Question

Women have a mean systolic blood pressure of 125.17 with a standard deviation of 10.34. Female blood pressure is known to be normally distributed.
(a) Find the probability that a randomly selected female has a blood pressure below 110.

(b) What female systolic blood pressure represents the 99th percentile?




(c) A group of 45 women who take an allergy drug have a mean systolic blood pressure under 120. Should the drug company include a warning about users having lower systolic blood pressure? Justify your answer using probability.

Answer 1

a)

X ~ N ( µ = 125.17 , σ = 10.34 )
We convert this to standard normal as
P ( X < x ) = P ( Z < ( X - µ ) / σ )
P ( ( X < 110 ) = P ( Z < 110 - 125.17 ) / 10.34 )
= P ( Z < -1.47 )
P ( X < 110 ) = 0.0708

b)

X ~ N ( µ = 125.17 , σ = 10.34 )
P ( X < x ) = 99% = 0.99
To find the value of x
Looking for the probability 0.99 in standard normal table to calculate critical value Z = 2.3263
Z = ( X - µ ) / σ
2.3263 = ( X - 125.17 ) / 10.34
X = 149.22

c)

X ~ N ( µ = 125.17 , σ = 10.34 )
P ( X < 120 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 120 - 125.17 ) / ( 10.34 / √45 )
Z = -3.35
P ( ( X - µ ) / ( σ/√(n)) = ( 120 - 125.17 ) / ( 10.34 / √(45) )
P ( X < 120 ) = P ( Z < -3.35 )
P ( X̅ < 120 ) = 0.0004

Since this probbaility is less than 0.05, the event is unusual.

The drug company should include a warning about users having lower systolic blood pressure.