Question

MR MARK MARKS HIS CLASS ON A NORMAL CURVE. THOSE WITH z-SCORES ABOVE 1.8 WILL RECEIVE AN A, THOSE

Answer 1

Using the normal distribution, it is found that the percentages are given as follows:

• 3.59% of the grades will be A.

• 9.98% of the grades will be B.

• 74.92% of the grades will be C.

• 8.64% of the grades will be D.

• 2.87% of the grades will be F.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The prorportion of students who receive an A is one subtracted by the p-value of Z = 1.8.

Looking at the z-table, Z = 1.8 has a p-value of 0.9641.

1 - 0.9641 = 0.0359.

3.59% of the grades will be A.

For B, it is the p-value of Z = 1.8 subtracted by the p-value of Z = 1.1, hence:

0.9641 - 0.8643 = 0.0998.

9.98% of the grades will be B.

For C, it is the p-value of Z = 1.1 subtracted by the p-value of Z = -1.2, hence:

0.8643 - 0.1151 = 0.7492.

74.92% of the grades will be C.

For D, it is the p-value of Z = -1.2 subtracted by the p-value of Z = -1.9, hence:

0.1151 - 0.0287 = 0.0864.

8.64% of the grades will be D.

For F, it is the p-value of Z = -1.9, hence 2.87% of the grades will be F.

More can be learned about the normal distribution at brainly.com/question/15181104

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