Question

Men’s heights in the USA are normally distributed with a mean of 69 inches and a standard deviation of 2.7 inches.

(a) What is the probability that a randomly selected man has a height of at least 68 inches?

(b) What height represents the 96^th percentile?

Answer 1

Part a)

X ~ N ( µ = 69 , σ = 2.7 )
P ( X ≥ 68 ) = 1 - P ( X < 68 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 68 - 69 ) / 2.7
Z = -0.3704
P ( ( X - µ ) / σ ) > ( 68 - 69 ) / 2.7 )
P ( Z > -0.3704 )
P ( X ≥ 68 ) = 1 - P ( Z < -0.3704 )
P ( X ≥ 68 ) = 1 - 0.3555
P ( X ≥ 68 ) = 0.6445

Part b)

X ~ N ( µ = 69 , σ = 2.7 )
P ( X < x ) = 96% = 0.96
To find the value of x
Looking for the probability 0.96 in standard normal table to calculate Z score = 1.7507
Z = ( X - µ ) / σ
1.7507 = ( X - 69 ) / 2.7
X = 73.7269 ≈ 74 inches
P ( X < 73.7269 ) = 0.96