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

Question

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

Redesign of Ejection Seats When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ACES-II ejection seats were designed for men weighing between 140 lb and 211 lb. Weights of women are now normally distributed with a mean of 171 lb and a standard deviation of 46 lb (based on Data Set 1 “Body Data” in Appendix B).

a. If 1 woman is randomly selected, find the probability that her weight is between 140 lb and 211 lb.

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The exam scores for a college entrance examination are normally distributed with a mean of 500 and a standard deviation of 100. If one applicant is randomly selected, find the probability that their score is between 400 and 600. Awesome. So it appears for this particular prong, we're asked to focus our attention on if one applicant is randomly selected. We're asked to determine ultimately what the probability is that this applicant will have a score that is between 400 and 600 for their entrance examination. So with that said, now that we know that we're ultimately trying to figure out the probability that their score is between 400 and 600, let's take a moment to read off our multiple choice answers to see what our final answer might be. A is 0.4772. B is 0.9545. C is 0.6826, and finally D is 0.8413. Awesome. So first off, in order to find the probability that a score will be between 400 and 600, we will need to recall and use the properties of normal distribution. So with that in mind, our first step, or major step towards solving this problem is we need to standardize the scores. So in order to do that, we need to recall a note that we need to convert the scores 400 and 600 into their corresponding Z scores by recalling and using the following formula for Z scores, which states that Z, the Z score, is equal to capital X. Minus m divided by sigma, where X is the value that we are interested in, and mu is the mean, and sigma is the standard deviation. So with that said, let's take a moment here to note that we're told that mu, or mean, is equal to 500. We're also told that sigma, our standard deviation, is equal to 100. And capital X subscript 1 is going to be equal to 400 because we're trying to find the probability that this randomly selected applicant scores between 400 and 600. So that will mean that capital X subscript 2 is going to be equal to 600. So now, with that said, now that we've wrote down all of our known variables that are given to us by the problem itself, our next step is we need to calculate the Z score for X subscript 1 is equal to 400. So with that said, we need to take Z1 and Z1 is equal to 400 minus 500 divided by 100, which when we plug that into our calculator, we will find that it's equal to -1. Fantastic. So now, our next step is we need to calculate the Z score for X subscript 2, which is equal to 600. So Z subscript 2 Z2 is equal to 600 minus 500 all divided by 100, which when we plug that into our calculator, we will find that it's equal to 1. So now, our next step is we need to use the Z scores to help us find the probability. So with that in mind, we need to recall that we need to look up the Z scores for -1 and 1. Using a standard normal distribution table, and a standard normal distribution table should be provided to you by your textbook. So using your textbook to look at your standard normal distribution table, you will find that for Z1 is equal to -1, that will mean that P1, where capital Ps are probability, so P1. is equal to P of Z is less than -1 will be equal to using your table. Once again, when you use your Once it like your standard normal distribution table, you will find that this is equal to 0.1587, and similarly for Z2, Z2 is going to be equal to 1, which will mean that P2 is equal to P of Z or Z. is less than 1, which is gonna be equal to and once again using your standard normal distribution table, you will find that it's equal to 0.8413. Fantastic. So, with that said, moving right along here, we now need to calculate the probability that a randomly selected applicant score is between 400 and 600. So as we should recall a note, the probability that a randomly selected applicant scores between 400 and 600 will be the difference between P2 and P1. So that will mean that P. Of -1 is less than Z and Z is less than 1. So once again, the probability of -1 is less than Z and Z is less than 1, is going to be equal to P2 minus P1. Which will be equal to 0.8413 minus 0.1587, which will be plugged that into our calculator. Will give us an answer of 0.6826, and that is our final answer. Hooray, we did it. So therefore, the probability that a randomly selected applicant score is between 400 and 600 is. 0.6826 and this corresponds to multiple choice answer letter C, which once again is 0.6826. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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Key Concepts

Normal Distribution

Z-Scores

Probability Calculation

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