Question

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 3.5. Suppose 49 golfers played the course today. Find the probability that the average score of the 49 golfers exceeded 76.

A) 0.1293
B) 0.4772
C) 0.3707
D) 0.0228

Answer 1

`P(\bar X >76)=P(Z>(76-75)/((3.5)/(√(49)))=2)`

And using the complement rule and a calculator, excel or the normal standard table we have that:

`P(Z>2)=1-P(Z<2)= 1-0.97725=0.0228`

And the correct answer would be:

D) 0.0228

Explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Solution to the problem

Let X the random variable who represent the score of all golfers. For this case we have the following info:

`\mu = 75, \sigma= 3.5`

We select a sample size of n =49>30. From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the probability of interest like this:

`P(\bar X >76)=P(Z>(76-75)/((3.5)/(√(49)))=2)`

And using the complement rule and a calculator, excel or the normal standard table we have that:

`P(Z>2)=1-P(Z<2)= 1-0.97725=0.0228`

And the correct answer would be:

D) 0.0228

Answer 2

D) 0.0228

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))`.

In this problem, we have that:

`\mu = 75, \sigma = 3.5, n = 49, s = (3.5)/(√(49)) = 0.5`

Find the probability that the average score of the 49 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 76. So

`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