Question

An art history professor assigns letter grades on a test according to the following scheme. A: Top 6% of scores B: Scores below the top 6% and above the bottom 63% C: Scores below the top 37% and above the bottom 16% D: Scores below the top 84% and above the bottom 8% F: Bottom 8% of scores Scores on the test are normally distributed with a mean of 77.4 and a standard deviation of 9.6. 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 92.5.

Explanation:

We are given that an art history professor assigns letter grades on a test according to the following scheme. A: Top 6% of scores B: Scores below the top 6% and above the bottom 63% C: Scores below the top 37% and above the bottom 16% D: Scores below the top 84% and above the bottom 8% E: Bottom 8% of scores

Also, Scores on the test are normally distributed with a mean of 77.4 and a standard deviation of 9.6.

Let X = Scores on a test

So, X ~ N(
`\mu=77.4,\sigma^(2) = 9.6^(2)`)

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean

`\sigma` = standard deviation

Now, we have to find the minimum score required for an A grade, i.e.; Top 6% of scores.

So, Probability that the test separate the top 6% of scores is given by;

P(X > x) = 0.06

P(
`(X-\mu)/(\sigma)` >
`(x-77.4)/(9.6)` ) = 0.06

P(Z >
`(x-77.4)/(9.6)` ) = 0.06

So, the critical value of x in z table which separate the top 6% is given as 1.5722, which means;

`(x-77.4)/(9.6)` = 1.3543

`x-77.4 = 9.6 * 1.5722`

`x` = 77.4 + 15.09312 = 92.5

Therefore, minimum score required for an A grade that represent top 6% of scores is 92.5.

Answer 2

The minimum score required for an A grade is 92.

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 = (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 = 77.4, \sigma = 9.6`

Find the minimum score required for an A grade.

Top 6%, so at least the 100-6 = 94th percentile, which is the value of X when Z has a pvalue of 0.94. So X when Z = 1.555. So

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

`1.555 = (X - 77.4)/(9.6)`

`X - 77.4 = 1.555*9.6`

`X = 92`

The minimum score required for an A grade is 92.