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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.

What percentage of pregnancies last beyond 248 days? Given: Mean = 262 days, Standard deviation = 17 days, P(X > 248 days) = ?

Calculate the percentage of pregnancies that last beyond 248 days in a normally distributed dataset with a mean of 262 days and a standard deviation of 17 days. Learn how to find the Z-score and determine the probability using standard normal distribution tables.

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