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.