Bios 603
Practice for Exam One
Answers
1. Suppose the probability of developing disease given exposure
equals .6, and the probability of developing disease given no
exposure equals .2. The probability of being exposed equals .4.
We can build a table of results for 100 hypothetical people.
Since P(E) = .4
Disease
Exposure Yes No
Yes 40
No 60
100
Now use the conditional probabilities
Disease
Exposure Yes No
Yes 24 16 40
No 12 48 60
36 64 100
Find the probability of
a. being exposed and developing disease.
From the table this is 24/100 = .24.
Using the laws of probability P(D and E) = P(D|E)P(E)
= .6 x .4 = .24.
b. being exposed if we know you have disease.
From the table 24/36 = 2/3
Using the laws P(D|E)P(E)/[P(D|E)P(E) + P(D|not E)P(not E)]
= .24/(.24 + .2 x .6) = .24/(.24 + .12) = 2/3.
c. being exposed.
This was given P(E) = .4
P(D) = 36/100 from the table
Using the laws P(D) = P(D|E)P(E) + P(D|not E)p(not E)
= .24 + .12 = .36
2. Suppose the probability that a teenage son smokes if his
father smokes equals .2, and the probability that he smokes if
his father does not smoke equals .1. Suppose that 30% of fathers
smoke.
Build a table for 100 father son pairs
Son
Father Yes No
Yes 6 24 30
No 7 63 70
13 87 100
a. What is the probability that the son smokes?
13/100 = .13 from the table
.2 x .3 + .1 x .7 = .06 + .07 = .13
b. What is the probability that the father and son both smoke?
6/100 from the table
.2 x .3 = .06
c. What is the probability that the father smokes if we know the
son does not smoke?
24/87 from the table
.8 x .3/(.8 x .3 + .9 x .7) = .24/(.24 + .63) = 24/87
We are using Bayes law
P(not S|F)P(F)/(P(not S|F)P(F) + P(not S|not F)P(not F))
3. Rate of Down's Syndrome in the area surrounding a nuclear
power plant was studied. The expected number of cases in a year
was 19. Using probability = .05 as the cutpoint suggesting
excessive risk, how many observed cases would it take to suggest
an increased risk?
Use the Poisson distribution. The rate is 19. If you have a talbe with
this rate, 27 and more sums to .049. If you approximate 19 with 20, 29
and more sums to .036.
4. There are twenty families in a neighborhood. Each family has a
mother, a father and two children. The probability of any individual
getting flu during the season has been estimated to be .2.
a. What is the expected number of mothers with flu?
There are 20 mothers. This is a binomial problem with n = 20, p = .2.
the expected number of mothers is 20 x .2 = 4.
b. Eight mothers had flu. Is this unusual?
The probability of eight or more is .0321. This is unusual.
c. In the eight families in which the mother had flu, three
fathers had flu. Is there any evidence that the mother's and
father's status are dependent?
This is a binomial problem with n = 8, and p still equals .2. The
probability that three or more fathers have flu is .2031. This is not
unusual so we do not have evidence that the events are dependent.
5. A variable is normally distributed with mean 50 and standard
deviation five.
a. What proportion of the population will have scores between 45
and 60?
Convert 60 to a z score. z = (60 - 50)/5 = 2. p= .0228.
b. A researcher wants to select the top ten percent for further
study. What will the cutpoint be?
The z score with 10% of the distribution above it is 1.28.
1.28 = (x - 50)/5
X = 50 + 5(1.28) = 56.4.
6. Suppose the probability that a condom fails is .05. A worker
in HIV suspected a higher than expected failure rate in a certain
brand of condoms. She tested 200 condoms.
a. What is the expected number of failures?
This is a binomial problem with n = 200 and p = .05. The expected
number of failures is 200 x .05 = 10.
7. Twenty people work in an office. The probability of any
individual having a certain trait is .25.
Find the probability of
a. No people having the trait.
This is a binomial problem with n = 20 and p = .25.
The probability is .0032.
b. Three or fewer people having the trait.
p = .2252
c. What is the expected number of people having the trait?
20 x .25 = 5
8. Suppose we expect 7 cases of cleft palate based on a birth
defect monitoring program. Would the occurrence of 12 cases
suggest that the rate has changed.
If you have a table with a rate of seven, the probability of twelve or
more is .027. If you approximate seven with five, the probability is
.005. In either case we have evidence that the rate has changed.
9. A variable is normally distributed with mean 100 and standard
deviation equal 10.
What is the probability that
a. X > 120?
z = (120 - 100)/10 = 2. p = .0228.
b. X is between 120 and 130
z = (130 - 100)/10 = 3.00. P(x > 3.0) = .0013. The proability of being
between 120 and 130 = .0228 - .0013 = .0215.