Question

Historically, the average score of PGA golfers for one round is 73.9 with a standard deviation of 1.1. A random sample of 99 golfers is taken. What is the probability that the sample mean is between 73.9 and 74.03

Answer 1

38.10% probability that the sample mean is between 73.9 and 74.03.

Explanation:

To solve this question, it is important to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 73.9, \sigma = 1.1, n = 99, s = (1.1)/(√(99)) = 0.1105`

What is the probability that the sample mean is between 73.9 and 74.03

This is the pvalue of Z when X = 74.03 subtracted by the pvalue of Z when X = 73.9. So

X = 74.03

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

By the Central Limit Theorem

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

`Z = (74.03 - 73.9)/(0.1105)`

`Z = 1.18`

`Z = 1.18` has a pvalue of 0.8810.

X = 73.9

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

`Z = (73.9 - 73.9)/(0.1105)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

0.8810 - 0.5 = 0.3810

38.10% probability that the sample mean is between 73.9 and 74.03.