Suppose that the heights of adult men in the United States are normally distributed with a mean of 71

inches and a standard deviation of 3.5

inches. What proportion of the adult men in the United States are at most 6

feet tall? (Hint: 6

feet =

72

inches.) Round your answer to at least four decimal places.

We are asked to find the proportion of adult men in the United States who are at most 6 feet tall, which is equivalent to 72 inches.

The heights are normally distributed with:
- Mean (\(\mu\)) = 71 inches
- Standard deviation (\(\sigma\)) = 3.5 inches

To find the proportion of men with heights at most 72 inches, we need to standardize this height using the Z-score formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \(X = 72\) inches (the height we are interested in),
- \(\mu = 71\) inches (the mean),
- \(\sigma = 3.5\) inches (the standard deviation).

Let's calculate the Z-score:

\[
Z = \frac{72 - 71}{3.5} = \frac{1}{3.5} \approx 0.2857
\]

Next, we look up the Z-score of 0.2857 in a standard normal distribution table or use a calculator to find the cumulative probability. This gives us the proportion of men who are at most 72 inches tall.

Using a standard normal table or a calculator, we find that the cumulative probability for a Z-score of 0.2857 is approximately 0.6129.

Thus, the proportion of adult men in the U.S. who are at most 6 feet tall is approximately **0.6129** or **61.29%**.

### Do you have any questions or need more details?

Here are five related questions for further exploration:
1. What proportion of adult men are taller than 6 feet?
2. What is the probability that a randomly selected man is shorter than 5 feet 10 inches (70 inches)?
3. If a man is 6 feet 2 inches tall, what is his corresponding Z-score?
4. How would the proportion change if the standard deviation were larger?
5. What is the range of heights that covers 95% of adult men?

**Tip**: Remember that the area under the normal distribution curve represents probabilities, and Z-scores help convert values to standardized probabilities.

Suppose that the heights of adult men in the United States are normally distributed with a mean of 71 inches and a standard deviation of 3.5 inches. What proportion of the adult men in the United States are at most 6 feet tall? (Hint: 6 feet = 72 inches.) Round your answer to at least four decimal places.

Learn how to calculate the proportion of adult men in the United States who are at most 6 feet tall using the normal distribution. This problem covers key statistical concepts like Z-scores and standard normal distribution with a mean of 71 inches and a standard deviation of 3.5 inches. Suitable for high school or college-level statistics students.

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