using r program answer this question 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 answer the question using R programming, we need to follow these steps:

1. **Calculate the z-scores** for 245 days and 308 days using the given mean of 280 days and standard deviation of 13 days.
2. **Find the probability** associated with those z-scores from the normal distribution table.
3. **Subtract those probabilities from 1** to determine the likelihood that the pregnancy falls outside the given range, which would imply that the alleged father is not the father.

Here's the process in R:

```r
# Given values
mean_pregnancy <- 280  # Mean of pregnancy in days
sd_pregnancy <- 13     # Standard deviation of pregnancy in days
lower_limit <- 245     # The lower limit (days the father was out of the country)
upper_limit <- 308     # The upper limit (days the father was out of the country)

# Calculate the z-scores
z_lower <- (lower_limit - mean_pregnancy) / sd_pregnancy
z_upper <- (upper_limit - mean_pregnancy) / sd_pregnancy

# Calculate the cumulative probabilities for the z-scores
p_lower <- pnorm(z_lower)
p_upper <- pnorm(z_upper)

# Calculate the probability that the pregnancy length is less than 245 or more than 308 days
prob_not_father <- p_lower + (1 - p_upper)

# Print the result
round(prob_not_father, 4)
```

### Explanation:
- We use the formula for z-score:  
  \[
  z = \frac{x - \mu}{\sigma}
  \]
  Where \(x\) is the value (245 and 308), \(\mu\) is the mean (280), and \(\sigma\) is the standard deviation (13).
- `pnorm()` calculates the cumulative probability up to a given z-score.
- We sum the probabilities for both tails of the distribution (less than 245 days or more than 308 days).

This will provide the probability that the alleged father is **not** the father, rounded to four decimal places.

You can copy and paste this code into an R environment to get the answer. Let me know if you need further clarification!

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.)

In this problem, we calculate the probability that an alleged father is not the biological parent, based on the normal distribution of pregnancy lengths. Using a mean of 280 days and a standard deviation of 13 days, we'll compute the z-scores for 245 and 308 days, and find the probability using R programming. Suitable for advanced high school or undergraduate students.

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