Question

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 day have the characteristics of the random sample from their customer base, find the mean and standard error 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 (Hint: Apply the central limit the sampling distribution in (a).)

Answer 1

a. Mean = $8.20, Standard Error = 0.3

b. 0.9938

Explanation:

Given

Mean = $8.20

Standard Deviation = 3

Charge = $8.95

a.

Given

n = number of customers = 100

The condition states that the 100 customers on a day have the characteristics of the random sample from their customer base.

This means that the mean remains unchanged

Mean = $8.2

And standard error is calculated as:

S.D/√n

Where S.D = Standard Deviation = 3

n =100

Standard Error = 3/√100

Standard Error = 3/10

Standard Error = 0.3

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

Given

X = $8.95

First, we'll Calculate the z value

z = (x - mean)/standard error

z = (8.95 - 8.2)/0.3

z = 0.75/0.3

z = 2.5

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

P(z<2.5)

P(z<2.5) = 0.9938 ----- z table