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 places. ### 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 dollarsThe 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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Answered: CNNBC recently reported that the mean…

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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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 places.

### 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}}\

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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