Question

Suppose that the heights of adult men in the United States are normally distributed with a mean of 69 inches and a standard deviation of

3 inches. What proportion of the adult men in United States are at least 6 feet tall? (Hint: 6 feet =72 inches.) Round your answer to at least four decimal places.

Answer 1

Solution :

Given that ,

mean = = 69 inches

standard deviation = = 3 inches

P(x 72) = 1 - P(x   72)

= 1 - P[(x - ) / (72 - 69) / 3]

= 1 -  P(z 1)   

= 1 - 0.8413

= 0.1587

proportion = 0.1587