To solve this problem, we need to calculate the probability that the length of the pregnancy was either less than 245 days or more than 308 days. We'll start by calculating the z-scores for both of these values using the formula for the z-score:

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

Where:

• $X$ is the value of interest (245 days or 308 days),
• $\mu$ is the mean (280 days),
• $\sigma$ is the standard deviation (13 days).

Step 1: Calculate the z-score for 245 days

For $X = 245$ days: $z_{245} = \frac{245 - 280}{13} = \frac{-35}{13} \approx -2.6923$

Step 2: Calculate the z-score for 308 days

For $X = 308$ days: $z_{308} = \frac{308 - 280}{13} = \frac{28}{13} \approx 2.1538$

Step 3: Use the z-scores to calculate probabilities

We now use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these z-scores.

• The probability that $z \leq -2.6923$ (for 245 days) is approximately 0.0036.
• The probability that $z \geq 2.1538$ (for 308 days) is approximately 1 - 0.9844 = 0.0156.

Step 4: Calculate the total probability

Since we are interested in the probability that the pregnancy duration was either less than 245 days or more than 308 days, we add the two probabilities:

$P(\text{pregnancy} < 245 \text{ or } > 308) = 0.0036 + 0.0156 = 0.0192$

Conclusion:

The probability that the alleged father was not the father, assuming the pregnancy duration was outside the 245-308 day range, is approximately 0.0192, or 1.92%.

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Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How would the result change if the standard deviation were 15 days instead of 13 days?
2. What is the probability that the pregnancy lasted between 245 and 308 days?
3. How can we use paternity testing probabilities to complement these calculations?
4. What assumptions are made about the normal distribution of pregnancy durations?
5. Can environmental or medical factors significantly affect the duration of pregnancy, and how would this change the analysis?

Tip: When using z-scores, remember that the area under the normal curve represents the probability, so combining areas from both tails can give you the desired total probability in cases like this one.