Suppose that the heights of adult men in the United States are normally distributed with a mean of 70.5 inches and a standard deviation of 3 inches. What proportion of the adult men in United States are at most 6 feet tall? (Hint: 6 feet = 72 inches.) Round your answer to at least four decimal places.
The Correct Answer and Explanation is:
To find the proportion of adult men in the United States who are at most 6 feet tall (72 inches), we can use the Z-score formula for a normal distribution: Z=X−μσZ = \frac{X – \mu}{\sigma}Z=σX−μ
Where:
- ZZZ is the Z-score,
- XXX is the value we are interested in (72 inches),
- μ\muμ is the mean (70.5 inches),
- σ\sigmaσ is the standard deviation (3 inches).
Step 1: Calculate the Z-score
Given:
- X=72X = 72X=72,
- μ=70.5\mu = 70.5μ=70.5,
- σ=3\sigma = 3σ=3,
Substitute these values into the Z-score formula: Z=72−70.53=1.53=0.5Z = \frac{72 – 70.5}{3} = \frac{1.5}{3} = 0.5Z=372−70.5=31.5=0.5
Step 2: Find the Proportion
The Z-score of 0.5 corresponds to the standard normal distribution. We now need to find the proportion of the population that lies to the left of this Z-score (i.e., at most 72 inches).
To find this, we can refer to a Z-table (or use a calculator). A Z-score of 0.5 corresponds to a cumulative probability of about 0.6915.
Step 3: Interpret the Result
This means that approximately 69.15% of adult men in the United States are at most 6 feet tall (72 inches).
Final Answer:
The proportion of adult men in the United States who are at most 6 feet tall is 0.6915 (or 69.15%).
