If the scores on an aptitude test are normally distributed with mean 500 and standard deviation 100,what proportion of the test scores are less than 585?
A).1977
B).8500
C).1500
D).8023

Question

Quizplus · Accepted Answer

To find the proportion of scores less than 585, we calculate the z-score: z = (585 - 500) / 100 = 0.85. Using the standard normal distribution table, the proportion corresponding to a z-score of 0.85 is approximately 0.8023.

If the Scores on an Aptitude Test Are Normally Distributed