Testing Goodness-of-Fit with a Normal Distribution Refer to Da...

Question

Testing Goodness-of-Fit with a Normal Distribution Refer to Data Set 1 “Body Data” in Appendix B for the heights of females.

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b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, use the methods of Chapter 6 to find the probability of a randomly selected height belonging to each class.

Accepted Answer

Hello 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. A factory produces light bulbs whose life spans are normally distributed with a mean of 1200 hours and a standard deviation of 100 hours. What is the probability that a randomly selected light bulb falls into each of the following intervals? Awesome. So that's our final answer that we're ultimately trying to solve or rather answers. We're asked to determine what is the probability that a randomly selected light bulb or bulb falls into each of the following intervals, and we're given a blank table with 12344 separate values that we're ultimately going to try to solve for. Starting with our column on the far right, we have greater than 1300. Then our next column we have 1200 through 1300. Our next column we have 1100 through 1200. And then our next column, we have less than 1100. And then our finally on the far left column, we have lifespan in units of hours, and then we have our row, which is the probability. Awesome. So now that we know what we're ultimately trying to solve for, let us note that our first step that we need to take in order to solve this particular problem is we need to standardize the interval boundaries using the Z score formula. So, as you should recall, the Z score formula states that Z is going to be equal to X minus mu divided by sigma. Where mu in this case, let us note that mu is going to be equal to 1000. 200 And once again, sigma, our standard deviation values given to us as 100. Fantastic. So, that means that when we plug in our known variables, so when X is equal to 1100, that will mean that Z 1100 is gonna be equal to, when we plug in our known variables, 1100 minus 1200. Divided by 100, it's going to be simply equal to -1. And in an effort to save time, I'm not going to write out the equation for the Z score every single time, so I'm just going to give you what the value of Z is for different values for X. So when X is equal to 1200, that will mean that Z subscript 1200 is going to be equal to simply zero. And when X is equal to 1300, that will mean that Z subscript 1300 will be simply equal to 1, which is a contrast to when X is equal to 1100, that would mean that Z subscript 1100 is equal to -1. Fantastic. So, now, our next step is we need to find the probabilities using the standard normal distribution table. So let us note that capital P denotes the probability. So P of Z is less than -1, and this is going to be for less than 1100 hours. And when P Of -1 is less than Z and Z is less than 0. This is for 1100 to 1200 hours, and for P of zero is less than Z and Z is less than 1 will be for the range of 1200 to 1300 hours. And finally, P of Z is greater than 1, and this value will correspond to those greater than 1300 hours. Awesome. So now, we can use the standard normal distribution table values, which when you look these up in your calculator, I mean, not your calculator, you not in your calculator. They'll actually be in your textbook, but you can use your calculator too to help calculate your values when you're plugging everything into your equations. But when you look up your, once again, the standard normal distribution table in your textbook, You will find that P, once again, so it says P of Z is less than -1 will be equal to 0.1587. And for P of Z is less than 0, will be equal to 0.5, and for P of Z is less than 1, it's going to be equal to, which is a contrast to P of Z is less than -1, will be, once again, to our value of P of Z is less than 1, will be equal to 0.8413. Fantastic. Moving right along here. So now, with that in mind, we need to calculate the probabilities for each interval. So for less than 1100 hours, we determined that P of Z is less than -1, will be equal to 0.1587. And from 1100 to 1200 hours, that will mean that. P of Z is less than 0, minus P of Z is less than -1, will be equal to 0.5 minus 0.1587, which will give us a value of 0. 341 3. And that is another one of our answers that we're trying to solve for. And we also need to solve for 1200 to 1300 hours, we need to take P of Z is less than 1. Minus P of Z is less than 0. And once again, I'm not going to plug in the numbers, you know, and show you, write it into our or I'm not going to write the equation again. So when you plug in your known variables, you will find that this is going to be equal to. So when you take p of Z is less than 1 minus P Z is less than 0. It will be equal to 0.3413. Fantastic. And finally, to solve for our fourth and final value, we need to figure out greater than $1300. We need to take 1 minus P of Z is less than 1, which will be equal to 0.1587. And that is it. We've officially solved for all of our final answers. Hooray, we did it. So going back to the top of our page here to look at our problem and to plug in all of our known variables into our provided table, or diagram, I should say, we should receive the following results. So the probability for less than 1100 is 0.1587. And for our probability for 1100 through 1200, the probability is 0.3413, and for the time for the, once again, our lifespan time and hours is 1200 to 1300 is 0.3413, and then finally greater than 1300. Our probability is 0.1587, and that's it. We've officially solved for this problem. 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

Goodness-of-Fit Test

Probability and Class Intervals

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