Question

A humanities professor assigns letter grades on a test according to the following scheme.
A: Top 8% of scores
B: Scores below the top 8% and above the bottom 62%
C: Scores below the top 38% and above the bottom 18%
D: Scores below the top 82% and above the bottom 9%
E: Bottom 9% of scores Scores on the test are normally distributed with a mean of 67 and a standard deviation of 7.3.
Find the numerical limits for a C grade.

Answer 1

Answer:

The numerical limits for a C grade are 60.6 and 69.1.

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 67 and a standard deviation of 7.3.

This means that $\mu = 67, \sigma = 7.3$

Find the numerical limits for a C grade.

Below the 100 - 38 = 62th percentile and above the 18th percentile.

18th percentile:

X when Z has a p-value of 0.18, so X when Z = -0.915.

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

$-0.915 = \frac{X - 67}{7.3}$

$X - 67 = -0.915*7$

$X = 60.6$

62th percentile:

X when Z has a p-value of 0.62, so X when Z = 0.305.

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

$0.305 = \frac{X - 67}{7.3}$

$X - 67 = 0.305*7$

$X = 69.1$

The numerical limits for a C grade are 60.6 and 69.1.