Question 20 1 pts According to government reports, the heights of adult male residents of the United States are approximately normally distributed with mean of 69.0 inches and a standard deviation of 2.8 inches. Find the probability of adult male residents in the U.S. whose heights are between 63 and 72 inches. (Round your answer to the nearest four decimal places) Question 21 1 pts

Step 1 ：The problem is asking for the probability of adult male residents in the U.S. whose heights are between 63 and 72 inches. This is a problem of normal distribution. We know that the mean (\(\mu\)) is 69.0 inches and the standard deviation (\(\sigma\)) is 2.8 inches.

Step 2 ：We can use the Z-score formula to standardize the heights and then use the standard normal distribution table (or a function that gives the cumulative probability for a standard normal distribution) to find the probabilities.

Step 3 ：The Z-score formula is: \(Z = \frac{X - \mu}{\sigma}\) where X is the value from the original distribution that we're interested in, \(\mu\) is the mean of the original distribution, and \(\sigma\) is the standard deviation of the original distribution.

Step 4 ：We need to find two Z-scores: one for X = 63 and one for X = 72. Then we can find the probabilities associated with these Z-scores and subtract the smaller probability from the larger one to get the probability of a height being between 63 and 72 inches.

Step 5 ：For X = 63, the Z-score is approximately -2.142857142857143 and the associated probability is approximately 0.016062285603828316.

Step 6 ：For X = 72, the Z-score is approximately 1.0714285714285714 and the associated probability is approximately 0.8580116141245442.

Step 7 ：Subtracting the smaller probability from the larger one, we get the probability of a height being between 63 and 72 inches is approximately 0.8419493285207159.

Step 8 ：Final Answer: The probability of adult male residents in the U.S. whose heights are between 63 and 72 inches is approximately \(\boxed{0.8419}\).