Ergonomics. Exercises 9–16 involve applications to ergonomics,...

Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.

Safe Loading of Elevators The elevator in the car rental building at San Francisco International Airport has a placard stating that the maximum capacity is “4000 lb—27 passengers.” Because 4000/27=148, this converts to a mean passenger weight of 148 lb when the elevator is full. We will assume a worst-case scenario in which the elevator is filled with 27 adult males. Based on Data Set 1 “Body Data” in Appendix B, assume that adult males have weights that are normally distributed with a mean of 189 lb and a standard deviation of 39 lb.

b. Find the probability that a sample of 27 randomly selected adult males has a mean weight greater than 148 lb.

Accepted Answer

All right. Hello, everyone. So this question says, a university shuttle has a sign indicating a maximum load of 3000 pounds or 20 passengers. This means the average passenger weight is 3000 pounds divided by 20, which equals 150 pounds when full. Suppose the shuttle is filled with 20 randomly selected college students. Assume student weights are normally distributed with a mean of 172 pounds. And a standard distribution or standard deviation, excuse me, of 30 pounds. What is the probability that the mean weight of the 20 students exceeds 150 pounds? And here we have 4 different answer choices labeled A through D. So first and foremost, what information are we given? Well, first and foremost, we're given the population mean of mu, which is equal to 172 pounds. We also given sigma the standard deviation which is equal to 30 pounds. And we're also given the sample size lowercase n, which is equal to 20. So what exactly is the probability that we are trying to find? In this case, we're trying to find the probability where X bar is less than 150 pounds, or excuse me, greater than. We want to find the probability. Of X bar being greater than 150, where X bar is the sample mean weight of 20 students. With this information in mind, we begin first with finding the standard error of the mean or sigma of X bar. Now, sigma of X bar is equal to sigma, the standard deviation, divided by the square root of the sample size. So that's equal to 30. Divided by the square root of 20. And now you can use this value to calculate a Z score. So the is equal to X bar, subtracted by mu, which is the population mean. And then divided by the standard error of the meat. So plugging in the information we found earlier, that's going to be 150, subtracted by 172. Divided by that. 30 divided by the square root of 20. Evaluating this expression gives you your Z score of approximately -3.28. So here Going back to the probability that we wanted to find earlier. The probability that where the sample mean exceeds 150 pounds is equal to the probability in which the Z score is greater than -3.28. So now using a standard normal table, you can find the probability. Matching it to your Z score. In this case, When referencing a standard normal table, you find that the probability where Z. Is actually less than. 3.28. Is equal to 0.005. So to find the probability. In which the Z score is actually greater than -3.28 instead of less than you would subtract 0.05 from one. So in this case, that's one subtracted by 0.005, which equals 0.9995. And that corresponds to option B in the multiple choice, making that your correct answer. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

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Key Concepts

Normal Distribution

Central Limit Theorem

Z-Score

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