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:

rCopy code
# 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!