Homework Assignment #5 Key
STAT 250
Question 1
The table below represents the results of a survey conducted of students at large university not located in
Pennsylvania. Each student was categorized by gender (M/F) and by whether or not the answered yes to the
question “have you be inebriated in the past 30 days?” (Yes/No).
Male
Female
Total
Yes
No
Total
22
10
32
32
36
68
54
46
100
a. What is the probability that a randomly selected student has been inebriated in the last 30 days?
P (I) = 32/100 = 32.0%
b. What is the probability that a randomly selected student is a female? P (F ) = 46/100 = 46.0%
c. What is the probability that a randomly selected student is a male, given that the have been inebriated
in the last 30 days? P (M |I) = 22/32 = 68.8%
d. What is the probability that a randomly selected student has not been inebriated in the last 30 days,
given that they are female? P (N I|F ) = 36/46 = 78.3%
Question 2
A survey was done on home pregnancy tests to determine the sensitivity and specificity of the test. Based
on the following table, please answer the questions below. “Tested Positive” means when taking the home
pregnancy test, they tested for being pregnant.
Tested Positive
Tested Negative
Total
Pregnant
Not Pregnant
Total
456
18
474
128
574
702
584
592
1176
a. What is the sensitivity of the test? That is, what is the probability of a positive test given that they
are pregnant? P (+|P ) = 456/474 = 96.2%
b. What is the specificity of the test? That is, what is the probability of a negative test given that they
are not pregnant? P (−|N P ) = 574/702 = 81.8%
c. What is the positive predictive value? That is, what is the probability that a woman is pregnant given
that they have test positive? P (P |+) = 456/584 = 78.1%
d. What is the negative predictive value? That is, what is the probability that a woman is not pregnant
given that they have tested negative? P (N P |−) = 574/592 = 97.0%
e. Based on all of these results, what can you say about the home pregnancy test? The specificity and
positive predictive values are too low.
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Question 3
In a marine sanctuary, 37% of coral reefs are getting smaller each year. Assume the reefs are independent. A
random sample of six reefs is taken.
1. What is the probability that all six reefs are getting smaller? 0.3%
dbinom(6,6,.37)
## [1] 0.002565726
2. What is the probability that none of the six reefs are getting smaller? 6.3%
dbinom(0,6,.37)
## [1] 0.0625235
3. What is the probability that at least one of the six reefs is getting smaller? 93.7%
1-dbinom(0,6,.37)
## [1] 0.9374765
sum(dbinom(1:6,6,.37))
## [1] 0.9374765
Question 4
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time
they reach adulthood. Assume that the people in the problem are independent.
1. Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution
appropriate for calculating the probability that exactly 89 had chickenpox before they reached adulthood?
Explain. (Hint: what are the four conditions needed for the binomial distribution?)
•
•
•
•
There are only two possible outcomes (yes/no)
There probability of success is constant (90%)
The sample size is fixed (100)
Independence
2. Calculate the probability that exactly 89 out of 100 randomly sampled American adults had chickenpox
during childhood. 12.0%
dbinom(89,100,.90)
## [1] 0.1198776
3. Calculate the probability that exactly 85 out of 100 randomly sampled American adults had chickenpox
during childhood. 3.3%
dbinom(85,100,.90)
## [1] 0.03268244
4. What is the probability that at least 89 out of 100 randomly sampled American adults have had
chickenpox? 70.3%
sum(dbinom(89:100,100,.9))
## [1] 0.7030331
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5. What is the probability that at most 85 out of 100 randomly sampled American adults have not had
chickenpox? 7.3% or 100%
This is the answer as written.
sum(dbinom(0:85,100,.1))
## [1] 1
I wanted to ask this:
sum(dbinom(0:85,100,.9))
## [1] 0.07257297
Question 5
A study has shown that 51% of patients use herbal medications before surgery, against their doctor’s advice.
A random sample of 25 patients was selected. Assume that the patients in the problem are independent.
1. What is the probability that exactly 8 patients have used herbal medications? 2.7%
dbinom(8,25,.51)
## [1] 0.02678869
2. What is the probability that exactly 15 patients have used herbal medications? 10.7%
dbinom(15,25,.51)
## [1] 0.1071264
3. What is the probability that at least 8 patients have used herbal medications? 98.3%
sum(dbinom(8:25,25,.51))
## [1] 0.9830246
4. What is the probability that at most 15 patients have used herbal medications? 86.5%
sum(dbinom(0:15,25,.51))
## [1] 0.864572
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