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

The minimum score required for an A grade is 83.

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 = (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 = (X - \mu)/(\sigma)`

`1.34 = (X - 72.3)/(8)`

`X - 72.3 = 1.34*8`

`X = 83`

The minimum score required for an A grade is 83.