The lengths of pregnancies are normally distributed with a mean of 270 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest \( 2 \% \), then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. a. The probability that a pregnancy will last 309 days or longer is (Round to four decimal places as needed.)

To solve part (a), we start by calculating the z-score for a pregnancy lasting 309 days. The formula for the z-score is: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value of interest (309 days), \( \mu \) is the mean (270 days), and \( \sigma \) is the standard deviation (15 days). Plugging in the values: \[ z = \frac{(309 - 270)}{15} = \frac{39}{15} = 2.6 \] Next, we look up the z-score of 2.6 in the standard normal distribution table (or use a calculator) to find the cumulative probability. This tells us the probability of a pregnancy lasting less than 309 days: Cumulative probability for \( z = 2.6 \) is approximately 0.9953. To find the probability of a pregnancy lasting 309 days or longer, we subtract this value from 1: \[ P(X \geq 309) = 1 - P(Z < 2.6) = 1 - 0.9953 = 0.0047 \] So, the probability that a pregnancy will last 309 days or longer is approximately **0.0047** (rounded to four decimal places). Now, for part (b), to find the length that separates the lowest 2% of pregnancies, we need to find the z-score that corresponds to the 2nd percentile, which is approximately -2.05 (from the z-table). Using the z-score formula rearranged: \[ X = \mu + z \sigma \] We plug in the values: \[ X = 270 + (-2.05)(15) \] \[ X = 270 - 30.75 \] \[ X = 239.25 \] Thus, the length that separates premature babies from those who are not premature is approximately **239.25 days**. If you have any questions or need further assistance, feel free to ask!