Question

An English 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 61% C: Scores below the top 39% and above the bottom 25% D: Scores below the top 75% and above the bottom 10% F: Bottom 10% of scores Scores on the test are normally distributed with a mean of 72.3 and a standard deviation of 8. Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.

Answer 1

Answer:

The minimum score required for an A grade is 83.

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.

Mean of 72.3 and a standard deviation of 8.

This means that $\mu = 72.3, \sigma = 8$

Find the minimum score required for an A grade.

This is the 100 - 9 = 91th percentile, which is X when Z has a pvalue of 0.91, so X when Z = 1.34.

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

$1.34 = \frac{X - 72.3}{8}$

$X - 72.3 = 1.34*8$

$X = 83$

The minimum score required for an A grade is 83.