Question

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.09 and a standard deviation of $0.08. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. What is the distribution to use for the average cost of gasoline for the 16 gas stations? X ~ N 4.09, 16 0.08 X ~ N 4.09, 0.08 16 X ~ N(4.09, 0.08) X ~ N 4.09, 16 0.08 Additional Materials

Answer 1

Answer:

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

$\mu_{\bar X} = 4.09$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{0.08}{\sqrt{16}}= 0.02$

So the best answer for this case would be:

$\bar X \sim N (4.09, \frac{0.08}{\sqrt{16}}= 0.02)$

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 costs of unleaded gasoline of a population, and for this case we know the distribution for X is given by:

$X \sim N(4.09,0.08)$  

Where $\mu=4.09$ and $\sigma=0.08$

We select a sample size of n =16. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

$\mu_{\bar X} = 4.09$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}= \frac{0.08}{\sqrt{16}}= 0.02$

So the best answer for this case would be:

$\bar X \sim N (4.09, \frac{0.08}{\sqrt{16}}= 0.02)$