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. a. What is the distribution of X? X N 5.82 ✓,2.5 OF OF b. What is the distribution of ? N 5.82 ✓0.3686 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 X d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314. 0.2315 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 places where possible. a. What is the distribution of X? X N 5.82 ✓,2.5 OF OF b. What is the distribution of ? N 5.82 ✓0.3686 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 X d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314. 0.2315 e. For part d), is the assumption that the distribution is normal necessary? No Yes O
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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I need help with c, and d
![**Understanding the Normal Distribution for Lunch Costs**
The average amount of money spent for lunch per person in the college cafeteria is $5.82, with a standard deviation of $2.5. We want to understand the distribution of costs observed from 46 randomly selected lunch patrons. Assume the distribution of money spent is normal. Here are the steps to solve the related questions:
### a. What is the distribution of \( X \)?
Given:
- Mean (\( \mu \)) = 5.82
- Standard deviation (\( \sigma \)) = 2.5
The distribution notation is:
\[ X \sim N(5.82, 2.5) \boxed{\checkmark} \]
### b. What is the distribution of \( \bar{X} \)?
When observing a sample of 46 patrons, the sample distribution mean remains the same (\( \mu \) = 5.82). However, the standard deviation of the sample mean (\( \sigma_{\bar{X}} \)) changes. It is calculated as:
\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{46}} = 0.3686 \]
The distribution notation is:
\[ \bar{X} \sim N(5.82, 0.3686) \boxed{\checkmark} \]
### c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314.
Using the normal distribution \( X \sim N(5.82, 2.5) \), you can find this probability using cumulative distribution functions (CDF). The result is:
\[ P(5.2871 < X < 5.6314) = 0.0514 \boxed{\text{Wrong}}\]
### d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314.
Using \( \bar{X} \sim N(5.82, 0.3686) \), you calculate:
\[ P(5.2871 < \bar{X} < 5.6314) = ? \]
The given incorrect answer is:
\[ 0.2315 \boxed{\text{Wrong}} \]
### e. For part](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba867a91-7946-4dc8-8b92-57b0d80fa10e%2F593838ab-a678-4f28-8f3b-3c36a610d75a%2Fhzbmahm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Normal Distribution for Lunch Costs**
The average amount of money spent for lunch per person in the college cafeteria is $5.82, with a standard deviation of $2.5. We want to understand the distribution of costs observed from 46 randomly selected lunch patrons. Assume the distribution of money spent is normal. Here are the steps to solve the related questions:
### a. What is the distribution of \( X \)?
Given:
- Mean (\( \mu \)) = 5.82
- Standard deviation (\( \sigma \)) = 2.5
The distribution notation is:
\[ X \sim N(5.82, 2.5) \boxed{\checkmark} \]
### b. What is the distribution of \( \bar{X} \)?
When observing a sample of 46 patrons, the sample distribution mean remains the same (\( \mu \) = 5.82). However, the standard deviation of the sample mean (\( \sigma_{\bar{X}} \)) changes. It is calculated as:
\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2.5}{\sqrt{46}} = 0.3686 \]
The distribution notation is:
\[ \bar{X} \sim N(5.82, 0.3686) \boxed{\checkmark} \]
### c. For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $5.2871 and $5.6314.
Using the normal distribution \( X \sim N(5.82, 2.5) \), you can find this probability using cumulative distribution functions (CDF). The result is:
\[ P(5.2871 < X < 5.6314) = 0.0514 \boxed{\text{Wrong}}\]
### d. For the group of 46 patrons, find the probability that the average lunch cost is between $5.2871 and $5.6314.
Using \( \bar{X} \sim N(5.82, 0.3686) \), you calculate:
\[ P(5.2871 < \bar{X} < 5.6314) = ? \]
The given incorrect answer is:
\[ 0.2315 \boxed{\text{Wrong}} \]
### e. For part
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