The length of human pregnancies is approximately normal with mean $\mu=266$ days and standard deviation $\sigma=16$ days. Complete parts (a) through (f). Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed) A. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 261 days. B. If 100 pregnant individuals were selected independently from this population, we would expect 38 pregnancies to last less than 261 days. C. If 100 pregnant individuals were selected independently from this population, we would expect $=$ pregnancies to last more than 261 days. (b) Suppose a random sample of 20 human pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. The sampling distribution of $\bar{x}$ is normal ${ }^{\top}$ with $\mu_{\bar{x}}=266$ and $\sigma_{\bar{x}}=3.5777^{\prime}$. (Type integers or decimals rounded to four decimal places as needed.) (c) What is the probability that a random sample of 20 pregnancies has a mean gestation period of 261 days or less? The probability that the mean of a random sample of 20 pregnancies is less than 261 days is approximately (Round to four decimal places as needed.)

Step 1 ：Calculate the z-score for 261 days using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in (261 days), \(\mu\) is the mean (266 days), and \(\sigma\) is the standard deviation (16 days).

Step 2 ：Substitute the values into the formula to get \(Z = \frac{261 - 266}{16} = -0.3125\).

Step 3 ：Look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability associated with a z-score of -0.3125 is approximately 0.3770, or 37.70%.

Step 4 ：If 100 pregnant individuals were selected independently from this population, we would expect about 38 pregnancies to last less than 261 days. Therefore, the final answer is \(\boxed{38}\).

Step 5 ：The sampling distribution of the sample mean length of pregnancies is normally distributed. The mean of this distribution, \(\mu_{\bar{x}}\), is equal to the population mean, which is 266 days. The standard deviation of this distribution, \(\sigma_{\bar{x}}\), is equal to the population standard deviation divided by the square root of the sample size (n). So, \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{20}} = 3.5777\).

Step 6 ：To find the probability that a random sample of 20 pregnancies has a mean gestation period of 261 days or less, we first calculate the z-score for 261 days using the mean and standard deviation of the sampling distribution: \(Z = \frac{X - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{261 - 266}{3.5777} = -1.3984\).

Step 7 ：Looking up this z-score in a standard normal distribution table or using a calculator gives a probability of approximately 0.0811, or 8.11%. So, the final answer is \(\boxed{0.0811}\).