Question

The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontraditional students are defined as 25 years or older and working. Based on the study results, we can assume the population mean and standard deviation for the GPA of nontraditional students is µ=3.5 and s=0.5. Suppose a random sample of 100 nontraditional students is selected and each student's GPA is calculated. What is the probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 3.5$ and $\sigma = 0.5$.

We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of $\sigma$ in the z score formula.

$s = \frac{\sigma}{\sqrt{100}} = \frac{0.5}{10} = 0.05$.

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 $X = 3.65$. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.65 - 3.50}{0.05}$

$Z = 3$

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.