If a man is randomly selected, find the probability that his height is more 72 inches.

Question

7. According to a National Health Survey, American men’s heights are normally distributed with a mean given by 𝜇 = 69.7 inches and a standard deviation given by 

𝜎 = 2.8 inches. 

a). If a man is randomly selected, find the probability that his height is more 72 inches.

Answer 1

a)

z = (x - µ)/σ = (72-69.7)/2.8 ≈ .82

If we look at a positive z-score table, where 0.8 and 0.02 intersect, we find:

0.79389

Therefore, 79.4% of men are shorter than 72 inches.

1 - .79389 = .20611 (or 100% - 79.4% = 20.6%)

Therefore, 20.6% of men are taller than 72 inches.

b)

z = (x - µ)/σ = (68-69.7)/2.8 ≈ -0.61

Looking at the negative z-score table where -0.6 and 0.01 intersect, we find:

0.27093

Subtracting the two, we have:

0.79389 - 0.27093 = 0.52296

Therefore, 52.3% of men are between 68 and 72 inches tall.