The average amount of money spent for lunch per person in the college cafeteria is $5.82 and the standard deviation is $2.5. Suppose that 46 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal placeswhere possible. a. What is the distribution of X? X N 5.82 b. What is the distribution of ? N 5.82 ✓.2.5 0.3686 O 0 c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314. 0.0514 d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314. 0.2315 X e. For part d), is the assumption that the distribution is normal necessary? No Yes o
The average amount of money spent for lunch per person in the college cafeteria is $5.82 and the standard deviation is $2.5. Suppose that 46 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal placeswhere possible. a. What is the distribution of X? X N 5.82 b. What is the distribution of ? N 5.82 ✓.2.5 0.3686 O 0 c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314. 0.0514 d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314. 0.2315 X e. For part d), is the assumption that the distribution is normal necessary? No Yes o
MATLAB: An Introduction with Applications
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Author:Amos Gilat
Publisher:Amos Gilat
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Help!!! With c and d
![### Distribution and Probability of Lunch Costs in a College Cafeteria
#### Problem Statement:
The average amount of money spent for lunch per person in the college cafeteria is $5.82, and the standard deviation is $2.5. Suppose that 46 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.
### Questions and Answers:
**a. What is the distribution of \(X\)?**
\[
X \sim N(5.82, 2.5^2)
\]
**b. What is the distribution of \(\bar{x}\)?**
\[
\bar{x} \sim N \left(5.82, \frac{2.5}{\sqrt{46}}\right) \sim N(5.82, 0.3686^2)
\]
**c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314.**
\[
P(5.2871 < X < 5.6314) = 0.0514
\]
**d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314.**
\[
P(5.2871 < \bar{x} < 5.6314) = 0.2315
\]
*Note: The provided answer in the image was incorrect as indicated by the red cross.*
**e. For part d, is the assumption that the distribution is normal necessary?**
\[
\text{Answer: Yes}
\]
**Hint:**
Click on "Hint" for additional hints and clarifications.
### Explanation of the Distributions and Probabilities:
- \(X\) represents the cost of lunch for a single patron, which follows a normal distribution with a mean (\(\mu\)) of $5.82 and a standard deviation (\(\sigma\)) of $2.5.
- \(\bar{x}\) is the sample mean cost of lunch for a group of 46 patrons. Its distribution is also normal but with a standard error of \(\frac{\sigma}{\sqrt{n}}\), where \(n = 46\).
By using the standard normal distribution and the properties of means and variances, these probabilities and distributions can be](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc3b72c7-eb99-406f-83fe-82fc41b2a8b5%2Fa19df9f7-892e-4056-977e-dcc292783b19%2Fltdqn8s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Distribution and Probability of Lunch Costs in a College Cafeteria
#### Problem Statement:
The average amount of money spent for lunch per person in the college cafeteria is $5.82, and the standard deviation is $2.5. Suppose that 46 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.
### Questions and Answers:
**a. What is the distribution of \(X\)?**
\[
X \sim N(5.82, 2.5^2)
\]
**b. What is the distribution of \(\bar{x}\)?**
\[
\bar{x} \sim N \left(5.82, \frac{2.5}{\sqrt{46}}\right) \sim N(5.82, 0.3686^2)
\]
**c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314.**
\[
P(5.2871 < X < 5.6314) = 0.0514
\]
**d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314.**
\[
P(5.2871 < \bar{x} < 5.6314) = 0.2315
\]
*Note: The provided answer in the image was incorrect as indicated by the red cross.*
**e. For part d, is the assumption that the distribution is normal necessary?**
\[
\text{Answer: Yes}
\]
**Hint:**
Click on "Hint" for additional hints and clarifications.
### Explanation of the Distributions and Probabilities:
- \(X\) represents the cost of lunch for a single patron, which follows a normal distribution with a mean (\(\mu\)) of $5.82 and a standard deviation (\(\sigma\)) of $2.5.
- \(\bar{x}\) is the sample mean cost of lunch for a group of 46 patrons. Its distribution is also normal but with a standard error of \(\frac{\sigma}{\sqrt{n}}\), where \(n = 46\).
By using the standard normal distribution and the properties of means and variances, these probabilities and distributions can be
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