Question

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4. Suppose 64 golfers played the course today. Find the probability that the average score of the 64 golfers exceeded 76. Round to four decimal places.

Answer 1

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

Explanation:

To solve this question, we need 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 normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 75, \sigma = 4, n = 64, s = (4)/(√(64)) = 0.5`

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

`Z = (76 - 75)/(0.5)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.