Question

6. A survey was conducted to measure the heights of U.S. men. In the survey, respondents were grouped by age. In the 20-29 age group, the heights were normally distributed, with a mean of 69.9 inches and a standard deviation of 3 inches. A study participant is randomly selected. a) Find the probability that his height is less than 66 inches. b) Find the probability that his height is between 66 and 72 inches. c) Find the probability that his height is more than 72 inches.

Answer 1

a] MEAN= 69.9 INCHES AND SD= 3 INCHES

P ( X<66 )=P ( X−μ<66−69.9 )=P ((X−μ)/σ<(66−69.9)/3)

Since (x−μ)/σ=Z and (66−69.9)/3=−1.3 we have:

P (X<66)=P (Z<−1.3)

Use the standard normal table to conclude that:

P (Z<−1.3)=0.0968

b] P ( 66<X<72 )=P ( 66−69.9< X−μ<72−69.9 )=P ((66−69.9)/3<(X−μ)/σ<(72−69.9)/3)

Since Z=(x−μ)/σ ,( 66−69.9)/3=−1.3 and (72−69.9)/3=0.7 we have:

P ( 66<X<72 )=P ( −1.3<Z<0.7 )

Use the standard normal table to conclude that:

P ( −1.3<Z<0.7 )=0.6612

c] P ( X>72 )=P ( X−μ>72−69.9 )=P ((X−μ)/σ>(72−69.9)/3)

Since Z=(x−μ)/σ and (72−69.9)/3=0.7 we have:

P ( X>72 )=P ( Z>0.7 )

Use the standard normal table to conclude that:

P (Z>0.7)=0.242