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.

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AbsorbingMango AbsorbingMango · 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 [tex]$D_1$[/tex] and the top limit for D grade be [tex]$D_2$[/tex].

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

[tex]$P(X<D_1)= 0.10$[/tex]

[tex]$P(X\leq D_1)= 0.10$[/tex]

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

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

From (1)

[tex]$\frac{D_1 - 72.8}{7.3} = -1.28$[/tex]

[tex]$D_1 = -1.28(7.3)+72.8$[/tex]

= 63.45

≈ 64

Now the top numerical limit for D grade :

[tex]$P(X>D_2)= 0.80$[/tex]

[tex]$1-P(X\leq D_2)= 0.80$[/tex]

[tex]$P(X\leq D_2)= 1-0.80$[/tex]

[tex]$P(X\leq D_2)= 0.20$[/tex]

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

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

From (2)

[tex]$\frac{D_2- 72.8}{7.3} = -0.84$[/tex]

[tex]$D_12= -0.84(7.3)+72.8$[/tex]

= 66.668

≈ 67

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