Question

The average ticket price for a Spring Training baseball game is $32.61, with a standard deviation of $8.62. In a random sample of 40 Spring Training tickets, find the probability that the mean ticket price exceeds $35. (Round your answer to three decimal places)

Answer 1

Using the normal probability distribution and the central limit theorem, it is found that there is a 0.04 = 4% probability that the mean ticket price exceeds $35.

------------------------------------------

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

------------------------------------------

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \frac{\sigma}{\sqrt{n}}$.

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

------------------------------------------

The average ticket price for a Spring Training baseball game is $32.61, with a standard deviation of $8.62.

This means that $\mu = 32.61, \sigma = 8.62$

------------------------------------------

Sample of 40:

This means that $n = 40, s = \frac{8.62}{\sqrt{40}}$

------------------------------------------

In a random sample of 40 Spring Training tickets, find the probability that the mean ticket price exceeds $35.

This is 1 subtracted by the p-value of Z when X = 35, so:

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

By the Central Limit Theorem

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

$Z = \frac{35 - 32.61}{\frac{8.62}{\sqrt{40}}}$

$Z = 1.754$

$Z = 1.754$ has a p-value of 0.96.

1 - 0.96 = 0.04

0.04 = 4% probability that the mean ticket price exceeds $35.

A similar question is given at brainly.com/question/24342706