Question

Suppose that the distribution of exam scores in a class of 200 student is a normal distribution. Also suppose that the​ distribution's mean is 60 points, and its standard deviation is 15.

Answer 1

Approximately 68% of the scores are expected to fall below 50. 136 students are expected to fall below 50. The z-score for a score of 50 is -1. Approximately 99.7% of scores fall within the range of 60 and 90. The z-score associated with an exam score of 81 is 2.1

Let's go through each part of the question:

a) Use the 68-95-99.7 rule to determine what percent of the scores do we expect to fall below 50?

According to the 68-95-99.7 rule for a normal distribution:

- Approximately 68% of the scores fall within 1 standard deviation of the mean,

- Approximately 95% fall within 2 standard deviations,

- Approximately 99.7% fall within 3 standard deviations.

Since the mean is 60 and the standard deviation is 10:

The scores below
`\(60 - 1 * 10 = 50\)`are within the first standard deviation.

Therefore, approximately 68% of the scores are expected to fall below 50.

b) How many scores do we expect to fall below 50?

If 68% of the scores fall within the first standard deviation, and there are 200 students in the class, then the number of scores expected to fall below 50 can be estimated as 68% of 200.

`\[ \text{Number of students} = 0.68 * 200 = 136 \text{ students} \]`

c) Which z-score corresponds to the exam score from parts a and b?

To find the z-score corresponding to a score of 50, we can use the formula for z-score:

`\[ z = \frac{(X - \text{Mean})}{\text{Standard Deviation}} \]`

`\[ z = ((50 - 60))/(10) = -1 \]`

So, the z-score for a score of 50 is -1.

d) What percent of scores do we expect to fall between 60 and 90?

Since 60 and 90 are 3 standard deviations above the mean, according to the 68-95-99.7 rule, approximately 99.7% of scores fall within this range.

e) Calculate the exam score associated with the z-score
`\(z = 1.2\)`.

`\[ X = \text{Mean} + z * \text{Standard Deviation} \]`

`\[ X = 60 + 1.2 * 10 = 72 \]`

So, the exam score associated with a z-score of 1.2 is 72.

f) Calculate the z-score associated with the exam score of 81.

`\[ z = \frac{(X - \text{Mean})}{\text{Standard Deviation}} \]`

`\[ z = ((81 - 60))/(10) = 2.1 \]`

So, the z-score associated with an exam score of 81 is 2.1.

The probable question may be:

"Suppose that the distribution of exam scores in a class of 200 student is a normal distribution. Also suppose that the distribution's mean is 60 points, and its standard deviation is 10. a) Use the 68-95-99.7 rule to determine what percent of the scores do we expect to fall below 50? % b) How many scores do we expect to fall below 50? c) Which z-score corresponds to the exam score from part a and b? d) What percent of scores do we expect to fall between 60 and 90? % e) Calculate the Exam score associated with the z -score z=1.2. f) Calculate the z -score associated with the Exam score 81. Round all answers to the second decimal place"