Foot Lengths of Women Assume that foot lengths of adult female...

Question

Foot Lengths of Women Assume that foot lengths of adult females are normally distributed with a mean of 246.3 mm and a standard deviation of 12.4 mm (based on Data Set 3 “ANSUR II 2012” in Appendix B).

d. Find the probability that 16 adult females have foot lengths with a mean greater than 250 mm.

d. Find the probability that 16 adult females have foot lengths with a mean greater than 250 mm.

Accepted Answer

Below there, today we're gonna 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. Suppose that the heights of a certain species of plants are normally distributed with a mean of 135.8 centimeters and a standard deviation of 9.6 centimeters. If a random sample of 25 plants is selected, what is the probability that their average height exceeds 139 centimeters? Awesome. So it appears for this particular problem we're asked to take a random sample of 25 plants, and we're asked to determine what the probability of their average heights exceeding 139 centimeters will be. So now that we know that we're ultimately trying to figure out what this probability value is, let's read off our multiple choice answers to see what our final answer might be. A is 0.0478. B is 0.0627. C is 0.0883. And finally, D is 0.0976. Awesome. So first off, let's take a moment to write down all of our given variables that are known to us. So all the variables that are given to us by the prom itself. We're told that the mean, which we're going to denote as mu, the mean is equal to 135.8 centimeters. We're also told the standard deviation, and the standard deviation, which we're going to denote as sigma is equal to 9.6 centimeters. And we're also told the number of the outcomes, which we're told that the random sample of 25 plants, so that means that N is going to be equal to 25. Awesome. So now, our next step is we need to find the probability that a sample mean X bar exceeds 139 centimeters. So that means we need to recall and use the equation to help us calculate the standard error of the mean. So that means that sigma subscript X bar is equal to sigma divided by the square root of N, which is going to be equal to 9.6 divided by the square root of 25, which when we plug that into our calculator, we will find that it's equal to 1.92. Awesome. So now, we need to calculate the Z score for X bar is equal to 139. So let's make a note. So when X bar is equal to 139 centimeters, So as we should recall, Z to Z score is equal to X minus mu divided by sigma X bar, which is going to be equal to let me plug in all of our known variables, 139 minus 135.8. Divided by 1.92, which is going to be approximately equal to when we plug this into our calculator, it's going to be approximately equal to 1.667 when we round to three decimal places. So now, our next step, therefore, is we need to find the probability P of Z that is greater than 1.667. And we can find the probability of Z is greater than 1.667, and this can be found by subtracting one from our probability of Z is is less than. So once again we can find this by subtracting one from P of Z is less than 1.667. So let's make a note of that. So now, with that said, we could go ahead and write that P of Z is greater than 1.667, is equal to 1 minus P of Z is less than 1.667. And we now need to refer to our table for positive Z scores, and I've gone ahead and copied a table to paste here for us to refer to. And when we use our table for positive Z scores, we could see that sense 1.667 is between 1.66 and 1.67. We need to perform linear interpolation. And this is where we we could see this from the highlighted part of our table where we have 1.6 and we have. 0, so 0.06 and 0.07, which gives us respectfully, 0.9515 and 0.9525. So, with that said, referring to our table, to let us guide us, we could say that P, the probability P minus 0.9515 divided by 5 divided by 3, which going back up above here for a moment here, this 1.667 is also equal to, if we just keep it in fraction form, 5 divided by 3. So let's make a note of that. So, going back, so we have P minus 0.9515 divided by 5 divided by 3 or 5/3 minus 1.66. which this will be equal to 0.9525 minus 0.9515 divided by 1.67 minus 1.66. So that will mean that the probability P is going to be equal to parentheses 0.9525, so 9525. -0.9515. Divided by 1.67 minus 1.66. Multiplied by 5/3 minus 1.66 plus 0.9515. Which, when we plug this into our calculator and we round to 4 decimal places, we will find that it's approximately equal to 0.9522. So once again, it's going to be approximately equal to 0.9522. So that will mean, without a doubt, that we could go ahead and write that P of Z is less than 1.667 is equal to 1 minus 0.9522, which will be equal to when we plug into our calculator and we round to four decimal places, 0.0478, and that is our final answer. Hooray, we did it. So therefore, the probability that the average height exceeds 139 centimeters for the random sample of 25 plants is approximately 0.0478, and this value corresponds to what appears to be multiple choice answer. Letter A, which once again is 0.0478. 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

Central Limit Theorem

Z-Score

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