Final answer:
The height of the shortest man in the top 10% of a normal distribution with a mean of 73.4 inches and a standard deviation of 2.7 inches is approximately 76.86 inches. This calculation is based on the z-score for the 90th percentile, which is 1.28.
Step-by-step explanation:
To find the height of the shortest man in the top 10% of a normal distribution, we should first use the z-score associated with the top 10%. The z-score that corresponds to the top 10% (90th percentile) is approximately 1.28. We then use the z-score formula to find this man's height:
z = (X - μ) / σ
Where:
- X is the height of the individual,
- μ is the mean height,
- σ is the standard deviation.
Given the survey's mean height (μ) is 73.4 inches and the standard deviation (σ) is 2.7 inches, we can substitute these values and the z-score into the formula to solve for X:
1.28 = (X - 73.4) / 2.7
Multiplying both sides by 2.7 gives:
1.28 * 2.7 = X - 73.4
Which simplifies to:
X = 1.28 * 2.7 + 73.4
X = 76.856 inches
The shortest man in the top 10% would be approximately 76.86 inches tall.