Question

On a standardized aptitude test, scores are normally distributed with a mean

of 72 and a standard deviation of 4. What is the % of scores > 76?
A) 16%
B) 21.5%
C) 55%
D) 29.5%

Answer 1

To calculate the percentage of scores greater than 76, we use the z-score, which is (76 - 72) / 4 = 1. A z-score of 1 corresponds to the 84th percentile, thus 16% of scores are above 76. The answer is A) 16%.

Step-by-step explanation:

To find the percentage of scores greater than 76 when the scores are normally distributed with a mean of 72 and a standard deviation of 4, we need to calculate the z-score for 76 and then find the corresponding percentile using the standard normal distribution table or a z-score calculator.

The z-score is calculated as follows:

Z = (X - μ) / σ

Where:

• X is the score of interest (76 in this case)
• μ (mu) is the mean of the distribution (72)
• σ (sigma) is the standard deviation of the distribution (4)

Substituting the given values:

Z = (76 - 72) / 4 = 1

A z-score of 1 corresponds to the 84th percentile of the standard normal distribution, which means that 84% of the scores are below 76. Therefore, the percentage of scores greater than 76 is 100% - 84% = 16%.

The correct answer to the question is A) 16%.