Question

1. Assume the cholesterol levels of adult American women are normally distributed with a mean of 190 mg/dL and a standard deviation of 26 mg/dL.
   1. What percent of adult women do you expect to have cholesterol levels of at least 200 mg/dL ? (Round to tenths)   
   2. b What percent of adult women do you expect to have cholesterol levels between 150 and 170 mg/dL ? (Round to tenths)
   3. Above what value are the top 15% of women’s cholesterol levels ? (Round to tenths)

4. What cholesterol level would represent the 10 th percentile ? (Round to tenths)

I am using A TI84. Can you explain how to plug into the calculator?

Answer 1

a)

probability =P(X>200)=P(Z>(200-190)/26)=P(Z>0.38)=1-P(Z<0.38)=1-0.6497=0.3503~ 35.0%

if using ti-84 press 2nd -vars -use command :normalcdf(200,1000000,190,26)

b)

probability =P(150<X<170)=P((150-190)/26)<Z<(170-190)/26)=P(-1.54<Z<-0.77)=0.2209-0.062=0.1589~15.9%

if using ti-84 2nd -vars - use command :normalcdf(150,170,190,26)

c)

for 85th percentile critical value of z=		1.04
therefore corresponding value=mean+z*std deviation=			216.9

(use invnorm(0.85,190,26)

d)

for 10th percentile critical value of z=		-1.28
therefore corresponding value=mean+z*std deviation=			156.7