b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 79 beats per minute. The probability is \( \square \). (Round to four decimal places as needed.)

To find the probability that the mean pulse rate of 16 adult females is less than 79 beats per minute, we can use the Central Limit Theorem (CLT) for the sampling distribution of the mean. Given: - Sample size (\( n \)) = 16 - Population mean (\( \mu \)) = 79 beats per minute - Population standard deviation (\( \sigma \)) = unknown The CLT states that the sampling distribution of the mean will be approximately normal with a mean equal to the population mean (\( \mu \)) and a standard deviation equal to the population standard deviation (\( \sigma \)) divided by the square root of the sample size (\( n \)). The probability that the mean pulse rate is less than 79 beats per minute can be calculated using the z-score formula: \[ z = \frac{\text{sample mean} - \text{population mean}}{\text{population standard deviation} / \sqrt{n}} \] Since we are looking for the probability that the mean is less than 79, we will use the z-score corresponding to the area to the left of 79 in a standard normal distribution. Let's calculate the z-score and then find the probability. Simplify the expression by following steps: - step0: Solution: \(\frac{\left(79-79\right)}{\left(\frac{\sigma }{\sqrt{16}}\right)}\) - step1: Remove the parentheses: \(\frac{79-79}{\frac{\sigma }{\sqrt{16}}}\) - step2: Subtract the terms: \(\frac{0}{\frac{\sigma }{\sqrt{16}}}\) - step3: Simplify the root: \(\frac{0}{\frac{\sigma }{4}}\) - step4: Divide the terms: \(0\) The z-score is 0, which means that the sample mean is equal to the population mean. This indicates that the probability of the mean pulse rate being less than 79 beats per minute is 0.5, as the area to the left of 79 in a standard normal distribution is 0.5. Therefore, the probability that the mean pulse rate of 16 adult females is less than 79 beats per minute is 0.5.