Answer:
There is a 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
For this problem, we have that:
Based on the study results, we can assume the population mean and standard deviation for the GPA of nontraditional students is
and
.
We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of
in the z score formula.
.
What is the probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65?
This is 1 subtracted by the pvalue of Z when
. So
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{3.65 - 3.50}{0.05}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3.65%20-%203.50%7D%7B0.05%7D)
![Z = 3](https://tex.z-dn.net/?f=Z%20%3D%203)
A zscore of 3 has a pvalue of 0.9987.
So, there is a 1-0.9987 = 0.0013 = 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.