Question

The heights of 20- to 29-year-old males in the United States are approximately normal, with mean 70.4 in. and standard deviation 3.0 in.

Round your answers to 2 decimal places.

a. If you select a U.S. male between ages 20 and 29 at random, what is the approximate probability that he is less than 69 in. tall?

The probability is about_______ %.

b. There are roughly 19 million 20- to 29-year-old males in the United States. About how many are between 68 and 69 in. tall?

About ________ million are between the two heights.

c. Find the height of 20- to 29-year-old males that falls at the 90th percentile.

The desired height is __________ inches.

Answer 1

a) Since μ=70.4 and σ=3.0 we have:
P ( X<69 )=P ( X−μ<69−70.4 )=P (X−μσ<69−70.43.0)
Since (x−μ)/σ=Z and (69−70.4)/3.0=−0.47 we have:
P (X<69)=P (Z<−0.47)
Use the standard normal table to conclude that:
P (Z<−0.47)=0.3192

The probability is about 31.92%.

b) Since μ=70.4 and σ=3.0 we have:
P ( 68<X<69 )=P ( 68−70.4< X−μ<69−70.4 )=P ( 68−70.43.0<X−μσ<69−70.43.0)
Since Z=(x−μ)σ , (68−70.4)/3.0=−0.8 and (69−70.4)/3.0=−0.47 we have:
P ( 68<X<69 )=P ( −0.8<Z<−0.47 )
Use the standard normal table to conclude that:
P ( −0.8<Z<−0.47 )=0.1073

c) ) What height represents the 90th percentile?
The probability closest to 0.90 on the table is 0.8997, which happens at z = 1.28.
Solving the equation 1.28 = x−70.4/3.0

x= 1.28*3+70.4

x= 3.84+70.4

x= 74.24
for x, we get that the 90th percentile is 74.24