An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 245 to 308 days before the birth of the child, so the pregnancy would have been less than 245 days or more than 308 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? Calculate the *z*-scores first, and then use those to calculate the probability. (Round your answer to four decimal places.)

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%.

---

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.

An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 245 to 308 days before the birth of the child, so the pregnancy would have been less than 245 days or more than 308 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? Calculate the z-scores first, and then use those to calculate the probability. (Round your answer to four decimal places.)

This problem involves calculating the probability that a pregnancy length falls outside the 245 to 308-day range, using the normal distribution with a mean of 280 days and a standard deviation of 13 days. Z-scores are used to determine the probability that the alleged father is not the father, rounding the final result to four decimal places.

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