Question

A humanities professor assigns letter grades on a test according to the following scheme. A: Top 12%12% of scores B: Scores below the top 12%12% and above the bottom 62%62% C: Scores below the top 38%38% and above the bottom 17%17% D: Scores below the top 83%83% and above the bottom 7%7% F: Bottom 7%7% of scores Scores on the test are normally distributed with a mean of 75.275.2 and a standard deviation of 9.89.8. 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 86.8.

Step-by-step explanation:

We are given that 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 62% C: Scores below the top 38% and above the bottom 17% D: Scores below the top 83% and above the bottom 7% E: Bottom 7% of scores

Scores on the test are normally distributed with a mean of 75.2 and a standard deviation of 9.8.

Let X = Scores on the test

SO, X ~ N($\mu = 75.2,\sigma^{2} = 9.8^{2}$)

The z-score probability distribution is given by ;

                  Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean score = 75.2

            $\sigma$ = standard deviation = 9.8

Now, the minimum score required for an A grade to be in the top 12% of scores is given by ;

              P(X $\geq$ $x$ ) = 0.12   {where $x$ is the minimum score required}

             P( $\frac{X-\mu}{\sigma}$ $\geq \frac{x-75.2}{9.8}$ ) = 0.12

             P(Z $\geq \frac{x-75.2}{9.8}$ ) = 0.12

Now, in z table we will find out that critical value of X for which the area is in top 12%, which comes out to be 1.1835.

This means;         $\frac{x-75.2}{9.8} = 1.1835$

                          $x-75.2=1.1835 \times 9.8$  

                              $x$ = 75.2 + 11.5983 = 86.8

Therefore, the minimum score required for the scholarship is 86.8.