Respuesta :
Answer:
94.52% probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
Normal probability distribution
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]
The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 3.5, \sigma = 0.5, n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05[/tex]
What is the probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42?
This is 1 subtracted by the pvalue of Z when X = 3.42.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.42 - 3.5}{0.05}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a pvalue of 0.0548.
So there is a 1-0.0548 = 0.9452 = 94.52% probability that the random sample of 100 nontraditional students has a mean GPA greater than 3.42.