The amount of money collected by a snack bar at a large university has been recorded daily for the past five years. Records indicate that the mean daily amount collected is $3650 and the standard deviation is $600. The distribution is skewed to the right due to several high volume days (including football game days). Suppose that 100 days were randomly selected from the five years and the average amount collected from those days was recorded. Describe the sampling distribution of the sample mean.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Solution to the problem

Let X the random variable that represent " the amount of money collected by a snack bar at a large university "

No matter if the distribution for the random variable X is skewed to the right. If we assume that the sample size is large (n ≥ 30) and on this case makes sense since they record info from the past 5 years, the central limit theorem guarantees that [tex]\bar x[/tex] is approximately normal.

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

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

And for this case the distribution would be:

[tex]\bar X \sim N(\mu=3650, \frac{\sigma}{\sqrt{n}}=\frac{600}{\sqrt{100}}=60)[/tex]