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.

Answers

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.


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Answers

Answer:

You get a D if you have a grade between 66 and 72.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

Scores on the test are normally distributed with a mean of 79.3 and a standard deviation of 8.4.

This means that [tex]\mu = 79.3, \sigma = 8.4[/tex]

D: Scores below the top 80% and above the bottom 6%

So between the 6th and the 20th percentile.

6th percentile:

X when Z has a pvalue of 0.06. So X when Z = -1.555.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.555 = \frac{X - 79.3}{8.4}[/tex]

[tex]X - 79.3 = -1.555*8.4[/tex]

[tex]X = 66.2[/tex]

Rounds to 66

20th percentile:

X when Z has a pvalue of 0.2. So X when Z = -0.84.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.84 = \frac{X - 79.3}{8.4}[/tex]

[tex]X - 79.3 = -0.84*8.4[/tex]

[tex]X = 72.2[/tex]

Rounds to 72

You get a D if you have a grade between 66 and 72.

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Answers

Answer:

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Answer:

3

Step-by-step explanation:

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у
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Answers

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