We are given a normal distribution with:

• Mean, $\mu = 265$ days
• Standard deviation, $\sigma = 13$ days

We need to find the percentage of pregnancies that last fewer than 299 days. To solve this, we'll calculate the z-score for 299 days and then use the standard normal distribution to find the corresponding probability.

Step 1: Calculate the z-score

The z-score formula is:

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

Where:

• $X = 299$ days
• $\mu = 265$ days
• $\sigma = 13$ days

Substitute the values into the formula:

$z = \frac{299 - 265}{13} = \frac{34}{13} \approx 2.615$

Step 2: Find the probability

Now, we need to look up the z-score of 2.615 in the standard normal distribution table, or use a calculator that gives the cumulative probability for the z-score.

For $z = 2.615$, the cumulative probability is approximately 0.9955.

Step 3: Convert to percentage

To convert this to a percentage, we multiply by 100:

$0.9955 \times 100 = 99.55\%$

Thus, the percentage of pregnancies that last fewer than 299 days is approximately 99.6%.

Final Answer:

$P(X < 299 \, \text{days}) = 99.6$

Let me know if you'd like further details or have additional questions!

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Here are five related questions that may help you explore this topic further:

1. How do you calculate the z-score for a different value of $X$?
2. How do you find the percentage of pregnancies lasting longer than a given number of days?
3. What is the significance of the standard deviation in this problem?
4. How does the normal distribution model real-world phenomena like pregnancy lengths?
5. How can you calculate probabilities for a range of values in a normal distribution?

Tip: When using a z-score table, remember that it gives the cumulative probability from the left, so it's useful for finding percentages less than a specific value.