Question

Adult male height (X) follows (approximately) a normal distribution with a mean of 69 inches and a standard deviation of 2.8 inches. Using a statistical package, we find the following the value for x that satisfies P(X < x)

P(X < x) = 0.005, x = 61.78768

P(X < x) = 0.9975, x = 76.8597

How tall must a male be, in order to be among the tallest 0.25% of males? Round your answer to ONE decimal place.

Answer 1

Solution :

mean = = 69

standard deviation = = 2.8

Using standard normal table,

P(Z < z) = 0.005

P(Z < -2.576) = 0.005

z = -2.576

Using z-score formula,

x = z * +

x = -2.576 * 2.8 + 69 = 61.7872

x = 61.7872

P(Z < z) = 0.9975

P(Z < 2.807) = 0.9975

z = 2.807

Using z-score formula,

x = z * +

x = 2.807 * 2.8 + 69 = 76.9596

x = 76.9596

P(Z > z) = 0.25%

1 - P(Z < z) = 0.0025

P(Z < z) = 1 - 0.0025 = 0.9975

P(Z < 2.807) = 0.9975

z = 2.807

Using z-score formula,

x = z * +

x = 2.807 * 2.8 + 69 = 76.9596 = 77.0

x = 77.0

Answer = 77.0