Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 3.5[/tex] and [tex]\sigma = 0.5[/tex].
We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of [tex]\sigma[/tex] in the z score formula.
[tex]s = \frac{\sigma}{\sqrt{100}} = \frac{0.5}{10} = 0.05[/tex].
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 [tex]X = 3.65[/tex]. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.65 - 3.50}{0.05}[/tex]
[tex]Z = 3[/tex]
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.
