Problem 5: Men’s Heights. The distribution of heights of adult American men is approximately Normal with mean 69 inches and standard deviation 2.5 inches. Use the Empirical Rule to answer the following questions. a) What percent of men are taller than 74 inches? b) Between what heights do the middle 95% of men fall? c) What percent of men are between 64 and 66.5 inches tall? d) A height of 71.5 inches corresponds to what percentile of adult male American heights.
The Correct Answer and Explanation is:
To answer these questions, we will use the Empirical Rule (also known as the 68-95-99.7 Rule), which applies to normal distributions. The rule states that for a normal distribution:
- 68% of the data falls within 1 standard deviation of the mean (i.e., between the mean minus 1 standard deviation and the mean plus 1 standard deviation).
- 95% of the data falls within 2 standard deviations of the mean.
- 99.7% of the data falls within 3 standard deviations of the mean.
Given:
- Mean (μ) = 69 inches
- Standard deviation (σ) = 2.5 inches
a) What percent of men are taller than 74 inches?
To find the percent of men taller than 74 inches, we first calculate how many standard deviations 74 inches is above the mean: z=74−692.5=52.5=2z = \frac{74 – 69}{2.5} = \frac{5}{2.5} = 2z=2.574−69=2.55=2
This means 74 inches is 2 standard deviations above the mean. From the Empirical Rule, we know that 95% of the data falls within 2 standard deviations of the mean, leaving 5% in the tails (2.5% in each tail). Since we are interested in the portion of men who are taller than 74 inches, we take the upper tail of 2.5%.
Answer: 2.5% of men are taller than 74 inches.
b) Between what heights do the middle 95% of men fall?
The middle 95% of men fall between 2 standard deviations below and 2 standard deviations above the mean: Lower bound=69−2(2.5)=69−5=64 inches\text{Lower bound} = 69 – 2(2.5) = 69 – 5 = 64 \, \text{inches}Lower bound=69−2(2.5)=69−5=64inches Upper bound=69+2(2.5)=69+5=74 inches\text{Upper bound} = 69 + 2(2.5) = 69 + 5 = 74 \, \text{inches}Upper bound=69+2(2.5)=69+5=74inches
Answer: The middle 95% of men fall between 64 inches and 74 inches.
c) What percent of men are between 64 and 66.5 inches tall?
To calculate this, we first find the z-scores for 64 inches and 66.5 inches.
- For 64 inches:
z=64−692.5=−52.5=−2z = \frac{64 – 69}{2.5} = \frac{-5}{2.5} = -2z=2.564−69=2.5−5=−2
- For 66.5 inches:
z=66.5−692.5=−2.52.5=−1z = \frac{66.5 – 69}{2.5} = \frac{-2.5}{2.5} = -1z=2.566.5−69=2.5−2.5=−1
From the Empirical Rule:
- 68% of data is within 1 standard deviation, so between the mean and 1 standard deviation below (between 69 and 66.5 inches) falls 34%.
- 95% of the data falls within 2 standard deviations, so between 64 and 69 inches falls 47.5% (half of the 95%).
The combined percentage between 64 and 66.5 inches is: Percentage=47.5%−34%=13.5%\text{Percentage} = 47.5\% – 34\% = 13.5\%Percentage=47.5%−34%=13.5%
Answer: 13.5% of men are between 64 and 66.5 inches tall.
d) A height of 71.5 inches corresponds to what percentile of adult male American heights?
We first calculate the z-score for 71.5 inches: z=71.5−692.5=2.52.5=1z = \frac{71.5 – 69}{2.5} = \frac{2.5}{2.5} = 1z=2.571.5−69=2.52.5=1
From the Empirical Rule, we know that 68% of men fall within 1 standard deviation of the mean (between 66.5 and 71.5 inches). Since 71.5 inches is 1 standard deviation above the mean, it corresponds to the 84th percentile (50% for the first standard deviation, plus 34% for the portion between the mean and 71.5 inches).
Answer: A height of 71.5 inches corresponds to the 84th percentile.
In summary:
- 2.5% of men are taller than 74 inches.
- The middle 95% of men fall between 64 inches and 74 inches.
- 13.5% of men are between 64 and 66.5 inches tall.
- A height of 71.5 inches corresponds to the 84th percentile.
