Questions: Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviation of σ=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 0.6255. (Round to four decimal places as needed.) b. If 25 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 .6255. (Round to four decimal places as needed.)

Solution Steps

To find the probability that a randomly selected adult female has a pulse rate less than 77 beats per minute, we first calculate the Z-score for \( X = 77 \) using the formula:

\[ Z = \frac{X - \mu}{\sigma} = \frac{77 - 73}{12.5} = \frac{4}{12.5} = 0.32 \]

Next, we use the cumulative distribution function \( \Phi(Z) \) to find the probability:

\[ P(X < 77) = \Phi(0.32) - \Phi(-\infty) = 0.6255 \]

Thus, the probability that one adult female has a pulse rate less than 77 bpm is:

\[ \boxed{P = 0.6255} \]

For 25 adult females, we need to find the probability that their mean pulse rate is less than 77 beats per minute. The standard error of the mean is calculated as:

\[ \sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}} = \frac{12.5}{\sqrt{25}} = \frac{12.5}{5} = 2.5 \]

Now, we calculate the Z-score for the sample mean:

\[ Z = \frac{\bar{X} - \mu}{\sigma_{\text{mean}}} = \frac{77 - 73}{2.5} = \frac{4}{2.5} = 1.6 \]

Using the cumulative distribution function \( \Phi(Z) \):

\[ P(\bar{X} < 77) = \Phi(1.6) - \Phi(-\infty) = 0.9452 \]

Thus, the probability that the mean pulse rate of 25 adult females is less than 77 bpm is:

\[ \boxed{P = 0.9452} \]

Final Answer