More Sample Exam Questions
Exam 2
Scott Stevens
COB 191
1.
The US Supreme Court once compared 6-person juries and 12-person juries, and declared that
“there is no discernable difference between the results reached by the two different sized juries”.
Since jurors are not permitted to discuss the case with anyone until all of the testimony has been
heard, we will look at an initial secret ballot conducted by the jury immediately after the testimony
has been completed. Assume that the jurors are a random sample from a population in which 90%
of all people would vote for conviction on this original ballot. We will examine the question:
How likely is a conviction verdict on this first ballot?
a.
b.
c.
d.
e.
Why is it appropriate to use the binomial distribution to solve this problem? Why would
the binomial distribution not necessarily be appropriate on a second (or later) ballot?
Many cases require a unanimous verdict by the jury members. What is the exact
probability of a unanimous guilty verdict for a six-person jury? A twelve-person jury?
Suppose that a simple majority of jury members is required for a guilty verdict. What is
the exact probability of a guilty verdict in a six-person jury? A twelve-person jury?
Suppose you were asked to approximate your answer to part c by using the normal
approximation to the sampling distribution. Why can you not justifiably do so?
Apply the normal approximation to the sampling distribution to the 12-person jury
problem anyway, even though it is inappropriate. (Look at how your answer compares to
the real one in part c.)
Answer:
Notes: On a real test, you would not need to do the work for both 6 and 12 member juries, but it’s
worthwhile to do so here, for purposes of comparison. You would also not be asked to solve problem c on
the in class exam, since doing so without BINOMDIST is quite time consuming.
My answers here are intended as explanation as well as solution, so they go on a bit. Key points are
boldfaced in the discussion below.
a.
Each person is either a "success" or "failure" in his or her initial vote, with success here
being defined as a conviction vote. Since we view the jury as being a random sample from a
population of which 90% would vote for conviction, we have met the requirements of the
binomial model. We have each "trial" (juror) being a "success" (convict vote) or "failure"
(acquit vote), all with the same probability of success (0.9), and all independent of one
another (since no discussion has yet taken place). Note that the assumptions of the binomial
are unlikely to be satisfied by later votes, since the votes of some jury members may alter the
probabilities of other jury members.
b.
We'll proceed by using the binomial model. Since the jury has less than 1000 people (!), Excel's
BINOMDIST function can handle it nicely.
For a six person jury, the chance of a unanimous conviction vote with p = 0.9 is
=BINOMDIST(6,6,0.9,false), while for a 12 person jury, the calculation is
=BINOMDIST(12,12,09.,false). (Remember the "false" tells Excel that you want exactly this
many successes. "true" means "this number of successes or less".) You should be able to do this
particular problem by hand, too. See why?
Anyway, the numbers come out to be 0.5314 for the six-person jury and 0.2824 for the twelveperson jury. That is, a conviction on the first ballot is almost twice as likely with a six-person
jury. You shouldn't be surprised. Since in either case, it takes only one "acquittal" vote to kill the
unanimous conviction vote, it should be clear that we're more likely to find an "acquit" vote in
twelve people than we are to find one in six people.
Moral: if most people think you're guilty, get as big a jury as you can!
To do the 12-person jury problem by hand, we use the binomial formula with p = 0.9, n = 12,
and k (the number of successes) = 12. The math gives
12 
 0.912 0.11212 = 1  0.912  1 = 0.912 = 0.5314
12 
as before.
Again, note that this problem can't answer the question of what the jury's final verdict will be. If
the first ballot is not unanimous, more ballots must take place after discussion. Since a person
who voted "convict" on the first ballot is more likely to vote "convict" again than a person who
originally voted "acquit" on the first ballot, later ballots no longer satisfy the assumption that each
"trial" has an equal chance of "success".
c.
If a simple majority is sufficient for conviction, we want (for the 12-person jury case) P(K > 7),
where K is the number of guilty votes. In Excel, the probability of 6 or less guilty votes is
=BINOMDIST(6,12,0.9,TRUE). The TRUE tells Excel to compute the probability of six or
less successes. This answer is 0.000541. The chance of 7 or more guilty votes is then 1 –
0.000541, or 0.999459. (This is because P(K>7) = 1 – P(K<6). Think about it.)
For a six person jury, we proceed similarly. We want P(K > 4) = 1 – P(K < 3)
= 1 – BINOMDIST(3,6,0.9,TRUE) = 0.98415. This time, if most people would convict you,
you’re much more likely to get off with a smaller jury. In our problem, a smaller jury means that
you’re more than 29 times as likely to not have a conviction on the first ballot!
This problem could be done by hand, using a formula similar to the one in part b. The problem is
that we would have to find the probabilities of: 12 guilty votes (as found in b), 11 guilty votes, 10
guilty votes, 9 guilty votes, 8 guilty votes, and 7 guilty votes, then add these six numbers together.
That’s a lot of work by hand!
d.
To use this approximation, the expected number of successes and failures must be at least 5.
But in our problem, p = 0.9. This means that the expected successes for a 12 person jury are 12 
0.9 = 10.8 (which is fine), and the expected failures are 12  0.1 = 1.2 (which is not fine). In the
case of a six person jury, it’s even worse. We have 6  0.9 = 5.4 expected successes, and 6  0.1
= 0.6 expected failures. So we cannot use the technique in either case. (Note that if the case were
one in which p = 0.5, then we could do the 12 person-jury case.
e.
In the approximation, we use  = p = 0.9. We use  = SQRT(0.9  (1 – 0.9)) = SQRT(0.09) = 0.3.
Then the sampling distribution of the sample proportion is distributed approximately as a normal
curve with a mean of 0.9 and a standard deviation of 0.3/SQRT(12) = 0.0866. This curve is
shown below, along with the actual sampling distribution for the proportion. We want to know:
How likely is it that a majority of the jury (sample) will vote “guilty”? That is, we want to know
P(ps > 0.5).
Actual and "Approximated" Distributions of
% of 12 Person Jury Voting "Guilty"
probability
0.4
0.3
0.2
0.1
1.33
1.25
1.17
1.08
1.00
0.92
0.83
0.75
0.67
0.58
0.50
0.42
0.33
0.25
0.17
0.08
0.00
0
% voting "guilty"
As you can see, the normal approximation (in blue) is just silly. It shows nonzero probability of more than
100% of the jury voting guilty! Still, as instructed, we’ll find the required probability. In Excel, it’s
=1 - NORMDIST(.5,0.9,0.0866, TRUE), which turns out to be 0.999998. Compare this to the true answer
of 0.999459. That is, the true chance of “getting off” on the first ballot is less than 1/10 as great at the
approximated value.
(You might not think this approximation is so bad—and it isn’t. But that’s only because we wanted to
know the probability of 7 or more guilty votes. As an example, suppose we tried to use it to predict the
probability of a unanimous vote. The approximation would be P(p s > 1.00)
= 1 – NORMDIST(1, 0.9, 0.0866, TRUE) = .1241. That is, the approximation says that there is a 12.41%
chance of a unanimous verdict. The actual answer was .5314!
Do to the calculation by hand (as you’d need to on a test) we’d need to find the z score for 0.5. This is z =
(0.5 – 0.9)/0.3