Computing Probabilities for Normal Distributions In Exercises ...

Question

Computing Probabilities for Normal Distributions In Exercises 1–6, the random variable x is normally distributed with mean mu=174 and standard deviation sigma=20. Find the indicated probability.

P(172 < x <192)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the test scores in a statistics class are normally distributed with a mean of 82 and a standard deviation of 7. What's the probability that a randomly chosen student scored between 75 and 90? And we're given four possible choices as our answers. For choice A, we have 0.1742. For choice B, we have 0.8142. For choice C, we have 0.4271, and for choice C, we have 0.7148. So the first thing we want to do is convert our two scores here of 75 and 90 into Z scores. And so we call your formula for finding Z scores, that's going to be Z is equal to the quantity of X minus u in quantity divided by stigma. Where Z here's our C score. X is going to be one of our test scores. Mu is going to be our mean and sigma is going to be our standard deviation. So we need to calculate two different Z scores. So the first Z score, we'll label as Z1. This is going to correspond to a test score of 75. So Z1 is going to be equal to the quantity of 75 minus and for mu are mean, we're told that that is 82 and the problem, so we'll have minus 82, and then that quantity is going to be divided by our standard deviation sigma, which we're given in the problem as 7. And so when we evaluate this expression, we see that Z1 is going to be equal to -1. We're gonna do the same thing for the other test score, so we'll call that Z2, and this is going to correspond to the test score of 90. So, Z2 is going to be equal to 90 minus 82. In quantity, divided by 7. And when we evaluate this expression, this is going to be approximately equal to 1.143. But now we need to do is to look up in our standard normal distribution table, the cumulative probabilities for each of these Z scores. So, for the first Z score, Z1, we have a cumulative probability of P of Z less than -1. And that is going to be equal to, when we look this up on our table, that's gonna be 0.1587. And if we try to do the same thing for the other Z score, we'll be looking for the cumulative probability of P of Z less than 1.143, we see that this actually falls in between two values in our table. So, we're actually going to have to use interpolation. And so the two values that are Z score falls between are gonna be Z equal to 1.14, and this has a corresponding um cumulative probability of 0.8729. And the other C score is going to be Z equal to 1.15. And this has a cumulative probability of 0.8749. So, in order to interpolate, Between these two values, we're going to look at basically finding this slope between these two points. So, our cumulative probability of Z less than 1.143 is going to be equal to the quantity and the numerator here will have the difference between the two areas. So we'll have 0.87. 49 minus 0.8729, and then that quantity is going to be divided by the quantity of the difference between the two z values, which is going to be 1.15 minus 1.14. And then we're going to need to multiply that fraction by the difference that our Z value is that we're looking for, which is the 1.143, is from the lower bound. So we'll have the quantity of 1.143 minus. 1.14 in quantity, and then we'll have to add to this the cumulative probability for the lower bound, which is the cumulative probability that corresponds to Z equal to 1.14. So this will be plus 0.8729. And so when we evaluate this expression, this is going to be approximately equal to. 0. 873 5. So now that we have calculated both of those cumulative probabilities to find the probability that we're between these two values of a test score of 90 and 75, we'll just need to take the difference of the cumulative probabilities that we actually found. So the probability. P That -1 is less than Z, which is less than 1.143. is going to be equal to. P of Z less than 1.143. Minus P of Z. Less than -1. And so substituting in the values that we found, this is going to be equal to. 0.8735 minus 0.1587. And when we evaluate the expression, this is going to be equal to. 0.7148. So this here is the answer that we're looking for. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice D. So, I hope that this explanation has been useful, and I'll see you in the next video.

Computing Probabilities for Normal Distributions In Exercises 1–6, the random variable x is normally distributed with mean mu=174 and standard deviation sigma=20. Find the indicated probability.

P(172 < x <192)

Key Concepts

Normal Distribution

Z-scores

Probability Calculation

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