Question

The heights of men in the United States are normally distributed with a mean of 69.1 inches and a standard deviation of 2.9 inches. What proportion of men are taller than 6 feet (72 inches)

Answer 1

Using the normal distribution, it is found that 0.1587 = 15.87% of men are taller than 6 feet (72 inches).

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Normal Probability Distribution  

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.
• After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 69.1 inches, thus $\mu = 69.1$.
• Standard deviation of 2.9 inches, thus $\sigma = 2.9$.
• The proportion above 72 inches is 1 subtracted by the p-value of Z when X = 72, then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{72 - 69.1}{2.9}$

$Z = 1$

$Z = 1$ has a p-value of 0.8413.

1 - 0.8413 = 0.1587.

0.1587 = 15.87% of men are taller than 6 feet (72 inches).

A similar problem is given at https://brainly.com/question/22934264