The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontraditional students are defined as 25 years old 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
σ
= 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 has a mean GPA greater than 3.42?

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.

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