Download 1. Shelia`s doctor is concerned that she may suffer from

1. Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood
glucose levels during pregnancy). There is variation both in the actual glucose level and
in the blood test that measures the level. A patient is classified as having gestational
diabetes if the glucose level is above 140 milligrams per deciliter (mg/dl) one hour after a
sugary drink. Shelia's measured glucose level one hour after the sugary drink varies
according to the Normal distribution with µ = 124 mg/dl and σ = 13 mg/dl.
If a single glucose measurement is made, what is the probability (±0.0001) that Shelia is
diagnosed as having gestational diabetes?
z=
140 −124
= 1.230769
13
P ( x > 140 ) = P ( z > 1.230769 ) = 0.1092
If measurements are made on 7 separate days and the mean result is compared with the
criterion 140 mg/dl, what is the probability (±0.0001) that Shelia is diagnosed as having
gestational diabetes?
z=
140 −124
= 3.256309
13 / 7
P ( x > 140 ) = P ( z > 3.256309 ) = 0.0006
2. Shelia's measured glucose level one hour after a sugary drink varies according to the
Normal distribution with µ = 125 mg/dl and σ = 10.8 mg/dl.
What is the level L (±0.1) such that there is probability only 0.03 that the mean glucose
level of 5 test results falls above L?
The z-value corresponding to a right tail area of 0.03 is z = 1.881
Then
L −125
= 1.881
10.8
L = 125 +10.8 (1.881)
L = 145.3
3. Andrew plans to retire in 30 years. He is thinking of investing his retirement funds in
stocks, so he seeks out information on past returns. He learns that over the 101 years from
1900 to 2000, the real (that is, adjusted for inflation) returns on U.S. common stocks had
mean 8.5% and standard deviation 20.8%. The distribution of annual returns on common
stocks is roughly symmetric, so the mean return over even a moderate number of years is
close to Normal.
What is the probability (assuming that the past pattern of variation continues) that the
mean annual return on common stocks over the next 30 years will exceed 12%?
(±0.0001)
z=
12 − 8.5
= 0.9216
20.8 / 30
P ( x > 12 ) = P ( z > 0.9216 ) = 0.1784
What is the probability (±0.0001) that the mean return will be less than 5%?
z=
5 − 8.5
= −0.9216
20.8 / 30
P ( x < 5) = P ( z < −0.9216 ) = 0.1784