Question

3. Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of 188 mg/dL and a standard deviation of 24. a. What percent of adult women do you expect to have cholesterol levels over 200 mg/dL? b. What percent of adult women do you expect to have cholesterol levels between 150 and 170 mg/dL? C. Estimate the interquartile range of the cholesterol levels. d. Above what value are the highest 15% of women's cholesterol levels?

Answer 1

Given data :

Mean=188

Std dev=24

Z score :

$z=\frac{X-\mu }{\sigma }$

(a)  percent of adult women do you expect to have cholesterol levels over 200

200 188 24 = 05

P(Z>0.5)=1-P(Z<0.5)

=1-0.6915

=0.3085

=30.85%

Percent of adult women do you expect to have cholesterol levels over 200 is 30.85%

(b) Cholesterol levle between 150 and 170

$=P(\frac{150-188}{24}<Z<\frac{170-188}{24})$

$=P(-1.5833<Z<-0.75)$

$=P(Z<-0.75)-P(Z<-1.58)$

$=0.2266-0.0571$

$=0.1695$

=16.95%

Percent of adult women do you expect to have cholesterol level between 15 and 170 is 16.95 %

(c) Interquartile range :

Z score for 0.25 = - 0.675 and for z score for 0.75 = 0.675

Value of X=24(-0.675)+188 =171.8

and X=24(0.675)+188=204.2

  Interquartile range : 204.2 - 171.8 = 32.4

(d)

Z score for 0.85 using Z table

Excel syntax

=NORMSINV(0.85)=1.036

Z=1.03

Calculate the value of x

X=Z(std dev)+mean

X=1.036(24)+188

X=212.86

The Cholestorel levels are above 212.86 mg/dL (Approximately 213 mg/dL)