Question

Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.6 and the standard deviation was 8.7. If 10 classes were selected at random, find the probability that the class average is greater than 90.
A. 0.0100



B. 0.5247



C. 0.1023



D. 0.0002

Answer 1

Correct Answer:
Option A. 0.01

Solution:
This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.

Converting 90 to z-score:

$z= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10}} }=2.33$

Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.

Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.