Counselor Education Comprehensive Exam (CECE) Practice Exam

If the mean of an achievement test is 31 and the standard deviation is 3, what is the percentile rank of a student who scored 37 on that test?

1. a. 97.5%

2. b. 50%

3. c. 83.7%

4. d. 99.75%

The correct answer is: a. 97.5%

To determine the percentile rank of a student who scored 37 on an achievement test with a mean of 31 and a standard deviation of 3, we utilize the concept of the z-score. The z-score measures how many standard deviations an element is from the mean and is calculated using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] Where: - \( X \) is the value of interest (37 in this case), - \( \mu \) is the mean (31), - \( \sigma \) is the standard deviation (3). Substituting the values into the formula: \[ Z = \frac{(37 - 31)}{3} = \frac{6}{3} = 2 \] A z-score of 2 indicates that the student scored 2 standard deviations above the mean. To find the corresponding percentile rank for a z-score of 2, we can refer to the standard normal distribution table, which shows the area under the curve to the left of a given z-score. For a z-score of 2, the cumulative area is approximately 0.9772, which corresponds to 97.72%. This means that the student scored better