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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 $8.50 and a standard deviation of $3. Complete parts (a) and (b). ….. 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. The mean is 8.50. (Round to the nearest hundredth as needed.) The standard deviation is 0.30. (Round to the nearest hundredth as needed.) b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than $9.10. (Hint: Apply the central limit theorem to the sampling distribution in (a).) (Round to the nearest thousandth as needed.)

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 $8.50 and a standard deviation of $3. Complete parts (a) and (b). ….. 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. The mean is 8.50. (Round to the nearest hundredth as needed.) The standard deviation is 0.30. (Round to the nearest hundredth as needed.) b. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than $9.10. (Hint: Apply the central limit theorem to the sampling distribution in (a).) (Round to the nearest thousandth as needed.)

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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**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 $8.50 and a standard deviation of $3. **Tasks:** **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. - **The mean is** 8.50. (Round to the nearest hundredth as needed.) - **The standard deviation is** 0.30. (Round to the nearest hundredth as needed.) **b.** **Find the probability** that the restaurant makes a profit that day, with the sample mean expense being less than $9.10. **(Hint: Apply the central limit theorem to the sampling distribution in (a).)** - (Round to the nearest thousandth as needed.)
Transcribed Image Text:**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 $8.50 and a standard deviation of $3. **Tasks:** **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. - **The mean is** 8.50. (Round to the nearest hundredth as needed.) - **The standard deviation is** 0.30. (Round to the nearest hundredth as needed.) **b.** **Find the probability** that the restaurant makes a profit that day, with the sample mean expense being less than $9.10. **(Hint: Apply the central limit theorem to the sampling distribution in (a).)** - (Round to the nearest thousandth as needed.)
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The mean is 8.50 and standard deviation of the sampling distribution is 0.30.

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