To find the probability that a pregnancy lasts at least 300 days, we'll use the standard normal distribution.

Step 1: Calculate the z-score

The z-score measures how many standard deviations a value (in this case, 300 days) is from the mean.

The z-score formula is:

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

where:

• $X = 300$ days (the value we are interested in),
• $\mu = 268$ days (the mean),
• $\sigma = 15$ days (the standard deviation).

Substitute the values:

$z = \frac{300 - 268}{15} = \frac{32}{15} \approx 2.13$

Step 2: Find the probability corresponding to the z-score

We need to find the probability that a standard normal variable $Z$ is greater than 2.13.

Using standard normal distribution tables or a calculator:

$P(Z > 2.13) \approx 0.0166$

Step 3: Interpretation

The probability that a pregnancy lasts at least 300 days is approximately 0.0166.

Answer:

The correct answer is a. 0.0166.

Would you like more details or have any other questions?

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Here are some related questions you might find interesting:

1. How do you calculate the z-score for any value in a normal distribution?
2. What does the z-score represent in a standard normal distribution?
3. How do you use a z-score table to find probabilities?
4. What is the probability that a pregnancy lasts between 250 and 280 days?
5. What is the probability that a pregnancy lasts less than 260 days?
6. How does changing the standard deviation affect the probability calculations?
7. How can you calculate the mean and standard deviation from raw data?
8. How do normal distributions apply to other real-world scenarios?

Tip: Always double-check the z-score formula and your calculations when solving problems involving the normal distribution!