Question

An art history professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top 13% and above the bottom 56% C: Scores below the top 44% and above the bottom 21% D: Scores below the top 79% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.

Answer 1

Answer:

The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4.

This means that $\mu = 79.7, \sigma = 8.4$

B: Scores below the top 13% and above the bottom 56%

So between the 56th percentile and the 100 - 13 = 87th percentile.

56th percentile:

X when Z has a p-value of 0.56, so X when Z = 0.15. Then

$Z = \frac{X - \mu}{\sigma}$

$0.15 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 0.15*8.4$

$X = 81$

87th percentile:

X when Z has a p-value of 0.87, so X when Z = 1.13.

$Z = \frac{X - \mu}{\sigma}$

$1.13 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 1.13*8.4$

$X = 89$

The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.