For the population of females between the ages of 3 and 74 who participated in t
ID: 2927798 • Letter: F
Question
For the population of females between the ages of 3 and 74 who participated in the Na-
tional Health Interview Survey, the distribution of hemoglobin levels has mean
(the greek letter mu) = 13.3g/100ml and standard deviation
sigma = 1.12g/100ml.
(a) What proportion of females in this population have a hemoglobin level between 13.0
and 13.5 g/100 ml?
(b) If repeated samples of size 15 are selected from this population, what proportion of
the samples will have a mean hemoglobin level between 13.0 and 13.5 g/100 ml?
(c) If the repeated samples are of size 30, what proportion will have a mean between
13.0 and 13.5 g/100 ml?
Explanation / Answer
ANSWERS
A.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 13.3
standard Deviation ( sd )= 1.12
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 13) = (13-13.3)/1.12
= -0.3/1.12 = -0.2679
= P ( Z <-0.2679) From Standard Normal Table
= 0.3944
P(X < 13.5) = (13.5-13.3)/1.12
= 0.2/1.12 = 0.1786
= P ( Z <0.1786) From Standard Normal Table
= 0.5709
P(13 < X < 13.5) = 0.5709-0.3944 = 0.1765
B.
Sample size (n) = 15
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 13) = (13-13.3)/1.12/ Sqrt ( 15 )
= -0.3/0.2892
= -1.0374
= P ( Z <-1.0374) From Standard Normal Table
= 0.1498
P(X < 13.5) = (13.5-13.3)/1.12/ Sqrt ( 15 )
= 0.2/0.2892 = 0.6916
= P ( Z <0.6916) From Standard Normal Table
= 0.7554
P(13 < X < 13.5) = 0.7554-0.1498 = 0.6056
C.
When sample size is 30
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 13) = (13-13.3)/1.12/ Sqrt ( 30 )
= -0.3/0.2045
= -1.4671
= P ( Z <-1.4671) From Standard Normal Table
= 0.0712
P(X < 13.5) = (13.5-13.3)/1.12/ Sqrt ( 30 )
= 0.2/0.2045 = 0.9781
= P ( Z <0.9781) From Standard Normal Table
= 0.836
P(13 < X < 13.5) = 0.836-0.0712 = 0.7648