Final answer:
The percent of adult males in the United States who are no more than 68 inches tall is approximately 2.28%, as calculated by the z-score and the standard normal distribution table.
Step-by-step explanation:
To find the percent of men who are no more than 68 inches tall, we use the properties of the normal distribution. Given that adult male heights are normally distributed with a mean of 72 inches and a standard deviation of 2 inches, we first calculate the z-score for a height of 68 inches.
The z-score formula is: z = (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation.
For a height of 68 inches, the z-score is:
z = (68 - 72) / 2
z = -4 / 2
z = -2
This means that a height of 68 inches is 2 standard deviations below the mean. We then consult a standard normal distribution table to find the area to the left of a z-score of -2, which gives us the percent of men who are no more than 68 inches tall. This value is approximately 2.28%, meaning that about 2.28% of adult males in the United States are 68 inches tall or below.