Assume that females have pulse rates that are normally distributed with a mean of $\mu=74.0$ beats per minute and a standard deviation of $\sigma=12.5$ beats per minute. Complete parts (a) through (c) below. a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 77 beats per minute. The probability is (Round to four decimal places as needed.) b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 77 beats per minute. The probability is (Round to four decimal places as needed.) c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30 ? A. Since the mean pulse rate exceeds 30 , the distribution of sample means is a normal distribution for any sample size. B. Since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size. C. Since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step 1 ：Given that the pulse rates of females are normally distributed with a mean of \(\mu = 74.0\) beats per minute and a standard deviation of \(\sigma = 12.5\) beats per minute.

Step 2 ：For part (a), we need to find the probability that a randomly selected female has a pulse rate less than 77 beats per minute. This is a standard normal distribution problem, where we need to find the z-score for 77 and then find the cumulative probability for that z-score. The z-score is calculated as \(z = \frac{X - \mu}{\sigma}\), where X is the value we are interested in, which is 77 in this case. So, \(z = \frac{77 - 74.0}{12.5} = 0.24\). The cumulative probability corresponding to z = 0.24 is approximately 0.5948.

Step 3 ：For part (b), we need to find the probability that the mean pulse rate of 16 randomly selected females is less than 77 beats per minute. This is a problem of sampling distribution of the mean. The standard deviation of the sampling distribution (standard error) is the standard deviation of the population divided by the square root of the sample size. So, \(\sigma_{b} = \frac{\sigma}{\sqrt{n}} = \frac{12.5}{\sqrt{16}} = 3.125\). We find the z-score for 77 with this new standard deviation and then find the cumulative probability for that z-score. So, \(z = \frac{77 - 74.0}{3.125} = 0.96\). The cumulative probability corresponding to z = 0.96 is approximately 0.8315.

Step 4 ：For part (c), we need to explain why we can use the normal distribution for part (b) even though the sample size is less than 30. The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size gets larger — usually considered as larger than 30. However, if the original population is normally distributed, then the sample means will also be normally distributed, regardless of sample size.

Step 5 ：\(\boxed{\text{Final Answer:}}\) The probability that a randomly selected female has a pulse rate less than 77 beats per minute is approximately \(\boxed{0.5948}\). The probability that the mean pulse rate of 16 randomly selected females is less than 77 beats per minute is approximately \(\boxed{0.8315}\). The normal distribution can be used in part (b) even though the sample size does not exceed 30 because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size. So, the correct answer for part (c) is option D.