The lengths of pregnancies are normally distributed with a mean of 269 days arid a standard deviation of 15 days a. In a letter to an advice column, a wife claimed to have given birth 309 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 309 days or longer. What does the result suggest? b. If the length of pregnancy is in the lowest $3 \%$, then the baby is premature. Find the length that separates premature babies from those who are not considered premature. (Round to four decimal places as needed.) What does the result suggest? A. The result suggests that the husband is the father. B. The result suggests the event did not occur C. The result suggests an uncommon but not significant event occurred. D. The result suggests that either a very rare event occurred or the husband is not the father b. Babies who are born on or before days are considered premature. (Round-to the nearest integer as needed.) rample Get more help. Clear all

Step 1 ：Calculate the z-score using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in (309 days), \(\mu\) is the mean (269 days), and \(\sigma\) is the standard deviation (15 days).

Step 2 ：Substitute the given values into the formula to get \(Z = \frac{309 - 269}{15} = \frac{40}{15} = 2.67\).

Step 3 ：Look up 2.67 in a standard normal distribution table to find the probability of a z-score being less than 2.67, which is 0.9962.

Step 4 ：Calculate the probability of a z-score being more than 2.67 by subtracting the value found from 1: \(P(Z > 2.67) = 1 - P(Z < 2.67) = 1 - 0.9962 = 0.0038\).

Step 5 ：\(\boxed{0.0038}\) or 0.38% is the probability of a pregnancy lasting 309 days or longer.

Step 6 ：To find the length of pregnancy that separates premature babies from those who are not considered premature, find the z-score that corresponds to the lowest 3% of the distribution. This z-score is approximately -1.88.

Step 7 ：Use this z-score to find the corresponding length of pregnancy using the formula \(X = \mu + Z\sigma\).

Step 8 ：Substitute the given values into the formula to get \(X = 269 + (-1.88)(15) = 269 - 28.2 = 240.8\).

Step 9 ：Round to the nearest integer to find that babies who are born on or before \(\boxed{241}\) days are considered premature.