Answer:
0.1% probability that the average score of a random sample of 20 students exceeds 59.5.
Step-by-step explanation:
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 = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12[/tex]
What is the probability that the average score of a random sample of 20 students exceeds 59.5?
This is 1 subtracted by the pvalue of Z when [tex]X = 59.5[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{59.5 - 56}{1.12}[/tex]
[tex]Z = 3.1[/tex]
[tex]Z = 3.1[/tex] has a pvalue of 0.9990.
So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.
