CNNBC recently reported that the mean annual cost of auto insurance is 1026 dollars. Assume the standard deviation is 252 dollars, and the cost is normally distributed. You take a simple random sample of 38 auto insurance policies. Round your answers to 4 decimal places. ### Statistical Analysis of Auto Insurance Costs**Scenario:**CNNBC recently reported that the mean annual cost of auto insurance is $1026. Assume the standard deviation is $252, and the cost is normally distributed. You take a simple random sample of 38 auto insurance policies.**Objective:**Analyze the distribution and calculate probabilities associated with the annual cost of auto insurance.**Instructions:**Round your answers to four decimal places where necessary.---**a. Distribution of $ X $**What is the distribution of $ X $? \[ X \sim N(\text{mean}, \text{standard deviation}) \]**b. Distribution of $ \bar{x} $**What is the distribution of $ \bar{x} $?\[ \bar{x} \sim N(\text{mean}, \text{standard error}) \]**c. Probability Calculation for One Policy**What is the probability that one randomly selected auto insurance cost is more than $1075?---**d. Probability Calculation for Sample Mean**A simple random sample of 38 auto insurance policies is taken. Find the probability that the average cost of these policies is more than $1075.\[ P(\bar{x} > 1075) = \]**e. Normality Assumption**For part (d), is the assumption of normality necessary?\[ \text{Yes } \bigcirc \text{ No } \bigcirc \]---**Explanation and Calculation Process:**1. **Understanding the given data:** - Mean ($ \mu $) = $1026 - Standard Deviation ($ \sigma $) = $2522. **Distribution of $ X $ (Annual cost for individual policies):** \[ X \sim N(1026, 252) \]3. **Distribution of $ \bar{x} $ (Mean of a sample of 38 policies):** - Mean of the sample distribution ($ \mu $) remains the same ($ = 1026 $). - Standard error of the mean ($ \sigma_{\bar{x}} $) is calculated as: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{252}{\sqrt{38}} \approx 40.8511 \] \[ \bar{x} \sim N(1026, 40

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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 1026 dollars. Assume the standard deviation is 252 dollars, and the cost is normally distributed. You take a simple random sample of 38 auto insurance policies. Round your answers to 4 decimal places.

### Statistical Analysis of Auto Insurance Costs

**Scenario:**
CNNBC recently reported that the mean annual cost of auto insurance is $1026. Assume the standard deviation is $252, and the cost is normally distributed. You take a simple random sample of 38 auto insurance policies.

**Objective:**
Analyze the distribution and calculate probabilities associated with the annual cost of auto insurance.

**Instructions:**
Round your answers to four decimal places where necessary.

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**a. Distribution of $ X $**

What is the distribution of $ X $?
\[ X \sim N(\text{mean}, \text{standard deviation}) \]

**b. Distribution of $ \bar{x} $**

What is the distribution of $ \bar{x} $?
\[ \bar{x} \sim N(\text{mean}, \text{standard error}) \]

**c. Probability Calculation for One Policy**

What is the probability that one randomly selected auto insurance cost is more than $1075?

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**d. Probability Calculation for Sample Mean**

A simple random sample of 38 auto insurance policies is taken. Find the probability that the average cost of these policies is more than $1075.

\[ P(\bar{x} > 1075) = \]

**e. Normality Assumption**

For part (d), is the assumption of normality necessary?
\[ \text{Yes } \bigcirc \text{ No } \bigcirc \]

---

**Explanation and Calculation Process:**

1. **Understanding the given data:**
- Mean ($ \mu $) = $1026
- Standard Deviation ($ \sigma $) = $252

2. **Distribution of $ X $ (Annual cost for individual policies):**
\[ X \sim N(1026, 252) \]

3. **Distribution of $ \bar{x} $ (Mean of a sample of 38 policies):**
- Mean of the sample distribution ($ \mu $) remains the same ($ = 1026 $).
- Standard error of the mean ($ \sigma_{\bar{x}} $) is calculated as:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{252}{\sqrt{38}} \approx 40.8511 \]
\[ \bar{x} \sim N(1026, 40

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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