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.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Chapter10: Statistics
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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.

---

**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 (\( \sigma \)) = $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
Transcribed Image Text:### 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 (\( \sigma \)) = $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
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