The performance of 20 students on their final exam on statistics in 1992 and for the second class of 20 students in 2017 is shown in the table below. Scores on Statistics Final Exam, 1992 and 2017 Class of 1992 Class of 2017 80 100 91 99 91 94 80 94 74 94 74 88 73 88 75 81 76 88 73 89 73 80 69 88 67 76 68 75 68 63 68 61 68 53 57 55 58 56 59 82 Use the data to answer the following questions. NOTE: For Parts B & C, use the z-score equation: and table located in Appendix A of the text (p. 697). Find the mean and standard deviation for the final exam scores in the two years, 1992 and 2017. Suppose you scored an 85 on this exam. In which distribution is your test score the farthest above the mean? How far above the mean is your score in both of these distributions? Admission to a (paid) summer internship program requires that students earn a C or better (70% or higher) on their statistics final exam. If we assume that scores (in both years) on this test are reasonably normally distributed, what percentage of students in 1992 and 2017 would qualify for this internship?

To calculate the z-score for a test score of 85 in both distributions, we will use the formula z = \frac{(X - \text{mean})}{\text{standard deviation}} .

For the class of 1992:
z_{1992} = \frac{(85 - 72.1)}{9.10} \approx 1.42

For the class of 2017:
z_{2017} = \frac{(85 - 80.2)}{15.06} \approx 0.32

Therefore, the z-score for a test score of 85 is farther above the mean in the 1992 distribution compared to the 2017 distribution.

To determine the percentage of students who would qualify for the internship program by scoring a 70% or higher, we will find the percentage of students whose scores are above or equal to 70.

Using the z-score formula, we calculate the z-score for a score of 70 for both years.

For the class of 1992:
z_{1992} = \frac{(70 - 72.1)}{9.10} \approx -0.23

For the class of 2017:
z_{2017} = \frac{(70 - 80.2)}{15.06} \approx -0.68

Using the standard normal distribution table, we find the percentage of students above these z-scores.

- For the class of 1992: Approximately 59.12% of students would qualify.
- For the class of 2017: Approximately 75.08% of students would qualify.

Therefore, a higher percentage of students in the 2017 distribution would qualify for the internship program by scoring 70% or higher.

\textbf{Answer:} The z-score for a test score of 85 is farther above the mean in the 1992 distribution, with a z-score of approximately 1.42. In the 2017 distribution, the z-score for 85 is approximately 0.32.