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).

c. Find P95.

c. Find P95.

Accepted Answer

Hello. In this video, we are told that the daily production of a factory is normally distributed with a mean of 1200 units and a standard deviation of 150 units. We want to determine what the ninety-fifth percentile of the daily production is. So just to write down some given information, we are told that we have a mean, which is new. Equal to 1200, and we are given a sample standard or a standard deviation, sigma equal to 150 units. Our goal is to find a 95 95th percentile. Or in other words, we want to determine the probability that x is less than some critical value which produces a percentile of 0.95. So, in order to approach this problem, let's just go ahead and convert this into a Z score. Now, to convert this into a Z score, we want to go ahead and rewrite this as the probability that capital Z is less than little Z and this should have the percentile of 0.95. So, the question is, what critical value for Z produces a percentile of 0.95? Well, that percentile from the Z table is going to be approximately 1.645. And now that we have this value, we want to go ahead and connect it to the raw score, which is X. Now, in order to do this, X is equal to U plus the Z value multiplied by the standard deviation. This is going to be 1200 plus 1.645 multiplied by a standard deviation, which is 150, and this is going to simplify to be 1,446.75. And so what this means is that the 9 and this is going to be the 95th percentile for the daily production. And this means that the solution to this problem is going to be a. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

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