Seat Designs. In Exercises 7–9, assume that when seated, adult...

Question

Seat Designs. In Exercises 7–9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats.

Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.

Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.

Accepted Answer

Hello. In this video, we are told that in a fitness study of adult males, hand grip strength scores are normally distributed with a mean of 45 pounds and a standard deviation of 8 pounds. We want to find the probability that the average grip strength of 12 randomly selected males will be greater than 47 pounds. So let's go ahead and list out our given information. First, we are told that we have a mean of 45 pounds, and we have a standard deviation of 8 pounds. Since we are working with 12 randomly selected males, we have a sample size of 12, and we want to find the probability that their grip strengths are greater than 47 pounds. Since the number of interest is 47, we are going to label a number X equal to 47. So what we want to do here is you want to go ahead and convert this X into a Z score. The Z score is defined as X minus U, which is our mean, divided by the standard error. But we don't have standard error. Let's go ahead and calculate the standard error for our Z score. Now, the standard error is defined as the following. It is our standard deviation divided by the square root of our sample size. That is going to be 8 divided by the square root of 12, and this will simplify to be 2.31. Now that we have all the information we need to find our Z score, let's go ahead and calculate our Z score. We get Z is equal to X, which is 47, minus the mean, which is 45, divided by our standard error, which is 2.31. And this is going to simplify to approximate Z, approximately to 0.866. Now what we want to do is we want to find the probability that Z is less than 0.866, but this is a bit complicated because 0.866 is not a normal value on the Z table. So what we are going to need to do is we are going to need to interpolate this Z value. Now, there are two Z values that we know of that surround 0.866. We have Z1, which is 0.86, and this has an approximate value or probability of 0.8051. We also have Z2, which is 0.87, and this has an approximate value of 0.8078. Now, the interpolation formula is as follows. We will want the probability that Z is less than Z1. Plus the product. Of Z2 minus Z1 multiplied by our desired Z. Which is the 0.866 that we're looking for. Minus Z1. And this will be multiplied by the probability. Of Z less than Z2. Minus the probability of Z less than Z1. So, let's go ahead and plug in the known information. So we know that the probability that Z is less than Z1 is 0.8051. Then we are going to add 0.87 minus 0.86 multiplied by 0.866, minus 0.86, multiplied by the difference of 0.8078 minus 0.8051. And once we simplify this, we get the approximate value of 0.8067. So just to recap, originally, we are looking for the probability that the grip strengths were greater than 47.0, which is the 47 pounds. This is equivalent to 1 minus the probability that Z is less than 0.866, and by using the value that we just calculated, we have 1 minus 0.8067. Which gives us a probability of 0.193. So the probability of the grip strength being greater than 47 pounds is approximately 19.3%. And with that being said, the solution to this problem is going to be C. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

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Key Concepts

Normal Distribution

Central Limit Theorem

Z-Score

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