Question

A PGA (Professional Golf Association) tournament organizer is attempting to determine whether hole (pin) placement has a significant impact on the average number of strokes for the 13th hole on a given golf course. Historically, the pin has been placed in the front right corner of the green, and the historical mean number of strokes for the hole has been 4.25, with a standard deviation of 1.6 strokes. On a particular day during the most recent golf tournament, the organizer placed the hole (pin) in the back left corner of the green. 64 golfers played the hole with the new placement on that day. Determine the probability of the sample average number of strokes exceeding 4.75

Answer 1

Answer:

$P(\bar X >4.75)=P(Z>\frac{4.75-4.25}{\frac{1.6}{\sqrt{64}}}=2.5)$

Using the complement rule and the normal standard table or excel we have this:

$P(Z>2.5)=1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we assume the distribution for X is given by:

$X \sim N(4.25,1.6)$  

Where $\mu=4.25$ and $\sigma=1.6$

And we select a sample of n =64. Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find the following probability:

$P(\bar X >4.75)$

And we can use the z score given by this formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

$P(\bar X >4.75)=P(Z>\frac{4.75-4.25}{\frac{1.6}{\sqrt{64}}}=2.5)$

Using the complement rule and the normal standard table or excel we have this:

$P(Z>2.5)=1-P(Z$