CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 242 dollars, and the cost is normally distributed. You take a simple random sample of 7 auto insurance policies. Round your answers to 4 decimal
CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 242 dollars, and the cost is normally distributed. You take a simple random sample of 7 auto insurance policies. Round your answers to 4 decimal
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.3: Measures Of Spread
Problem 26PFA
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CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 242 dollars, and the cost is
![### Understanding Distributions and Probabilities in Auto Insurance Costs
**Scenario:** CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 242 dollars, and the cost is normally distributed. You take a simple random sample of 7 auto insurance policies.
**Objective:** Round your answers to 4 decimal places.
---
#### Questions:
**a. What is the distribution of \( X \)?**
\[ X \sim \mathcal{N}(\square, \square) \]
**b. What is the distribution of \( \overline{x} \)?**
\[ \overline{x} \sim \mathcal{N}(\square, \square) \]
**c. What is the probability that one randomly selected auto insurance is less than $848?**
\[ \boxed{\square} \]
**d. For a simple random sample of 7 auto insurance policies, find the probability that the average cost is less than $848.**
\[ \boxed{\square} \]
**e. For part d), is the assumption of normality necessary?**
\[ \begin{array}{c} \text{( ) No} \\ \text{( ) Yes} \end{array} \]
---
#### Detailed Explanations:
**a. Distribution of \( X \)**
Given:
- Mean (\( \mu \)) = 980 dollars
- Standard deviation (\( \sigma \)) = 242 dollars
The distribution of \( X \) is \( \mathcal{N}(980, 242^2) \).
\[ X \sim \mathcal{N}(980, 242^2) \]
**b. Distribution of \( \overline{x} \)**
For a sample size (\( n \)) of 7:
- The mean of the sampling distribution of \( \overline{x} \) remains the same: 980 dollars.
- The standard error (SE) is given by \( \frac{\sigma}{\sqrt{n}} = \frac{242}{\sqrt{7}} \).
Standard error calculation:
\[ \text{SE} = \frac{242}{\sqrt{7}} \]
Therefore, the distribution of \( \overline{x} \) is:
\[ \overline{x} \sim \mathcal{N}\left(980, \left(\frac{242}{\sqrt{7}}\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b94522a-bae3-4964-ba8a-526370d903b8%2F0948def3-77f0-4fa7-a4a2-8d22a5a7d361%2Fi4gq2be_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Distributions and Probabilities in Auto Insurance Costs
**Scenario:** CNNBC recently reported that the mean annual cost of auto insurance is 980 dollars. Assume the standard deviation is 242 dollars, and the cost is normally distributed. You take a simple random sample of 7 auto insurance policies.
**Objective:** Round your answers to 4 decimal places.
---
#### Questions:
**a. What is the distribution of \( X \)?**
\[ X \sim \mathcal{N}(\square, \square) \]
**b. What is the distribution of \( \overline{x} \)?**
\[ \overline{x} \sim \mathcal{N}(\square, \square) \]
**c. What is the probability that one randomly selected auto insurance is less than $848?**
\[ \boxed{\square} \]
**d. For a simple random sample of 7 auto insurance policies, find the probability that the average cost is less than $848.**
\[ \boxed{\square} \]
**e. For part d), is the assumption of normality necessary?**
\[ \begin{array}{c} \text{( ) No} \\ \text{( ) Yes} \end{array} \]
---
#### Detailed Explanations:
**a. Distribution of \( X \)**
Given:
- Mean (\( \mu \)) = 980 dollars
- Standard deviation (\( \sigma \)) = 242 dollars
The distribution of \( X \) is \( \mathcal{N}(980, 242^2) \).
\[ X \sim \mathcal{N}(980, 242^2) \]
**b. Distribution of \( \overline{x} \)**
For a sample size (\( n \)) of 7:
- The mean of the sampling distribution of \( \overline{x} \) remains the same: 980 dollars.
- The standard error (SE) is given by \( \frac{\sigma}{\sqrt{n}} = \frac{242}{\sqrt{7}} \).
Standard error calculation:
\[ \text{SE} = \frac{242}{\sqrt{7}} \]
Therefore, the distribution of \( \overline{x} \) is:
\[ \overline{x} \sim \mathcal{N}\left(980, \left(\frac{242}{\sqrt{7}}\
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