Chapter 2.1, Problem 17E

(a)

To determine

To write:a sentence comparing S’s percentile among the national group

Answer to Problem 17E

S did better among all test takers than he did among the 50 boys at his school.

Explanation of Solution

Given:

1.5 million college-bound high school juniors take the test.

Mean score = 46.9

Standard deviation = 10.9

5.2% of test takers earned a score of 65.

Since 5.2% of test takers earned a score of 65 or higher on the Critical Reading test's 20 to 80 scale. In the national group, about $100-5.2=94.8\%$ of the test takers scored below 65. S's percentiles, 94.8th among the national group and 68th within the school, indicate that he did better among all test takers than he did among the 50 boys at his school.

(b)

To determine

To calculate: S’s z-score

Answer to Problem 17E

The standardized score for S among the national group is greater than the standardized score among the 50 boys at his school.

Explanation of Solution

Calculation:

The standardized score for S among the national group is obtained below.

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{64-46.9}{10.9}\\ \approx 1.57\end{array}$

The standardized score for S among the 50 boys at his school is obtained below.

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{64-58.2}{9.4}\\ \approx 0.62\end{array}$

The standardized score for S among the national group is greater than the standardized score among the 50 boys at his school. This indicates that he did better among all test takers than he did among the 50 boys at his school.

Conclusion:

Therefore, the standardized score for S among the national group is greater than the standardized score among the 50 boys at his school.