The Average Amount Of Money Spent For Lunch Per Person In The College Cafeteria Is $$ 7.44$ And The Standard Deviation Is $$ 2.79$. Suppose That 11 Randomly Selected Lunch Patrons Are Observed. Assume The Distribution

Mar 10, 2025 by ADMIN 224 views

The Average Amount of Money Spent for Lunch per Person in the College Cafeteria

Introduction

When it comes to understanding the behavior of a population, one of the most effective ways to do so is by analyzing a sample of that population. In this case, we are interested in understanding the average amount of money spent for lunch per person in the college cafeteria. Given that the average amount of money spent for lunch per person in the college cafeteria is $7.44 and the standard deviation is $2.79, we can use this information to make predictions about the behavior of the population.

Understanding the Problem

To begin with, let's understand the problem at hand. We are given the average amount of money spent for lunch per person in the college cafeteria, which is $7.44. This is a measure of the central tendency of the population, and it gives us an idea of what the average person spends on lunch. However, we are also given the standard deviation, which is $2.79. This is a measure of the variability of the population, and it gives us an idea of how spread out the data is.

The Central Limit Theorem

One of the most important concepts in statistics is the Central Limit Theorem (CLT). The CLT states that the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is sufficiently large. In this case, we are given a sample size of 11, which is a relatively small sample size. However, we can still use the CLT to make predictions about the behavior of the population.

Calculating the Standard Error

To calculate the standard error, we need to divide the standard deviation by the square root of the sample size. In this case, the standard deviation is $2.79 and the sample size is 11. Therefore, the standard error is:

\frac{2.79}{\sqrt{11}} = 0.83

Calculating the Margin of Error

The margin of error is a measure of the maximum amount by which the sample mean can differ from the population mean. To calculate the margin of error, we need to multiply the standard error by the critical value from the standard normal distribution. In this case, we are 95% confident that the sample mean will be within 2 standard errors of the population mean. Therefore, the margin of error is:

2 \times 0.83 = 1.66

Calculating the Confidence Interval

The confidence interval is a range of values within which we are confident that the population mean lies. To calculate the confidence interval, we need to add and subtract the margin of error from the sample mean. In this case, the sample mean is $7.44 and the margin of error is 1.66. Therefore, the confidence interval is:

7.44 - 1.66 = 5.78

7.44 + 1.66 = 9.10

Conclusion

In conclusion, we have used the given information to make predictions about the behavior of the population. We have calculated the standard error, the margin of error, and the confidence interval, and we have used these values to make predictions about the population mean. While the sample size is relatively small, we can still use the Central Limit Theorem to make predictions about the behavior of the population.

References

[1] Moore, D. S., & McCabe, G. P. (2013). Introduction to the practice of statistics (6th ed.). New York: W.H. Freeman and Company.
[2] Larson, R. E., & Farber, B. (2018). Elementary statistics: Picturing the world with statistics (5th ed.). Boston: Pearson Education.

Discussion

The average amount of money spent for lunch per person in the college cafeteria is $7.44 and the standard deviation is $2.79. Suppose that 11 randomly selected lunch patrons are observed. Assume the distribution is normal. What is the probability that the sample mean will be within 1 standard error of the population mean?

Solution

To solve this problem, we need to calculate the z-score of the sample mean. The z-score is a measure of how many standard errors away from the population mean the sample mean is. In this case, we are 1 standard error away from the population mean. Therefore, the z-score is:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{7.44 - 7.44}{2.79 / \sqrt{11}} = 0

Since the z-score is 0, we know that the sample mean is exactly 1 standard error away from the population mean. Therefore, the probability that the sample mean will be within 1 standard error of the population mean is:

P(-1 < z < 1) = 0.6827

This means that there is a 68.27% chance that the sample mean will be within 1 standard error of the population mean.

Discussion

Solution

To solve this problem, we need to calculate the z-score of the sample mean. The z-score is a measure of how many standard errors away from the population mean the sample mean is. In this case, we are 2 standard errors away from the population mean. Therefore, the z-score is:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{7.44 - 7.44}{2.79 / \sqrt{11}} = 0

Since the z-score is 0, we know that the sample mean is exactly 2 standard errors away from the population mean. Therefore, the probability that the sample mean will be within 2 standard errors of the population mean is:

P(-2 < z < 2) = 0.9544

This means that there is a 95.44% chance that the sample mean will be within 2 standard errors of the population mean.

Discussion

Solution

To solve this problem, we need to calculate the z-score of the sample mean. The z-score is a measure of how many standard errors away from the population mean the sample mean is. In this case, we are 3 standard errors away from the population mean. Therefore, the z-score is:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{7.44 - 7.44}{2.79 / \sqrt{11}} = 0

Since the z-score is 0, we know that the sample mean is exactly 3 standard errors away from the population mean. Therefore, the probability that the sample mean will be within 3 standard errors of the population mean is:

P(-3 < z < 3) = 0.9974

This means that there is a 99.74% chance that the sample mean will be within 3 standard errors of the population mean.

Discussion

Solution

To solve this problem, we need to calculate the z-score of the sample mean. The z-score is a measure of how many standard errors away from the population mean the sample mean is. In this case, we are 4 standard errors away from the population mean. Therefore, the z-score is:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{7.44 - 7.44}{2.79 / \sqrt{11}} = 0

Since the z-score is 0, we know that the sample mean is exactly 4 standard errors away from the population mean. Therefore, the probability that the sample mean will be within 4 standard errors of the population mean is:

P(-4 < z < 4) = 0.9999

This means that there is a 99.99% chance that the sample mean will be within 4 standard errors of the population mean.

Discussion

Solution

To solve this problem, we need to calculate the z-score of the sample mean. The z-score is a measure of how many standard errors away from the population mean the sample mean is. In this case, we are 5 standard errors away from the population mean. Therefore, the z-score is:

z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{7.44 - 7.44<br/> # The Average Amount of Money Spent for Lunch per Person in the College Cafeteria: Q&A

Introduction

In our previous article, we discussed the average amount of money spent for lunch per person in the college cafeteria, which is $7.44 and the standard deviation is $2.79. We also used the Central Limit Theorem to make predictions about the behavior of the population. In this article, we will answer some of the most frequently asked questions about the average amount of money spent for lunch per person in the college cafeteria.

Q: What is the average amount of money spent for lunch per person in the college cafeteria?

A: The average amount of money spent for lunch per person in the college cafeteria is $7.44.

Q: What is the standard deviation of the average amount of money spent for lunch per person in the college cafeteria?

A: The standard deviation of the average amount of money spent for lunch per person in the college cafeteria is $2.79.

Q: What is the Central Limit Theorem?

A: The Central Limit Theorem states that the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is sufficiently large.

Q: How do we calculate the standard error?

A: To calculate the standard error, we need to divide the standard deviation by the square root of the sample size. In this case, the standard deviation is $2.79 and the sample size is 11. Therefore, the standard error is: