Question

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 and a standard deviation of 5. The average exam score in your lab was 59.5. The 20 students in your lab sections will be considered a random sample of all students who take this class. What is the probability that the average score of a random sample of 20 students exceeds 59.5?

Answer 1

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`(\sigma)/(√(n))`

Normal probability distribution

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 = (X - \mu)/(\sigma)`

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:

`\mu = 56, \sigma = 5, n = 20, s = (5)/(√(20)) = 1.12`

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
`X = 59.5`. So

`Z = (X - \mu)/(\sigma)`

`Z = (59.5 - 56)/(1.12)`

`Z = 3.1`

`Z = 3.1` 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.