Question

Jan's All You Can Eat Restaurant charges $9.40 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 $8.40 and a standard deviation of $4. 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 8.40. (Round to the nearest hundredth as needed.) The standard deviation is 0.4. (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 $9.40. (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 $9.40 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 $8.40 and a standard deviation of $4. 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 8.40. (Round to the nearest hundredth as needed.)
The standard deviation is 0.4. (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 $9.40. (Hint: Apply the central limit theorem to the sampling distribution in (a).)
(Round to the nearest thousandth as needed.)

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Jan's All You Can Eat Restaurant charges 9.40 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 of8.40 and a standard deviation of 4. 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 8.40. (Round to the nearest hundredth as needed.)
The standard deviation is 0.4. (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 than9.40. (Hint: Apply the central limit theorem to the sampling distribution in (a).)
(Round to the nearest thousandth as needed.)

Added by Jonathan F.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert David Nguyen

06/30/2022

Step 1

40. - Since the mean expense is $59.40 and the standard deviation is 34, we need to find the z-score for $59.40. - z = ($59.40 - $58.40) / 34 = 0.0294 - Using the z-table, the probability of a customer spending more than $59.40 is 0.5122. Show more…

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Jan's All You Can Eat Restaurant charges $59.40 per customer. The restaurant manager finds that expenses that exceed the average of $58.40 and standard deviation of 34 are considered abnormal. (a) Calculate the probability that a customer spends more than $59.40. (b) Calculate the probability that a customer spends less than $59.40. If the 100 customers are a random sample from their customer base and have a mean and standard deviation of the sampling distribution, find the mean expense per sample. Round the nearest hundredth. The standard deviation needed is rounded to the nearest hundredth. Find the probability that the restaurant makes a profit that day, with the sample mean expense being less than $59.40. (Hint: Apply the central limit theorem to the sampling distribution.) (Round to the nearest thousandth needed.)

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Transcript

00:01 Hi, i'm david and i'm here to have you answering your question.

00:03 In the question here, we are going to discuss about the assembly distribution of the sample mean.

00:08 By the central limit theorem, it says that if the n -quieter equal to the 30, then the sample mean x -par will be approximately to the normal, in such a way that the mean of the x -bar equal to the mean in the population, standard division x -bar equal to the sigma of the square of n.

00:27 And because it follows by the normal, it will turn the x -bar xxxx we minus the mean over standard deviation.

00:33 We obtain the standard normal.

00:35 In the question here, we're given the x that followed by the normal.

00:39 With the mean, it will equal to the 8 .4, and the standard division equal to the 4.

00:49 He will given the n equal to 100, so it's greater than 30 already.

00:53 In the question a, ask not you find the x bar with the mean.

00:57 So the mean by the formula just equal to the mule and equal to the 8 .4, standard division x par by the formula here, equal to the sigma over square of n, equal to the 4 over square root of the hundred, equal to the 4 over 10, and equal to the 0 .4.

01:16 And i see you got the answer correctly, so good job for your work.

01:20 Now for the question b, want to find the probability that the sample mean x par will be less than the 9 .40.

01:29 Now to find this probability, i need to convert the xxpang to the z.

01:33 To do it, i need to apply this formula here.

01:36 So i will turn the 9 .40 and minus the mean 8 .4 divided by the standard division 0 .4...

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