Using the Central Limit Theorem. In Exercises 5–8, assume that...

Question

Using the Central Limit Theorem. In Exercises 5–8, assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of 1.2 kg and a standard deviation of 4.9 kg (based on Data Set 13 “Freshman 15” in Appendix B).
a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year.

a. If 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year.

Accepted Answer

Hello. In this video, we are told that the time it takes for a student to complete a standardized math test is normally distributed with a mean of 48 minutes and a standard deviation of 9 minutes. If one student is chosen at random, what is the probability that the student finishes the test between 45 and 60 minutes? So let's go ahead and write down our given information. First, we are told that we have a mean of 48 minutes with a standard deviation of 9 minutes, and our goal is to find the probability that a student finishes an exam between 45 and 60 minutes. So, in order to approach this problem, let's go ahead and convert this probability into a Z score. Now, starting with the endpoint, X is equal to 45, in order to convert this into a Z score, we will take our value of interest, which is 45, subtract the mean from it, and divide by the standard deviation. And this is going to simplify to give us -0.333. So our goal here is to find the probability that Z is less than negative 0.333. Now, we know that this data is normally distributed, so an equivalent way of writing this probability is 1 minus the probability that Z is less than 0.333. So now what we need to do is we need to find this probability. Now, the hard part about this is that we don't have a critical point, 0.333 on the Z table. So what we're going to need to do is we're going to need to interpolate this critical point. Now, one value that we do have is that Z is equal to 0.33. Which has an approximate value. Of 0.62930. And we also know That Z equal to 0.34 has an approximate value of 0.63307. So if we interpolate 0.333 using these two values, we end up getting 0.62930 plus the product that is defined. As 0.333. Minus 0.33. Divided by 0.34 minus 0.33, multiplied by the difference of the probability at 0.34, which is 0.63307, minus the probability at 0.33, 0.62930. This is going to simplify to give us 0.670544 and 1. And finally, if we plug this in to the probability, we get 1 minus this value. Which approximates to 0.3694559. Now we want to repeat this process with X's equal to 60. So, we are going to convert this into a Z score, and that is going to be 60 minus the mean, which is 48 divided by 9. And this simplifies to 1.333. So we need to find the probability that Z is less than 1.333. And again, we are going to have to interpolate this. So we have the probability. Of Z equal to 1.33, which approximates to 0.90824. And we have the probability of Z equal to 1.34, which approximates to 0.90988. So again, we are going to interpolate this, and this is going to give us 0.90824, plus, The quotient 1.333 minus 1.33. Divided by 1.34 minus 1.33. Multiplied by the difference of 0.90988 minus 0.90824. And this is going to simplify to give us a probability of 0.9087812. So, to combine all these probabilities together, we want to find the probability. That Z is between -0.333. And 1.333. And that is going to be 0.9087812 minus 0.3694559. And this will simplify to be 0.5393, which is going to be the probability that a student finishes an exam between 45 and 60 minutes, and the solution to this problem is going to be B. 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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Key Concepts

Central Limit Theorem

Normal Distribution

Probability Calculation

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