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(x > 182)

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says the scores on a college entrance exam are normally distributed with a mean of 600 and a standard deviation of 90. What is the probability that a randomly selected student scores above 720? And we're given four possible choices as our answers. For choice A, we have 0.0198. For choice B, we have 0.0912. For choice C, we have 0.9082, and for choice D, we have 0.0918. Now we're asked to find the probability that a randomly selected student scores above 720. So the first thing we want to do is convert our test score 1720 into a Z. And to recall your formula for the Z score, that is, Z is going to be equal to the quantity of X minus mu in quantity divided by sigma. Where Z here is our Z score, X here is our test score, mu is going to be our mean, and sigma is our standard deviation. And we have all those quantities given to us in the problem, so we can actually calculate our Z score. So that means Z is going to be equal to, or our test score, we're gonna use 720. Then we'll have minus our mean, which is 600. And then we'll take that quantity and divide it by our standard deviation, which we're told the problem is 90. And when we evaluate this expression, this is going to be equal to 4 divided by 3, which is approximately equal to 1.3333. So here we just rounded it to 4 decimal places. And so now what we want to do is to find the cumulative probability for this Z score, and that will tell us the probability of finding a student that scores less than 720. We'll then use that information to determine the probability that the randomly selected student scores above 720. So, when we look in our standard normal distribution tables, we see that our Z score here actually falls in between two different values in our tables. It falls between Z is equal to 1.33. Which has a cumulative probability of. 0.9082, and the other value is Z is equal to 1.34, and that has a cumulative probability of 0.9099. So since our value falls in between two of our Z values in our table, we'll actually need to use interpolation to find our cumulative probability. So, we're looking for the cumulative probability P of Z is less than or divided by 3. And so this is going to be equal to, and here we call your formula 4 interpolation. So, we'll first have The change in our cumulative probabilities divided by our change in our Z scores from our table, so that will be the quantity of 0.9099 minus 0.9082. And that quantity, it's going to be divided by. The quantity of 1.34 minus 1.33 in quantity. We then need to multiply this by the difference between the Z score that we're looking for and our smaller of our two Z scores from the table. So, we'll multiply this fraction here by the quantity of our Z score that we're looking for, which is going to be 4 divided by 3. They'll subtract from that, the smaller of our two Z scores, which is going to be 1.33. In quantity, and then we'll have to add to that the cumulative probability of Z equal to 1.33. So we'll have plus 0.9082. And when we evaluate this expression, we get that this is going to be approximately equal to. 0.9088. So here we've just rounded to 4 decimal places. So now we want to do is to use this information to find the probability that randomly selected students scored above 720. So what we have here is the cumulative probability that the randomly selected student score is below 720. So to find The cumulative probability P, that our student scores above 720, which would correspond to a Z score that is greater than or defined by 3. That this is going to be equal to, and here we're just going to take the compliment. So we'll have 1 minus our probability P of Z less than or divided by 3. And so calculating this, this is going to be equal to 1 minus, and here we'll substitute in the value that we just um calculated. So it's gonna be 0.9088. And when we evaluate this expression, this is going to be equal to. 0.0912. So this here is the answer that we're looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice actually corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

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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(x > 182)

Key Concepts

Normal Distribution

Z-Score

Probability Calculation

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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(x > 182)