The average age of vehicles registered in the United States is 96 months. Assume the population is normally distributed with a standard deviation of 15 months. Find the probability that the mean age of a sample of 36 vehicles is between 98 and 100 months?

To solve this problem, we need to use the concept of the **sampling distribution of the sample mean** and apply the **Central Limit Theorem**.

### Given dataTo solve this problem, we need to use the concept of the **sampling distribution of the sample mean** and apply the **Central Limit Theorem**.

### Given 
- Population mean (\(\mu\)) = 96 months
- Population standard deviation (\(\sigma\)) = 15 months
- Sample size (\(n\)) = 36
- We want to find the probability that the sample mean is between 98 and 100 months.

### Steps:

1. **Standard Error of the Mean (SEM)**:
   The standard error of the mean is calculated using the formula:
   \[
   \text{SEM} = \frac{\sigma}{\sqrt{n}}
   \]
   Where:
   - \(\sigma\) is the population standard deviation
   - \(n\) is the sample size

   Substituting the values:
   \[
   \text{SEM} = \frac{15}{\sqrt{36}} = \frac{15}{6} = 2.5
   \]

2. **Z-scores**:
   We now calculate the **Z-scores** for the given sample means (98 and 100 months) using the formula:
   \[
   Z = \frac{X - \mu}{\text{SEM}}
   \]
   Where:
   - \(X\) is the sample mean value
   - \(\mu\) is the population mean
   - SEM is the standard error of the mean

   - For \(X = 98\):
     \[
     Z_{98} = \frac{98 - 96}{2.5} = \frac{2}{2.5} = 0.8
     \]
   - For \(X = 100\):
     \[
     Z_{100} = \frac{100 - 96}{2.5} = \frac{4}{2.5} = 1.6
     \]

3. **Find the probability**:
   We now use the **standard normal distribution table** (Z-table) to find the probabilities corresponding to these Z-scores.
   
   - For \(Z = 0.8\), the cumulative probability from the Z-table is approximately **0.7881**.
   - For \(Z = 1.6\), the cumulative probability is approximately **0.9452**.

4. **Calculate the probability between the two Z-scores**:
   The probability that the sample mean is between 98 and 100 months is:
   \[
   P(98 < \bar{X} < 100) = P(Z_{100}) - P(Z_{98}) = 0.9452 - 0.7881 = 0.1571
   \]

Thus, the probability that the mean age of the sample is between 98 and 100 months is approximately **0.1571** or **15.71%**.

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Would you like more details or have any questions on this?

Here are 5 related questions you can explore:
1. What is the probability that the sample mean age is greater than 100 months?
2. What is the probability that the sample mean age is less than 98 months?
3. How would the probability change if the sample size were increased to 64 vehicles?
4. What assumptions are we making by using the normal distribution in this case?
5. How do we interpret the Z-scores in real-world terms?

**Tip**: When working with sample means, the larger your sample size, the smaller the standard error, leading to more precise estimates of the population mean.

Learn how to find the probability that the mean age of a sample of vehicles is between 98 and 100 months. Given a population mean of 96 months, standard deviation of 15 months, and a sample size of 36, apply the Central Limit Theorem and Z-scores to solve this probability problem. Suitable for college-level statistics students.

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