This question: 1 point(S) possible Submit test The lengths of pregnancies are normally distributed with a mean of 267 days and 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 $4 \%$, then the baby is premature. Find the length that separates premature babies from those who are not considered premature. a. The probability that a pregnancy will last 309 days or longer is $\square$. (Round to four decimal places as needed.) What does the result suggest? A. The result suggests that either a very rare event occurred or the husband is not the father. B. The result suggests an uncommon but not significant event occurred. C. The result suggests the event did not occur. D. The result suggests that the husband is the father. b. Babies who are born on or before $\square$ days are considered premature. (Round to the nearest integer as needed.) (1) Time Remaining: 01:25:06 Next

Step 1 ：First, we need to convert 309 days into a z-score. The z-score is calculated by subtracting the mean from the value and then dividing by the standard deviation. The formula for z-score is \(Z = \frac{X - \mu}{\sigma}\)

Step 2 ：Substitute the given values into the formula: \(X = 309\) days, \(\mu = 267\) days (mean), \(\sigma = 15\) days (standard deviation). So, \(Z = \frac{309 - 267}{15} = 2.8\)

Step 3 ：Now, we look up the z-score in the standard normal distribution table. The value for 2.8 is 0.9974, which represents the probability of a pregnancy lasting 309 days or less. However, we want the probability of a pregnancy lasting 309 days or more, so we subtract the value from 1.

Step 4 ：Calculate the probability: \(P(X \geq 309) = 1 - P(X \leq 309) = 1 - 0.9974 = 0.0026\)

Step 5 ：\(\boxed{0.0026}\) is the probability of a pregnancy lasting 309 days or longer.

Step 6 ：To find the length that separates premature babies from those who are not considered premature, we need to find the z-score that corresponds to the lowest 4% of the distribution. From the standard normal distribution table, we find that the z-score for the lowest 4% is approximately -1.75.

Step 7 ：We then use the z-score formula to find the corresponding length of pregnancy: \(X = \mu + Z\sigma\)

Step 8 ：Substitute the given values into the formula: \(X = 267 + (-1.75)(15) = 240.75\)

Step 9 ：Rounding to the nearest integer, we get 241 days. So, babies who are born on or before 241 days are considered premature.

Step 10 ：So, the final answer is \(\boxed{241}\) days.