Question

A humanities professor assigns letter grades on a test according to the following scheme. A: Top 12% of scores B: Scores below the top 12% and above the bottom 61% C: Scores below the top 39% and above the bottom 21% D: Scores below the top 79% and above the bottom 6% F: Bottom 6% of scores Scores on the test are normally distributed with a mean of 67.7 and a standard deviation of 7.8. 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 77.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 67.7, \sigma = 7.8$

Find the minimum score required for an A grade.

Top 12% of scores get an A.

100-12 = 88th percentile.

The 88th percentile of scores is the minimum required for an A grade. This score is X when Z has a pvalue of 0.88. So X when Z = 1.175.

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

$1.175 = \frac{X - 67.7}{7.8}$

$X - 67.7 = 7.8*1.175$

$X = 76.865$

Rounding to the nearest whole number:

The minimum score required for an A grade is 77.