Assume that females have pulse rates that are normally distributed with a mean of \( \mu=72.0 \) beats per minute and a standard deviation of \( o =12.5 \) 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 78 beats per minute. The probability is \( \square \) (Round to four decimal places as needed.)

To find the probability that a randomly selected adult female has a pulse rate of less than 78 beats per minute, we first need to transform the pulse rate into a z-score using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \(X\) is the value of interest (78), \(\mu\) is the mean (72), and \(\sigma\) is the standard deviation (12.5). Calculating the z-score: \[ z = \frac{(78 - 72)}{12.5} = \frac{6}{12.5} = 0.48 \] Next, we will use a standard normal distribution table or calculator to find the probability associated with this z-score. Looking up \(z = 0.48\), we find: \[ P(Z < 0.48) \approx 0.6844 \] So, the probability that her pulse rate is less than 78 beats per minute is approximately **0.6844** (rounded to four decimal places). Now, moving on to parts (b) and (c) would require more details from the original prompt to continue. However, isn’t it interesting how pulse rate can be indicative of overall health, and knowing the average can guide you in many real-life scenarios, like fitness assessments?