Question

An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores below the top 13%13% and above the bottom 56%56% C: Scores below the top 44%44% and above the bottom 21%21% D: Scores below the top 79%79% and above the bottom 9%9% F: Bottom 9%9% of scores Scores on the test are normally distributed with a mean of 79.779.7 and a standard deviation of 8.48.4. 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 is between 81 and 89.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 79.7, \sigma = 8.4$

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

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

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

$1.127 = \frac{X - 79.7}{8.4}$

$X - 79.7 = 8.4*1.127$

$X = 89$

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

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

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

$X - 79.7 = 8.4*0.15$

$X = 81$

The numerical limits for a B grade is between 81 and 89.