Suppose that cars arrive at the Burger King’s drive-through at the rate of 20 cars every hour between noon and 1:00 PM. A random sample of 40 one-hour time periods between noon and 1:00 PM is selected and has a mean of 22.1 cars arriving. Assume that µ = 20 and that σ = √20 for the population of cars arriving at Burger King. What is the probability that a simple random sample of 40 one-hour time periods results in a mean of at least 22.1 cars?

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 variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(20,\sqrt{20})[/tex]

Where [tex]\mu=20[/tex] and [tex]\sigma=\sqrt{20}=4.472[/tex]

Since the distribution for X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

We want to find this probability:

[tex]P(\bar X > 22.1)[/tex]

And we can use the z score given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And replacing we got

We can find the individual probabilities like this:

[tex]P(\bar X >22.1)=P(Z>\frac{22.1-20}{\frac{4.472}{\sqrt{40}}}=2.970)[/tex]

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

[tex]P(Z>2.970)=1-P(Z<2.970) = 1-0.9985=0.0015[/tex]