Question

A study of college football games shows that the number of holding penalties assessed has a mean of 2.2 penalties per game and a standard deviation of 0.8 penalties per game. What is the probability that, for a sample of 40 college games to be played next week, the mean number of holding penalties will be 2 penalties per game or less

Answer 1

Answer:

5.71% probability that, for a sample of 40 college games to be played next week, the mean number of holding penalties will be 2 penalties per game or less

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 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.

In this question, we have that:

$\mu = 2.2, \sigma = 0.8, n = 40, s = \frac{0.8}{\sqrt{40}} = 0.1265$

What is the probability that, for a sample of 40 college games to be played next week, the mean number of holding penalties will be 2 penalties per game or less

This is the pvalue of Z when X = 2. So

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

By the Central Limit Theorem

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

$Z = \frac{2 - 2.2}{0.1265}$

$Z = -1.58$

$Z = -1.58$ has a pvalue of 0.0571

5.71% probability that, for a sample of 40 college games to be played next week, the mean number of holding penalties will be 2 penalties per game or less