Question

The heights of American women 18 to 29 are normally distributed with a mean of 65 inches and a standard deviation of 3.5 inches. An American woman in this age bracket is chosen at random.

What is the probability that she is more than 68 inches stall?

What is the probability that she is less than 64 inches tall?

What is the probability that she is between 63 and 67 inches tall?

What is the probability that an American woman's height differs from the mean by more than 2.5 inches?

Answer 1

a)

X ~ N ( µ = 65 , σ = 3.5 )
We covert this to standard normal as
P ( X < x) = P ( (Z < X - µ ) / σ )
P ( X > 68 ) = P(Z > (68 - 65 ) / 3.5 )
= P ( Z > 0.86 )
= 1 - P ( Z < 0.86 )
= 1 - 0.8051
= 0.1949

b)

P ( ( X < 64 ) = P ( Z < 64 - 65 ) / 3.5 )
= P ( Z < -0.29 )
P ( X < 64 ) = 0.3859

c)

P ( 63 < X < 67 ) = P ( Z < ( 67 - 65 ) / 3.5 ) - P ( Z < ( 63 - 65 ) / 3.5 )
= P ( Z < 0.57) - P ( Z < -0.57 )
= 0.7157 - 0.2843
= 0.4313

d)

P( | x - | > 2.5) = ?

P( | x - | > 2.5) = 1 - P( | x - | <= 2.5)

P( | x - | <= 2.5) = P( -2.5 < x - < 2.5)

= P(-2.5 + < X < 2.5 + )

= P(-2.5 + 65 < X < 2.5 + 65 )

= P( 62.5 < X < 67.5)

= P ( Z < ( 67.5 - 65 ) / 3.5 ) - P ( Z < ( 62.5 - 65 ) / 3.5 )
= P ( Z < 0.71) - P ( Z < -0.71 )
= 0.7611 - 0.2389
= 0.5223

P( | x - | > 2.5) = 1 - 0.5223

= 0.4777