Restaurant profit? Jan's All You Can Eat Restaurant charges $8.95 per customer to eat at the restaurant. Restaurant management finds that its expense per customer, based on how much the customer eats and the expense of labor, has a distribution that is skewed to the right with a mean of \$8.20 and a standard deviation of \$3

a. If the 100 customers on a particular day have the characteristics of a random sample from their customer base, find the mean and standard deviation of the sampling distribution of the restaurant's sample mean expense per customer.

b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than $8.95.

#### Solution

a) Using the central limit theorem:

The sampling distribution of $\overline{X}$ is $N(\mu,\sigma^2/n)$.

The mean of the sampling distribution of the restaurant's sample mean expense per customer is

$E(\overline{X}) = \mu = 8.20$.

The standard deviation of the sampling distribution of the restaurant's sample mean expense per customer is

Standard Error $= \sigma^2/\sqrt{n} = 3/\sqrt{100} = 0.30$.

b) The probability that the restaurant makes a profit that day, with the sample mean expense being less than $8.95 is

` $$ \begin{aligned} P(\overline{X} < 8.95)&=P\big(\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}<\frac{8.95-8.20}{0.30}\big)\\ &= P(Z < 2.5)\\ & = 0.9938 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators