We are given a normal distribution of pregnancy lengths with:

• Mean ($\mu$) = 262 days
• Standard deviation ($\sigma$) = 17 days

The problem asks for the percentage of pregnancies that last beyond 248 days. We are looking for:

$P(X > 248)$

Step 1: Find the Z-score

The Z-score is calculated using the formula:

$Z = \frac{X - \mu}{\sigma}$

Substitute the given values: $Z = \frac{248 - 262}{17} = \frac{-14}{17} \approx -0.8235$

Step 2: Find the probability

Using the Z-score table or a normal distribution calculator, find the cumulative probability for $Z = -0.8235$.

The cumulative probability for $Z = -0.8235$ is approximately 0.21. This represents the percentage of pregnancies lasting less than 248 days.

To find the percentage of pregnancies lasting beyond 248 days: $P(X > 248) = 1 - P(X < 248) = 1 - 0.21 = 0.79$

Step 3: Convert to percentage

The percentage is $0.79 \times 100 = 79\%$.

Thus, 79% of pregnancies last beyond 248 days.

Would you like any further details or clarifications?

Here are some follow-up questions:

1. What is the importance of using the Z-score in a normal distribution?
2. How can you interpret the result of 79% in the context of this problem?
3. How would the result change if the standard deviation increased?
4. Can we use the same method for non-normally distributed data? Why or why not?
5. How would you calculate the probability for a range of days, e.g., between 248 and 280 days?

Tip: Z-scores help standardize values in a normal distribution, allowing easy lookup of probabilities from standard tables.