Question

An English professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top 13% and above the bottom 55% C: Scores below the top 45% and above the bottom 20% D: Scores below the top 80% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 78.8 and a standard deviation of 9.8. Find the numerical limits for a C grade. Round your answers to the nearest whole number, if necessary.

Answer 1

Scores between 71 and 80 give a C grade.

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.

Scores on the test are normally distributed with a mean of 78.8 and a standard deviation of 9.8.

This means that
`\mu = 78.8, \sigma = 9.8`

Find the numerical limits for a C grade.

Above the bottom 20%(20th percentile) and below the top 45%(below the 100 - 45 = 55th percentile).

20th percentile:

X when Z has a p-value of 0.2, so X when Z = -0.84.

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

`-0.84 = (X - 78.8)/(9.8)`

`X - 78.8 = -0.84*9.8`

`X = 70.57`

So it rounds to 71.

55th percentile:

X when Z has a p-value of 0.55, so X when Z = 0.125.

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

`0.125 = (X - 78.8)/(9.8)`

`X - 78.8 = 0.125*9.8`

`X = 80`

Scores between 71 and 80 give a C grade.