History Grades In a history class, the grades for various asse...

Question

History Grades In a history class, the grades for various assessments are all positive numbers and have different distributions. Determine whether the grades for each assessment could be normally distributed. Explain your reasoning.

b. a final with a mean of 72, standard deviation of 9, and 90th percentile score of 93

b. a final with a mean of 72, standard deviation of 9, and 90th percentile score of 93

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says in a math class, you take several tests with scores that were all positive numbers and vary across assessments. Determine if the test scores for a final exam could reasonably follow a normal distribution. Provide your reasoning. For this final exam, U is equal to 65, sigma is equal to 10, and the 90th percentile score is 83. Now we're given 4 possible choices as our answers. For choice A, the grades are likely not normally distributed because they mean a standard deviation are the same as the 90th percentile. For choice B, the grades are likely. Normally distributed because the 90th percentile is much lower than expected. For choice C, the grades are likely not normally distributed because the 90th percentile is much higher than expected for a normal distribution and for choice of D, the grades perfectly match a normal distribution because the 9th percentile exactly matches the expected value. So, in this problem, we're given several pieces of information about the final exam. We're told that the mean, mu is equal to 65, the standard deviation sigma is equal to 10, and that the 90th percentile score is 83. So, to check to see if we have a normal distribution or not, we're gonna see if the 90th percentile score that is um reported here, matches what is expected for a normal distribution. Now, for a normal distribution, The cumulative probability P of Z less than lowercase z for the 90th percentile is going to be equal to 0.9. And so we need to find the corresponding Z score for this um cumulative probability. And looking this up in our standard normal distribution tables, we see that we get a Z score Z that's going to be approximately equal to 1.28. So now we just need to convert the C score back into a test score. And recall your formula to do that, we'll have our test score X, and since this is for the 9th percentile, we'll call this X90, and recall that this is going to be equal to mu, plus Z multiplied by sigma. Where mu here is our mean, Z is our Z score, and sigma is our standard deviation. So substituting in our values, those quantities. X90 is going to be equal to 65. Plus 1.28. Multiplied by 10. And evaluating this expression, this is going to be approximately equal to. 78. So here we just rounded to the nearest whole number. And if we compare that to the 9th percentile score that was recorded for the test, we see that 83. is actually greater than 78. And so, since these two values do not match up. That means that our distribution is not likely to be normally distributed. And so that is going to be our answer for this problem. And so if we compare what we have looked at here to our answer choices, we see that the correct answer choice corresponds to answer choice Z, which says that the grades are likely not normally distributed because the 9th percentile is much higher than expected for a normal distribution. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Normal Distribution

Mean and Standard Deviation

Percentiles

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