Question

A philosophy 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 58% C: Scores below the top 42% and above the bottom 25% D: Scores below the top 75% and above the bottom 8% F: Bottom 8% of scores Scores on the test are normally distributed with a mean of 78.9 and a standard deviation of 9.3. 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 90.

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:

$\mu = 78.9, \sigma = 9.3$

Find the minimum score required for an A grade.

Top 12%, so at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.

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

$1.175 = \frac{X - 78.9}{9.3}$

$X - 78.9 = 1.175*9.3$

$X = 89.8$

Rounding to the nearest whole number, the answer is 90.

The minimum score required for an A grade is 90.