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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 73.9, \sigma = 1.1, n = 99, s = \frac{1.1}{\sqrt{99}} = 0.1105$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{74.03 - 73.9}{0.1105}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 73.9

$Z = \frac{X - \mu}{s}$

$Z = \frac{73.9 - 73.9}{0.1105}$

$Z = 0$

$Z = 0$ 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.