Question

In a recent study, the Centers for Disease Control and Prevention reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.3 and standard deviation 10.1. (a) What proportion of women have blood pressures lower than 73? (b) What proportion of women have blood pressures between 74 and 90? (c) A diastolic blood pressure greater than 90 is classified as hypertension (high blood pressure). What proportion of women have hypertension? (d) Is it unusual for a woman to have a blood pressure lower than 66? Round the answers to four decimal places.

Answer 1

Part a)

X ~ N ( µ = 80.3 , σ = 10.1 )
P ( X < 73 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 73 - 80.3 ) / 10.1
Z = -0.7228
P ( ( X - µ ) / σ ) < ( 73 - 80.3 ) / 10.1 )
P ( X < 73 ) = P ( Z < -0.7228 )
P ( X < 73 ) = 0.2349

Part b)

X ~ N ( µ = 80.3 , σ = 10.1 )
P ( 74 < X < 90 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 74 - 80.3 ) / 10.1
Z = -0.6238
Z = ( 90 - 80.3 ) / 10.1
Z = 0.9604
P ( -0.62 < Z < 0.96 )
P ( 74 < X < 90 ) = P ( Z < 0.96 ) - P ( Z < -0.62 )
P ( 74 < X < 90 ) = 0.8316 - 0.2664
P ( 74 < X < 90 ) = 0.5652

Part c)

X ~ N ( µ = 80.3 , σ = 10.1 )
P ( X > 90 ) = 1 - P ( X < 90 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 90 - 80.3 ) / 10.1
Z = 0.9604
P ( ( X - µ ) / σ ) > ( 90 - 80.3 ) / 10.1 )
P ( Z > 0.9604 )
P ( X > 90 ) = 1 - P ( Z < 0.9604 )
P ( X > 90 ) = 1 - 0.8316
P ( X > 90 ) = 0.1684

Part d)

X ~ N ( µ = 80.3 , σ = 10.1 )
P ( X < 66 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 66 - 80.3 ) / 10.1
Z = -1.4158
P ( ( X - µ ) / σ ) < ( 66 - 80.3 ) / 10.1 )
P ( X < 66 ) = P ( Z < -1.4158 )
P ( X < 66 ) = 0.0784

No, since the probability is greater than 0.05.