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 18 classes were selected at random, find the probability that the class average is less than 80.
A. 0.0235



B. 0.7427



C. 0.1730



D. 0.0396

Answer 1

Correct Ans:
Option D

Solution.
This is a problem of statistics and uses the concepts of normal distribution. First step is to convert the given score into z score. 80 converted to z score for given mean, sample size and standard deviation will be:

$z= \frac{80-83.6}{ \frac{8.7}{18} } =-1.756$

We are to find the probability that the class average is less than 80. Or in terms of z score, we have to find the score is less than -1.756. 

Using the standard normal distribution table this probability comes out to be 0.0396.

Therefore, the probability that the class average is less than 80 is 0.0396

Answer 2

The correct answer is:

D) 0.0396.

Explanation:

The formula for a z-score is:
$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$

where X is the score we're using, μ is the mean, σ is the standard deviation, and n is the sample size.

In our problem, X is 80, μ is 83.6, σ is 8.7, and n is 18:

$z=\frac{80-83.6}{\frac{8.7}{\sqrt{18}}}=\frac{-3.6}{\frac{8.7}{\sqrt{18}}}=-1.756\approx -1.76$

Using a z-table, we see that the area to the left of this (less than) is 0.0392.  This is closest to choice D, 0.0396.