Finding Probabilities for Normal Distributions In Exercises 7–...

Question

Finding Probabilities for Normal Distributions In Exercises 7–12, find the indicated probabilities. If convenient, use technology to find the probabilities.

MCAT Scores In a recent year, the MCAT total scores were normally distributed, with a mean of 500.9 and a standard deviation of 10.6. Find the probability that a randomly selected medical student who took the MCAT has a total score that is (b) between 490 and 510. Identify any unusual events in parts (a)–(c). Explain your reasoning. (Source: Association of American Medical Colleges)

Accepted Answer

Hello everyone. Let's take a look at this question together. The test scores for a class are normally distributed with a mean of 82 and a standard deviation of 6. What is the probability that a randomly chosen student scored between 73 and 88, is this considered an unusual event? So, in order to solve this question, we have to recall how to find the probability that a randomly chosen student scored between 73 and 88 if the test scores for the class are normally distributed with a mean of 82 and a standard deviation of 6. And we also have to determine if this is considered an unusual event. So the first step in solving this problem is to standardize the scores using the Z score formula, which we know the Z score is calculated as Z equals X minus the mean divided by the standard deviation, where we know that X is the score, mu is the mean, and sigma is the standard deviation. And Using the formula, we can plug in our X1 value, which is equal to 73, to get our first Z score, which substituting the values into the formula, we get -1.5. So our first Z score is -1.5. And then we do the same thing for our second X value of 88. By plugging it into our Z score formula, giving us a value of 1 for our second Z score, and then we find the probabilities corresponding to those Z scores using the standard normal distribution table, and the probability of Z being less than -1.5 is approximately 0.0668. And the probability of Z being less than 1 is approximately 0.8413. So then we can calculate the probability that a score is between 73 and 88 by subtracting the lower probability from the higher probability, which results in the probability of X being between 73 and 88, is equal to the. of Z being less than 1, subtracted by the probability of Z being less than -1.5, and substituting the values for those probabilities, we get 0.8413, subtracted by 0.0668, which equals 0.7745. Therefore, the probability that a randomly chosen student scored between 73 and 80. 8 is 0.7745, and based on this information, we know a probability of 0.7745 means that it is quite common for a student to score between 73 and 88, so this is not considered an unusual event, as unusual events typically have probabilities less than 0.05. So we have a probability of 0.7745, and this is not considered an unusual event, which makes answer choice A the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Normal Distribution

Z-Scores

Probability Calculation

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