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 61% C: Scores below the top 39% and above the bottom 20% D: Scores below the top 80% and above the bottom 10% F: Bottom 10% of scores Scores on the test are normally distributed with a mean of 72.8 and a standard deviation of 7.3. Find the numerical limits for a D grade. Round your answers to the nearest whole number, if necessary.

Answer 1

Solution :

The test is distributed normally with mean of 72.8 and the standard deviation of 7.3

Finding numerical limits for the D grade.

D grade : Scores below the top 80% and above the bottom 10%.

Let the bottom limit for D grade be $D_1$ and the top limit for D grade be $D_2$.

First find the bottom numerical limit for a D grade is :

$P(X$

$P(X\leq D_1)= 0.10$

$P\left(\frac{X-\mu}{\sigma} \leq \frac{D_1-\mu}{\sigma}\right) = 0.10$

$P\left(Z \leq \frac{D_1-72.8}{7.3}\right) = 0.10$    ..........(1)

From (1)

$\frac{D_1 - 72.8}{7.3} = -1.28$

$D_1 = -1.28(7.3)+72.8$

      = 63.45

       ≈ 64

Now the top numerical limit for D grade :

$P(X>D_2)= 0.80$

$1-P(X\leq D_2)= 0.80$

$P(X\leq D_2)= 1-0.80$

$P(X\leq D_2)= 0.20$

$P\left(\frac{X-\mu}{\sigma} \leq \frac{D_2-\mu}{\sigma}\right) = 0.20$

$P\left(Z \leq \frac{D_2-72.8}{7.3}\right) = 0.20$    ..........(2)

From (2)

$\frac{D_2- 72.8}{7.3} = -0.84$

$D_12= -0.84(7.3)+72.8$

      = 66.668

       ≈ 67

Therefore, the numerical limit for a D grade is 64 to 67.