Step 1 :We are given that females have pulse rates that are normally distributed with a mean of 75.0 bpm and a standard deviation of 12.5 bpm.
Step 2 :We are asked to find the probability that a randomly selected adult female has a pulse rate less than 82 beats per minute.
Step 3 :To solve this, we use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.
Step 4 :We calculate the Z-score for 82 beats per minute using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are interested in (82 bpm), \(\mu\) is the mean (75.0 bpm), and \(\sigma\) is the standard deviation (12.5 bpm).
Step 5 :Substituting the given values into the formula, we get a Z-score of approximately 0.56.
Step 6 :We then look up this Z-score in the standard normal distribution table to find the probability. The probability corresponding to a Z-score of 0.56 is approximately 0.7123.
Step 7 :This means that there is a 71.23% chance that a randomly selected adult female will have a pulse rate of less than 82 beats per minute.
Step 8 :Final Answer: The probability that the pulse rate of a randomly selected adult female is less than 82 beats per minute is approximately \(\boxed{0.7123}\).