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To solve this problem, we need to calculate the probability that the average pregnancy length for a sample of six randomly chosen women exceeds 270 days. 

### Step-by-Step Solution:

1. **Identify Given Information:**
   - Mean pregnancy length (\(\mu\)) = 266 days
   - Standard deviation (\(\sigma\)) = 16 days
   - Sample size (\(n\)) = 6
   - Target average pregnancy length we are comparing to = 270 days

2. **Determine the Distribution of the Sample Mean:**
   Since we have a sample, the distribution of the sample mean \(\overline{X}\) will follow a normal distribution with:
   - Mean of the sample mean, \(\mu_{\overline{X}} = \mu = 266\) days
   - Standard deviation of the sample mean (standard error), \(\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{6}}\)

   Let's calculate the standard error:
   \[
   \sigma_{\overline{X}} = \frac{16}{\sqrt{6}} \approx 6.53 \text{ days}
   \]

3. **Calculate the Z-Score:**
   The Z-score helps us determine how far 270 days is from the mean in terms of standard deviations. The Z-score formula is:
   \[
   Z = \frac{X - \mu_{\overline{X}}}{\sigma_{\overline{X}}}
   \]
   where \(X = 270\) days.

   Substituting the values:
   \[
   Z = \frac{270 - 266}{6.53} \approx \frac{4}{6.53} \approx 0.61
   \]

4. **Find the Probability:**
   We now need to find the probability that the Z-score is greater than 0.61. Using a standard normal distribution table, the probability that \(Z \leq 0.61\) is approximately 0.7291. Therefore, the probability that \(Z > 0.61\) is:
   \[
   P(Z > 0.61) = 1 - 0.7291 = 0.2709
   \]

5. **Conclusion:**
   The probability that the average pregnancy length for six randomly chosen women exceeds 270 days is approximately **0.27**.

### Answer:
The correct answer is:
- **0.27**

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Would you like further details on any steps, or do you have any questions?

### Related Questions:
1. How would the probability change if the sample size were increased to 10 women?
2. What is the probability that a single pregnancy exceeds 270 days?
3. How does the standard deviation of the sample mean change with different sample sizes?
4. Why is the sample mean approximately normally distributed even with a small sample size?
5. What is the significance of the Central Limit Theorem in this context?

### Tip:
When dealing with averages of samples, remember that the standard deviation of the sample mean decreases as the sample size increases, making the distribution narrower around the mean.

The probability that the average pregnancy length for six randomly chosen women exceeds 270 days is about:
- 0.40
- 0.07
- 0.27

Learn how to calculate the probability that the average pregnancy length for six randomly chosen women exceeds 270 days, using a normal distribution with a mean of 266 days and a standard deviation of 16 days. This solution covers key statistical concepts, including Z-scores and the Central Limit Theorem.

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