Question

The scores on a physics exam are normally distributed with a mean of 81 and a standard deviation of 1.5. What is the minimum score a student would need to earn to be in the top 17 percent of scores? Enter your answer with accuracy to two decimal places.

Answer 1

To be in the top 17% of the scores on the physics exam, a student would need to score at least 82.43, using the z-score for the 83rd percentile which is approximately 0.95.

Step-by-step explanation:

To find the minimum score a student would need to earn to be in the top 17 percent of scores in a normally distributed set with a mean of 81 and a standard deviation of 1.5, we need to find the z-score that correlates with the top 17% of the distribution.

Consulting a standard normal distribution (z-score) table or using a statistical calculator, we find that the z-score corresponding to the top 17% (or the 83rd percentile, since 100% - 17% = 83%) is approximately 0.95.

Using the z-score formula:

1. Z = (X - mean) / standard deviation
2. X = Z * standard deviation + mean
3. X = 0.95 * 1.5 + 81
4. X = 1.425 + 81
5. X = 82.43

Therefore, a student would need to score at least 82.43 to be in the top 17 percent of scores on this physics exam.