Low values HDL (High density lipoprotein) have been found to be associated with
ID: 3074782 • Letter: L
Question
Low values HDL (High density lipoprotein) have been found to be associated with increased risk of heart disease. Assume that HDL levels in a population of women (ages 20 years or more) are approximately normally distributed with mean mu= 55 mg/dL and standard deviation sigma=15.5 mg/dL.
a) Find the probability that a woman in this population chosen at random will have HDL less than 40 mg/dL? (This is below the recommended level of HDL)
b) Find the percent of women who are considered to have a protective level of HDL which is 60 mg/dL or greater.
c) Compute the proportion of women who have HDL levels between 40 and 60 mg/dL.
d) A patient’s HDL level is higher than 75% of the values in this population. Find that patient’s HDL level.
e) A patient’s HDL level is in the lowest 10th percentile. Find that patient’s HDL level.
Explanation / Answer
a)
mu= 55
sigma= 15.5
X= 40
Z=(X-mu)/sigma
=-0.96774
probability =P(Z<-0.96774)
=0.16658
b)
mu= 55
sigma= 15.5
X= 60
Z=(X-mu)/sigma
=0.322580
probability =P(Z>0.32258)
=0.373506
in percentage = 37.35%
c)
mu= 55
sigma= 15.5
X= 60
Z=(X-mu)/sigma
0.322580
probability P1=P(Z<0.32258)
p1= 0.626493
mu= 55
sigma= 15.5
X= 40
Z=(X-mu)/sigma
-0.96774
probability P2=p(Z<-0.966774)
p2= 0.166586634
Required probability is (P1-P2)=0.459906
proportion is 0.4599
d)
mu= 55
sigma= 15.5
X= unknown
probability= 0.75
z-score for probability=0.67448
Z=(X-mu)/sigma
x=(mu+Z*sigma)
x= 65.45459
HDL level =65.45 mg/dL
e)
mu= 55
sigma= 15.5
X= unknown
probability= 0.1
z-score for probability=-1.281551
Z=(X-mu)/sigma
x=(mu+Z*sigma)
x= 35.1359
HDL level= 35.1359 mg/dL
.....................
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