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It looks like you're working on a probability problem involving the normal distribution, where you need to calculate the probability that the sample mean falls between 956 and 974 dollars. Let's walk through the steps to calculate the probability correctly.

### Problem Summary:
- **Population mean** (\(\mu\)): 954 dollars
- **Standard deviation** (\(\sigma\)): 119 dollars
- **Sample size** (\(n\)): 141
- **Range**: You're looking for the probability that the sample mean falls between 956 and 974 dollars.

### Steps to Solve:

1. **Find the standard error (SE)** for the sample mean:
   \[
   SE = \frac{\sigma}{\sqrt{n}} = \frac{119}{\sqrt{141}} \approx 10.021
   \]

2. **Convert the values (956 and 974) to Z-scores**:
   The Z-score is calculated as:
   \[
   Z = \frac{X - \mu}{SE}
   \]

   - For \( X = 956 \):
     \[
     Z = \frac{956 - 954}{10.021} \approx 0.1996
     \]
   
   - For \( X = 974 \):
     \[
     Z = \frac{974 - 954}{10.021} \approx 1.9962
     \]

3. **Find the probabilities for these Z-scores** using a standard normal distribution table or calculator:
   - The probability corresponding to \( Z = 0.1996 \) is approximately 0.5791.
   - The probability corresponding to \( Z = 1.9962 \) is approximately 0.9767.

4. **Calculate the probability**:
   The probability that the sample mean is between 956 and 974 dollars is the difference between these probabilities:
   \[
   P(956 < M < 974) = P(Z = 1.9962) - P(Z = 0.1996) = 0.9767 - 0.5791 = 0.3976
   \]

Thus, the correct probability is approximately **0.3976**.

It seems that you entered 0.3968, which was very close, but rounding to four decimal places might have caused the small error.

---

Do you want to go over any details or need further clarification?

Here are 5 related questions to deepen your understanding:

1. How would the result change if the sample size increased to 200?
2. What is the Z-score formula for a single value in a normal distribution?
3. How does the standard error affect the shape of the distribution for larger samples?
4. How do you interpret the probability in this context (in terms of insurance policies)?
5. What if the range changed to 950 to 970? How would the calculation differ?

### Tip:
Always check your final answer with the correct number of decimal places, especially when dealing with probability calculations where rounding can affect the final result.

CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 119 dollars. You will use a simple random sample of 141 auto insurance policies. Find the probability that a random sample of size n = 141 has a mean value between 956 and 974 dollars.

Find the probability that the mean of a random sample of 141 auto insurance policies falls between 956 and 974 dollars. Learn how to use normal distribution, Z-scores, and the central limit theorem in this step-by-step solution. Population mean: 954 dollars, Standard deviation: 119 dollars.

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