Scores on a test have a mean of 70 and q3 is 83. the scores have a distribution that is approximately normal. find p90. (you will need to first find the standard deviation.) a. 94.84 b. 85.08 c. 85.31 d. 83.7

Q3 represents 75%, meaning a z of ~0.67 80 - 70 is 10, so the standard deviations is ~14.9. 10 / 0.67 = 14.9 now find the z that represents a score of 90 90 - 70 is 20 20 / 14.9 = 1.34 from a z-table, a z of 1.34 represents a probability of ~90.99% meaning that there is about a 9.01% chance of getting a 90 or better.

The answer is part a: use