We are asked to find the probability of a pregnancy lasting 309 days or longer, given that the lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. Here's the step-by-step solution:

Step 1: Standardize the value using the Z-score formula

To find the probability of a pregnancy lasting 309 days or longer, we first convert the number of days (309 days) into a Z-score. The Z-score formula is:

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

Where:

• $X$ is the value (309 days),
• $\mu$ is the mean (268 days),
• $\sigma$ is the standard deviation (15 days).

Substituting the values into the formula:

$Z = \frac{309 - 268}{15} = \frac{41}{15} = 2.73$

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

Next, we need to find the probability that a Z-score is greater than 2.73. We can use standard normal distribution tables or a calculator to find this probability.

From the Z-table or a standard normal distribution calculator, the cumulative probability of a Z-score of 2.73 is approximately 0.9968. This means the probability of a Z-score less than 2.73 is 99.68%.

To find the probability of a Z-score greater than 2.73, we subtract this value from 1:

$P(Z > 2.73) = 1 - 0.9968 = 0.0032$

Step 3: Interpret the result

The probability of a pregnancy lasting 309 days or longer is 0.32% (0.0032). This is an extremely small probability, which suggests that such a long pregnancy is very rare.

What does this suggest?

Given that the likelihood of a pregnancy lasting 309 days or longer is only 0.32%, the result suggests that it is highly unlikely that the wife gave birth 309 days after the husband's visit. Therefore, the claim might raise questions about the timeline she provided.

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Would you like further details or clarifications? Here are some related questions:

1. How do you interpret the standard deviation in this context?
2. What does the Z-score represent in normal distribution problems?
3. Could a different standard deviation change the outcome significantly?
4. What are other real-life situations where the Z-score is used?
5. How does this normal distribution apply to everyday occurrences?

Tip: In normal distributions, data values far from the mean (in terms of Z-scores) are considered outliers or rare events.