The length of a human pregnancy is normally distributed with a mean of 270 days with a standard deviation of 8 days. How many days would a pregnancy last for the shortest 15%? Round answer to 2 decimal places.

To find the duration of pregnancy that lasts for the shortest 15%, we need to determine the 15th percentile of the normal distribution with a mean (\(\mu\)) of 270 days and a standard deviation (\(\sigma\)) of 8 days.

1. **Find the z-score** for the 15th percentile using a standard normal distribution table or calculator. The z-score for the 15th percentile is approximately \(-1.036\).

2. **Use the z-score formula**:
   \[
   X = \mu + z \cdot \sigma
   \]
   Substituting the values:
   \[
   X = 270 + (-1.036) \cdot 8
   \]
   \[
   X = 270 - 8.288 = 261.71
   \]

Thus, a pregnancy would last approximately **261.71 days** for the shortest 15%.

Would you like more details or have any questions? Here are some related questions you might consider:

1. How would changing the mean affect the 15th percentile?
2. What is the probability of a pregnancy lasting less than 250 days?
3. How do you calculate percentiles for other normal distributions?
4. What is the z-score for the 85th percentile in this context?
5. How does the standard deviation influence the distribution shape? 

**Tip:** Understanding how to use z-scores can help with various statistical problems involving normal distributions.

Explore how to calculate the duration of a human pregnancy for the shortest 15% using a normal distribution with a mean of 270 days and a standard deviation of 8 days. This problem demonstrates key statistical concepts and includes a detailed solution.

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