Question

your turn
suppose the heights (in inches) of men (ages 20 - 29) in the united states are normally distrib a mean of 69.3 inches and a standard deviation of 2.92 inches. find each of the following.

1. the percent of men who are between 60.54 inches and 78.06 inches tall.
2. the percent of men who are shorter than 60.54 inches.

Explanation:

Step1: Calculate the number of standard deviations from the mean

For \(60.54\): \(\frac{69.3 - 60.54}{2.92}=\frac{8.76}{2.92} = 3\)
For \(78.06\): \(\frac{78.06 - 69.3}{2.92}=\frac{8.76}{2.92}=3\)

Step2: Use the empirical rule for normal distribution

The empirical rule states that for a normal distribution, about \(99.7\%\) of the data lies within \(3\) standard deviations of the mean.

Step3: Find the percentage for part 2

Since \(99.7\%\) of the data is within \(z=- 3\) and \(z = 3\), the percentage of data outside of this range (in both tails) is \(100 - 99.7=0.3\%\). Since the normal distribution is symmetric, the percentage of data less than \(z=-3\) (i.e., shorter than \(60.54\) inches) is \(\frac{0.3\%}{2}=0.15\%\)

Answer:

1. \(99.7\%\)
2. \(0.15\%\)