Question

The cost of unleaded gasoline in the Bay Area once followed a normal distribution with a mean of $4.74 and a standard deviation of $0.16. Sixteen gas stations from the Bay area are randomly chosen. We are interested in the average cost of gasoline for the 15 gas stations. What is the approximate probability that the average price for 15 gas stations is over $4.99?

Answer 1

Answer:

Approximately 0% probability that the average price for 15 gas stations is over $4.99.

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:

$\mu = 4.74, \sigma = 0.16, n = 16, s = \frac{0.16}{\sqrt{16}} = 0.04$

What is the approximate probability that the average price for 15 gas stations is over $4.99?

This is 1 subtracted by the pvalue of Z when X = 4.99. So

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

By the Central Limit Theorem

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

$Z = \frac{4.99 - 4.74}{0.04}$

$Z = 6.25$

$Z = 6.25$ has a pvalue very close to 1.

1 - 1 = 0

Approximately 0% probability that the average price for 15 gas stations is over $4.99.