Download HW 2 SOLUTIONS 1. 3.11. Suppose that a medical test has a 92

HW 2 SOLUTIONS
1. 3.11. Suppose that a medical test has a 92% chance of detecting a disease if the person has it
(i.e. 92% sensitivity) and a 94% chance of correctly indicating that the disease is absent if the
person really does not have the disease (i.e. 94% specificity). Suppose that 10% of the
population has the disease.
What is the probability that a randomly chosen person will test positive?
Correct:
What is the probability of having the disease given a positive test?
Correct: By the definition of conditional probability
2. 3.13. The following data table is taken from a study of the relationship between health risk
and income using a large group of people living in Massachusetts. Here "stressed" means the
person reported that most days are extremely stressful or quite stressful; "not stressed" means
that the person reported that most days are a bit stressful, not very stressful, or not at all
stressful.
Low
Stressed
526
Not Stressed 1954
2480
Total
Medium
274
1680
1954
High
216
1899
2115
Total
1016
5533
6549
Use the table to answer the following questions.
a. What is the probability that someone in this study is stressed?
Correct:
1016 / 6549 = 0.155
b. What is the probability that someone is stressed and has low income?
Correct:
526 / 6549 = 0.080
c. What is the probability that someone with a low income in this study is stressed?
Correct:
Pr{stressed|low income} =
Pr{stressed AND low income} / Pr{low income} =
(526/6549) / (2480/6549) =
526 / 2480 =
0.212
d. What is the probability that someone in this study is stressed or has low income?
Correct:
Pr{stressed OR low income) =
Pr{stressed} + Pr{low income) - Pr{stressed AND low income} =
1016/6549 + 2480/6549 - 526/6549 =
0.454
3. 3.18. In a certain population of the European starling, there are 5,000 nests with young. The
distribution of brood size (number of young in a nest) is given in the accompanying table.
Brood Number
Size of Broods
1 90
2 230
3 610
4 1400
5 1760
6 750
7 130
8 26
9 3
10 1
_____________
Total 5,000
a. Calculate the mean brood size for this population of the European starling.
Correct:
μY = [(1)(90) + (2)(230) + (3)(610) + (4)(1400) + (5)(1760) + (6)(750) + (7)(130) + (8)(26) +
(9)(3) + (10)(1)]/5000 = 22435/5000 = 4.487 young
Note that the number of broods in this population had to be turned into probabilities to compute
the population mean (i.e. we need 90/5000 as the probability of a brood of size 1), so rather than
dividing each by 5,000, this calculation divides by 5,000 at the end (just factored it out).
b. Calculate the standard deviation for brood size in this population of the European starling.
Correct:
σY2 = [(1-4.487)2  90 + (2-4.487)2  230 + (3-4.487)2  610 + (4-4.487)2  1400 + (5-4.487)2  1760 +
(6-4.487)2  750 + (7-4.487)2  130 + (8-4.487)2  26 + (9-4.487)2  3 + (10-4.487)2  1]/5000 =
[1094.33 + 1422.59 + 1348.81 + 332.04 + 463.18 + 1716.88 + 820.87 + 320.87 + 61.10 +
30.39]/5000 = 7611.06 / 5000 = 1.522212
Then, σY = √σY2 = √1.522212 = 1.234 young
TI 83/84
STAT -> ENTER to go to Stats List editor.
Enter the values and their probabilities into your STATS List editor.
Enter the brood sizes in List 1. Enter the corresponding numbers of broods in List 2.
Now convert the numbers of broods into probabilities by dividing List 2 by 5,000. Scroll up to
highlight “L2” and hit ENTER. The cursor should be at the bottom of the screen now. Scroll over to
the end of the parenthesis. Hit the “divided by” button and enter the number 5000. Hit ENTER. Your
List 2 should now be turned into probabilities (you just divided each number in List 2 by 5,000).
Compute the population mean and population standard deviation.
STAT -> scroll to the right to highlight CALC -> ENTER. The 1-Var Stats option should be on your
screen. Enter “L1,L2” at the cursor (by hitting 2nd 1 comma 2nd 2) -> ENTER
x-bar is the sample or population mean
sigma x is the population standard deviation (s sub x is the sample standard deviation, which we do not
want here).
c. Find the probability that a randomly selected brood from this population will have more than
7 young.
Correct:
Let Y be the size of the chosen brood. Then
Pr{there are more than 7 in the brood} =
Pr{Y > 7} =
Pr{Y ≥ 8} =
Pr{Y=8} + Pr{Y=9} + Pr{Y=10} =
26/5000 + 3/5000 + 1/5000 =
0.006
4. 3.43. A certain drug causes kidney damage in 1% of patients. Suppose the drug is to be tested
on 50 patients.
a. 3.43a. What is the probability that none of the patients will experience kidney damage?
Correct:
Pr{none will experience kidney damage} =
TI-84 binompdf(50, 0.01, 0) = 0.605
b. What is the probability that fewer than 2 patients will experience kidney damage?
Correct:
Letting Y=number patients who experience kidney damage, we have
Pr{fewer than 2 patients} =
Pr{1 or less patients} =
Pr{Y  1} =
TI-84 binomcdf(50, 0.01, 1) = 0.9105
c. What is the probability that at least one patient will experience kidney damage?
Correct:
Using the complement rule, we have Pr{at least one} =
1 - Pr{none will experience kidney damage} =
1 - Pr{Y=0} =
1 - 0.605 =
0.395
d. What is the probability that more than 2 but at most 10 patients will experience kidney
damage?
Pr {2 < Y  10} =
TI-84 binomcdf(50, 0.01, 10) - binomcdf(50, 0.01, 2) = 0.0138
e. What is expected number of patients to experience kidney damage?
E[patients with damage] = E[Y] = Y = np = 50 x 0.01 = 0.5 patients
5. 3.46. The density curve shown here represents the distribution of systolic blood pressures in a
population of middle-aged men. Areas under the curve are shown in the figure. Suppose a man
is selected at random from the population, and let Y be his blood pressure.
a. Find Pr{Y < 120}
Correct:
0.01 + 0.2 = 0.21
b. Find Pr{Y ≤ 120}
Correct:
This is a continuous distribution, so Pr{Y ≤ 120} = Pr{Y < 120}, which we found to be 0.21 in
part (a).