Question

Suppose that the heights of adult men in the United States are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What proportion of the adult men in United States are less than 6 feet tall? (Hint: 6 feet 72 inches) Round your answer to at least four decimal places. X 5 ?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 70

standard deviation = $\sigma$ = 3

P(x < 72) = P[(x - $\mu$ ) / $\sigma$ < (72 - 70) / 3]

              = P(z < 0.6667)

Probability =0.7475

Proportion of the adult men is United States = 0.7475