Question

Suppose that the heights of adult women in the United States are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. Jennifer is taller than 90% of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place. inches x 6

Answer 1

Given that,

mean = $\mu$ = 65

standard deviation = $\sigma$ = 2.5

Using standard normal table,

P(Z > z) = 90%

= 1 - P(Z < z) = 0.90  

= P(Z < z ) = 1 - 0.90

= P(Z < z ) = 0.10

= P(Z < z ) = 0.10  

z = -1.28

Using z-score formula  

x = z $\sigma$ + $\mu$

x = -1.28 *2.5+65

x=61.8