Question

Men’s heights are normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches.

1. What percent of men are taller than 5 feet?

2. What percent of men can be a flight attendant if you must be between 5’2” and 6’1”?

3. Find the 45th Percentile of men’s heights.

Answer 1

Part a)

X ~ N ( µ = 69.5 , σ = 2.4 )
P ( X > 60 ) = 1 - P ( X < 60 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 60 - 69.5 ) / 2.4
Z = -3.9583
P ( ( X - µ ) / σ ) > ( 60 - 69.5 ) / 2.4 )
P ( Z > -3.9583 )
P ( X > 60 ) = 1 - P ( Z < -3.9583 )
P ( X > 60 ) = 1 - 0
P ( X > 60 ) = 1
Percentage is 1 * 100 = 100%

Part b)

X ~ N ( µ = 69.5 , σ = 2.4 )
P ( 62 < X < 73 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 62 - 69.5 ) / 2.4
Z = -3.125
Z = ( 73 - 69.5 ) / 2.4
Z = 1.4583
P ( -3.12 < Z < 1.46 )
P ( 62 < X < 73 ) = P ( Z < 1.46 ) - P ( Z < -3.12 )
P ( 62 < X < 73 ) = 0.9276 - 0.0009
P ( 62 < X < 73 ) = 0.9267
Percentage is 0.9267 * 100 = 92.67%

Part c)

X ~ N ( µ = 69.5 , σ = 2.4 )
P ( X < x ) = 45% = 0.45
To find the value of x
Looking for the probability 0.45 in standard normal table to calculate critical value Z = -0.1257
Z = ( X - µ ) / σ
-0.1257 = ( X - 69.5 ) / 2.4
X = 69.1983 ≈ 69.2 inches
P ( X < 69.2) = 0.45