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.

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 and standard deviation , the z-score of a measure X is given by:

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

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.

The minimum score required for an A grade is 83.

step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.