Question

A psychology professor assigns letter grades on a test according to the following scheme. A: Top 9% of scores B: Scores below the top 9% and above the bottom 57% C: Scores below the top 43% and above the bottom 18% D: Scores below the top 82% and above the bottom 5% F: Bottom 5% of scores Scores on the test are normally distributed with a mean of 76.9 and a standard deviation of 9.6. Find the numerical limits for a D grade. Round your answers to the nearest whole number, if necessary.

Answer 1

Answer:

Grades between 61 and 68 are the numerical limits for a D grade.

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:

$\mu = 76.9, \sigma = 9.6$

D: Scores below the top 82% and above the bottom 5%

So scores between 5th and the 100-82 = 18th percentile.

5th percentile:

value of X when Z has a pvalue of 0.05. So X when Z = -1.645.

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

$-1.645 = \frac{X - 76.9}{9.6}$

$X - 76.9 = -1.645*9.6$

$X = 61$

18th percentile:

value of X when Z has a pvalue of 0.18. So X when Z = -0.925.

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

$-0.925 = \frac{X - 76.9}{9.6}$

$X - 76.9 = -0.925*9.6$

$X = 68$

Grades between 61 and 68 are the numerical limits for a D grade.