Jan's All You Can Eat Restaurant charges $9.10 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 $8.10 and a standard deviation of $4. 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.

Step-by-step explanation:

From the given question;

Given that:

Jan's All You Can Eat Restaurant charges $9.10 per customer to eat at the restaurant.

Distribution  is skewed and and has a mean of $8.10 and a standard deviation of $4.

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.

the mean by using the central limit theorem is 8.10

the standard error of the sampling distribution  =

the standard error of the sampling distribution =

= 4/10

= 0.4

B.  

P(X > $8.95) = P (Z > 8.95 - 8.10/0.4)

P(X > $8.95) = P (Z > 2.1)

P(X > $8.95) = 1 - P (Z < 2.1)

P(X > $8.95) = 1 - 0.9821

P(X > $8.95) = 0.0179

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%