We are dealing with a normally distributed variable for the lengths of pregnancies, with:

• Mean ($\mu$) = 266 days
• Standard deviation ($\sigma$) = 15 days

Part 1: Probability of a Pregnancy Lasting 309 Days or Longer

To find the probability that a pregnancy lasts 309 days or longer, we need to calculate the Z-score for 309 days. The Z-score is given by the formula:

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

where:

• $X = 309$ (the given pregnancy length),
• $\mu = 266$ (mean),
• $\sigma = 15$ (standard deviation).

Substitute the values:

$Z = \frac{309 - 266}{15} = \frac{43}{15} = 2.87$

Now, using the Z-score of 2.87, we can find the corresponding probability from the standard normal distribution table (or use a calculator). The area to the left of $Z = 2.87$ is approximately 0.9979.

Since we are looking for the probability of the pregnancy lasting 309 days or longer, we want the area to the right of $Z = 2.87$, which is:

$P(X \geq 309) = 1 - 0.9979 = 0.0021$

Thus, the probability of a pregnancy lasting 309 days or longer is approximately 0.0021, or 0.21%.

Interpretation

This result suggests that a pregnancy lasting 309 days or longer is highly unusual. In fact, the probability of such an event is only about 0.21%, which implies that the claim of the wife giving birth after 309 days is statistically very unlikely.

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Part 2: Length That Separates Premature Babies

To find the length of pregnancy that separates premature babies (lowest 4%), we need to find the pregnancy length corresponding to the 4th percentile of the normal distribution.

From standard normal distribution tables, the Z-score corresponding to the lowest 4% is approximately -1.75.

Now, we use the Z-score formula to find the corresponding length of pregnancy:

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

Substitute the known values:

$-1.75 = \frac{X - 266}{15}$

Solve for $X$:

$X - 266 = -1.75 \times 15 = -26.25$

$X = 266 - 26.25 = 239.75$

Thus, the pregnancy length that separates premature babies is approximately 239.75 days.

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Summary

a. The probability that a pregnancy lasts 309 days or longer is approximately 0.0021 (or 0.21%).

b. The length of pregnancy that separates premature babies from those who are not is approximately 239.75 days.

Would you like more details on any of the steps or explanations?

Related Questions:

1. What are other factors that can influence pregnancy duration aside from biological variation?
2. How would the probability change if the mean and standard deviation were different for a specific population?
3. What is the significance of calculating percentiles in normally distributed data?
4. How would the interpretation change if the probability of a 309-day pregnancy were higher?
5. Can the Z-score approach be used to estimate probabilities in other biological or medical contexts?

Tip:

When working with normal distributions, understanding the Z-score is key, as it allows you to standardize any normal variable and use standard normal tables for probability calculations.