In a survey of men in the United
States (ages 20-29), the mean height was 69.6 inches with a standard deviation of 3.0 inches. Find the minimum height in the top 16%.
μ=69.6σ=3.0P(X>x)=0.16P(X<x)=1−0.16=0.84P(Z<x−μσ)=0.84x−69.63.0=0.995x−69.6=3.0×0.995x=69.6+2.985=72.585\mu=69.6 \\ \sigma=3.0 \\ P(X>x) = 0.16 \\ P(X<x) = 1-0.16 = 0.84 \\ P(Z< \frac{x-\mu}{\sigma}) = 0.84 \\ \frac{x-69.6}{3.0}=0.995 \\ x-69.6= 3.0 \times 0.995 \\ x = 69.6+2.985 = 72.585μ=69.6σ=3.0P(X>x)=0.16P(X<x)=1−0.16=0.84P(Z<σx−μ)=0.843.0x−69.6=0.995x−69.6=3.0×0.995x=69.6+2.985=72.585
Answer: 72.58 inches
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