The length of human pregnancies is approximately normal with mean $\mu=266$ days and standard deviation $\sigma=16$ days. Complete parts (a) through (i) A. If 100 independent random samples of size $n=20$ pregnancies were obtained from this population, we would expect 1 sample(s) to have a sample mean of 261 days or more. B. If 100 independent random samples of size $n=20$ pregnancies were obtained from this population, we would expect 8 sample(s) to have a sample mean of 261 days or less: C. If 100 independent random samples of size $n=20$ pregnancies were obtained from this population, we would expect (d) What is the probability that a random sample of 66 pregnancies has a mean gestation period of 261 days or less? The probability that the mean of a random sample of 66 pregnancies is less than 261 days is approximately 00056 (Round to four decimal places as needed.) 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 independent random samples of size $n=66$ pregnancies were obtained from this population, we would expect $\square$ sample(s) to have a sample mean of exactly 261 days. B. If 100 independent randorn samples of size $n=66$ pregnancies were obtained from this population, we would expect $\square$ sample(s) to have a sample mean of 261 days or less. C. If 100 independent random samples of size $n=66$ pregnancies were obtained from this population we would expect lill sample(s) to have a sample mean of 261 days or mote.

Step 1 ：Given that the length of human pregnancies is approximately normal with mean \(\mu=266\) days and standard deviation \(\sigma=16\) days.

Step 2 ：We are asked to find the number of samples we would expect to have a sample mean of 261 days or more if 100 independent random samples of size 20 pregnancies were obtained from this population.

Step 3 ：To solve this, we need to calculate the z-score for a sample mean of 261 days, and then use the standard normal distribution to find the probability of obtaining a z-score greater than or equal to this value.

Step 4 ：The z-score is calculated as \((\text{sample mean} - \text{population mean}) / (\text{standard deviation} / \sqrt{\text{sample size}})\).

Step 5 ：Substituting the given values, we get \(z = -1.3975424859373686\).

Step 6 ：The probability of obtaining a z-score greater than or equal to a given value is found by looking up the value in a standard normal distribution table or using a function that calculates this probability.

Step 7 ：Using the calculated z-score, we find that the probability is \(p = 0.9188747500763841\).

Step 8 ：Once we have the probability, we can multiply it by the number of samples (100) to find the expected number of samples with a mean of 261 days or more.

Step 9 ：Doing this, we get \(\text{expected samples} = 91.88747500763841\).

Step 10 ：The expected number of samples with a mean of 261 days or more is approximately 92. However, the question asks for the number of samples we would expect to have a sample mean of 261 days or more, so we should round this number to the nearest whole number.

Step 11 ：Final Answer: \(\boxed{92}\)