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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. (Hint: Apply the central limit theorem to the sampling distribution in part a.)

Step by step solution

01

Identify Known Values

From the problem statement, we know the following: the charge per customer is \( \\(8.95 \), the mean expense per customer is \( \\)8.20 \), and the standard deviation of the expense is \( \$3.00 \). The sample size is \( n = 100 \) customers.

02

Determine the Mean of the Sample Distribution

The mean of the sampling distribution of the sample mean expense per customer is equal to the population mean expense per customer. Thus, \( \mu_{\bar{x}} = 8.20 \).

03

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation for the sampling distribution, known as the standard error, is calculated using the formula \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. So, \( \sigma_{\bar{x}} = \frac{3}{\sqrt{100}} = 0.30 \).

04

Assess Probability Using the Central Limit Theorem

According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normally distributed. To find the probability that the sample mean is less than \( \$8.95 \), compute the \( z \)-score using \( z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \), where \( \bar{x} = 8.95 \). Therefore, \( z = \frac{8.95 - 8.20}{0.30} = 2.50 \).

05

Determine the Probability from the Z-Table

Using a standard normal distribution table, find the probability that corresponds to \( z = 2.50 \). This probability is approximately 0.9938. Consequently, the probability that the mean expense per customer is less than \$8.95 is 0.9938.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Sampling Distribution

Sampling distribution refers to the probability distribution of a statistic, like the mean, that is obtained through repeated sampling from a larger population. Imagine if we took many random samples of 100 customers from Jan's restaurant and calculated the average expense per customer for each sample. The sampling distribution consists of these average expenses from each sample.

• As the sample size increases, the sampling distribution tends to become more normal (bell-shaped) in form due to the Central Limit Theorem.
• The mean of the sampling distribution (ex) is equal to the population mean (pop).
• The spread or variability of the sampling distribution is measured by its standard deviation, known as the standard error.

Understanding this concept helps us analyze how sample statistics relate to population parameters, which is essential in inferential statistics.

Decoding Standard Error

Standard error is a vital concept in statistics, especially when dealing with sample data. It provides an estimate of the standard deviation of the sampling distribution of a statistic. In simpler terms, standard error tells us how much the sample mean might vary from the true population mean.

• It's calculated using the formula: \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
• A smaller standard error indicates the estimate of the population mean is more accurate and reliable.

In our case with Jan's restaurant, the calculation gives a standard error of 0.30, which means there's relatively low variability in the sample means we might expect to calculate from different samples of 100 customers.

The Role of Z-score

A z-score is a numerical measurement that describes a value's relation to the mean of a group of values. When dealing with sample means, the z-score helps us understand how far, and in what direction, a sample mean is from the population mean, measured in standard errors.

• The z-score formula is: \( z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \), which shows the standardized distance of the sample mean from the population mean.
• Z-scores are crucial in probability assessments and hypothesis testing as they allow comparison across different datasets.

In the exercise, the z-score helps in determining the likelihood that the average expense remains below the charge of $8.95 per customer, contributing to evaluating profitability.

Exploring the Normal Distribution

The normal distribution, often expressed as a bell curve, is a key concept in statistics representing how data points are likely to accumulate around the mean. Many natural phenomenon, including earnings and expenses related to customer behavior at a restaurant, tend to align with this type of distribution when sample sizes are large enough.

• It is symmetrical and has most data points clustering around a central peak, representing the mean.
• The Central Limit Theorem aids in ensuring that the sampling distribution of the sample mean approximates normality, especially with larger samples.

In context, even though the expense distribution per customer at Jan's restaurant is originally skewed, the sampling distribution of the mean expenses among a sample of 100 will approximate a normal distribution. This allows us to confidently apply the z-score and probability calculations.