Question

The average score of all students who took a particular statistics class last semester has a mean of 70 and a standard deviation of 3.0. Suppose 36 students who are taking the class this semester are selected at random. Find the probability that the average score of the 36 students exceeds 71.

Answer 1

2.28% probability that the average score of the 36 students exceeds 71.

Explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 70, \sigma = 3, n = 36, s = (3)/(√(36)) = 0.5`

Find the probability that the average score of the 36 students exceeds 71.

This is 1 subtracted by the pvalue of Z when X = 71. So

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

By the Central Limit Theorem

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

`Z = (71 - 70)/(0.5)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that the average score of the 36 students exceeds 71.