The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of students?
A) 81.48
B) 90.00
C) 83.72
D) 78.96

Question

Quizplus · Accepted Answer

To find the score that separates the top 5% from the lower 95%, we need to find the z-score that corresponds to the 95th percentile, which is approximately 1.645. Using the formula: score = mean + (z-score * standard deviation), we get 70 + (1.645 * 7) = 81.48.

The Average Score of 100 Students Taking a Statistics Final