Step 1 :Calculate the z-score using the formula \(Z = \frac{X - \mu}{\sigma}\) where \(X\) is the value we are interested in (109 beats per minute), \(\mu\) is the mean (72 beats per minute), and \(\sigma\) is the standard deviation (25 beats per minute).
Step 2 :Substitute the given values into the formula to get \(Z = \frac{109 - 72}{25} = 1.48\). This means that a pulse rate of 109 beats per minute is 1.48 standard deviations above the mean.
Step 3 :Next, find the percentage of healthy adults who have a resting pulse rate greater than 109 beats per minute. This is equivalent to finding the area under the normal distribution curve to the right of the z-score of 1.48.
Step 4 :Use a standard normal distribution table (z-table) to find this area. However, the z-table gives the area to the left of the given z-score. So, subtract the area to the left of the z-score from 1 to get the area to the right of the z-score.
Step 5 :Looking up 1.48 in the z-table, we find that the area to the left of this z-score is approximately 0.9306.
Step 6 :Calculate the area to the right of the z-score (which is the percentage of healthy adults with a resting pulse rate greater than 109 beats per minute) by subtracting the area to the left of the z-score from 1: \(1 - 0.9306 = 0.0694\).
Step 7 :To express this as a percentage, multiply by 100: \(0.0694 * 100 = 6.94\%\).
Step 8 :So, approximately \(\boxed{6.94\%}\) of healthy adults have a resting pulse rate greater than 109 beats per minute.