Question

9. High density lipoprotein (HDL) in healthy males follows a normal distribution with a mean of 50 and a standard deviation of 8. a. What proportion of healthy males has HDL exceeding 60? b. What proportion of healthy males has HDL lower than 40? What is the 90th percentile of HDL in healthy males

Answer 1

Solution:

Given,

μ=50, σ=8

a) We need to compute P( x > 60 ) = _____________?

The corresponding z-value needed to be computed is:

GT'I = 08 - 097-x=2

P(> 60) = P(2 > 1.25)

Using excel, =1-NORMSDIST(1.25)

P( x > 60 ) = 0.1056

b) We need to compute P( x < 40 )= __________ ?

The corresponding z-value needed to be computed:

Z = = = 40 - 50 = -1.25

P<40) = P(Z < -1.25)

P( x < 40 )= 0.1056

c) We need to find 90th percentile of HDL in healthy males

That is we are given

P( x < x ) = 0.9 for which we have to find value of x.

For this we first need to find z for area 0.9

Using excel,  =NORMSINV(0.9)

z = 1.282

Formula for x is,

2*0+1 =

x = 50+ 1.282 *8

x = 60.256 $\approx$ 60

90th percentile of HDL in healthy males = 60

Done