Question

Each year, about $1.5$million college-bound high school juniors take the PSAT. In a recent year, the mean score on the Critical Reading test was $46.9$and the standard deviation was$10.9$. Nationally, $5.2\%$of test takers earned a score of 65 or higher on the Critical Reading test’s $20$to $80$scale.9

PSAT scores Scott was one of $50$junior boys to take the PSAT at his school. He scored $64$on the Critical Reading test. This placed Scott at the 68th percentile within the group of boys. Looking at all $50$boys’ Critical Reading scores, the mean was $58.2$and the standard deviation was $9.4$

(a) Write a sentence or two comparing Scott’s percentile among the national group of test-takers and among the $50$boys at his school.

(b) Calculate and compare Scott’s z-score among these same two groups of test-takers.

Short answer

Part (a) S did better among all test takers than he did among the $50$ boys at his school.

Part (b) The standardized score for S among the national group is greater than the standardized score among the $50$ boys at his school.

Step-by-step solution

Part (a) Step 1. Given

Ms. Martin's statistics quiz had a mean score of $12$ points.

On Ms. Martin's statistics quiz, the standard deviation was $3$

Transform the results to a mean of $75$

Transform the results so that the standard deviation is equal to $12$

Part (a) Step 2. Concept

The formula used: $z=\frac{x-\mu }{\sigma }$

Part (a) Step 3. Calculation

The test is taken by $1.5$ million college-bound high school juniors.

$46.9\%$ is the average score.

$10.9$ standard deviation

A score of $65$ was achieved by $5.2$ percent of test-takers.

Because$5.2\%$ of test-takers received a $65$ or higher on the Critical Reading test's $20$ to $80$ range. About $100-5.2=94.8\%$ of test-takers in the national group scored below $65$, indicating that S's percentiles, $94.8^{th}$ in the national group and ${68}^{th}$ inside the school, suggest that he performed better among all test takers than among the $50$ males at his school.

Part (b) Step 1. Calculation

The national group's standardized score for S can be found below.

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

The standardized score for S among his school's 50 lads is shown below.

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

S's standardized score in the national group is higher than S's standardized score among his school's $50$lads. This suggests that he performed better across the board than among the $50$boys at his school. Therefore, the standardized score for S among the national group is greater than the standardized score among the $50$boys at his school.