds. Show your work. (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

Answer

a) Distribution of population unknown.
b) $P(\overline{x}>1.56)=0.0594$

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Video Transcript

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