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\rightline{November 13, 2007}

\centerline{\bf Math 102 / Core 143 DX --- Exam II}

\vspace*{0.5cm}
\noindent
Put your answers to all the problems on the numbered sheet, except
for the essays (\#6 and \#7), which go on the back.  The blue book is
for your scratch work and will NOT be graded.

\begin{enumerate}

\item[1.] \textit{(25 points)} A regular dodecahedron (a 12-sided solid)
has its sides numbered 1 through 12.
\begin{itemize}
\item[(a)] What is the conditional probability, given that a roll of this
12-sided die is at least 6, that it is even?
\item[(b)] What is the probability that, on 2 rolls of this die,
the first is even and the second is less than 3 (i.e., 1 or 2)?
\item[(c)] What is the probability that, on 2 rolls of this die,
the first is even \underbar{or} the second is less than 3 (or both)?
\item[(d)] What is the probability that, on 5 rolls, at least one is less
than 3?
\item[(e)] What is the probability that, on 8 rolls, at least 7 are less
than 3?
\end{itemize}

\item[2.] \textit{(10 points)} This question refers to the same dodecahedral
die as in Question~1. The standard deviation of the numbers from 1 to 12 is
about 3.45.  Using this value, find the approximate probability that the
average of 100 rolls of this die will lie between 6 and 7.

\item[3.] \textit{(15 points)} The deuces, treys, and 4's through 10's
are selected from a staight ($=$bridge$=$poker) deck, making a 36-card
deck in which the average card value is 6.
We could (I) pick 20 cards from the deck (replacing the card and shuffling
between picks) or (II) pick 40 cards (under the same conditions).  In
which of these two scenarios, I or II, would the following be more likely?
\begin{itemize}
\item[(a)] The average value of the cards chosen is at least 8.
\item[(b)] The average value of the cards chosen is at least 5.
\item[(c)] The sum of the cards chosen is at least 8 more than
6 times the number of cards chosen.
\item[(d)] The sum of the cards chosen is at least 8 times the number of
cards chosen.
\item[(e)] More than 20\% of the cards chosen are 6's (exactly). (Warning:
This should change your ``box model'' from the one you were using in (a)
through (d).)
\end{itemize}

\item[4.] \textit{(6 points)} The test for the disgusting disease lantzeria
has a sensitivity of 85\% (i.e., if you have it, the test will say you have
it 85\% of the time) and a specificity of 90\% (i.e., if you don't have it,
the test will say you don't 90\% of the time).  Suppose that 5\% of the
population have this disease, and that you test positive
for it: what is the probability that you actually have it? (Steps in the
solution: 1.~Percent of the population who are sick and test positive.
2.~Percent who are well and still test positive. 3.~Percent who test
positive. 4.~Percent who are sick and test positive divided by percent
who test positive.)

\item[5.] \textit{(24 points)} For each of the following data sources,
should we expect the data to follow a normal distribution, a uniform
distribution (all outcomes equally frequent), or neither.  For the
``neither'' responses, give a rough sketch.
\begin{itemize}
\item[(a)] salaries in a large population
\item[(b)] averages salaries of large samples from a large population
\item[(c)] rolls of an icosahedral (20-sided) die, with the numbers
1 through 20 on its faces
\item[(d)] heights in a large population
\end{itemize}

\newpage

\item[6.] \textit{(10 points)} Relative to the article ``Cancer coincidence?
An overwhelmed state office tries to connect disease clusters to pollution
-- but some say the agency is wasting its time'', by Beth Daley: If each
census tract is thought of as one survey, what is the proportion of tracts
that you would expect to be significantly higher than normal? (Later in
the course, ``significantly'' becomes a technical term, but here it means
only your general understanding of the word.)

\item[7.] \textit{(10 points)} Relative to the article ``What is the chance
of your being guilty?'' by John Kay:  What does Kay mean by the ``prosecutor's
fallacy'' and how can this mislead jurors?


\end{enumerate}

\vskip 12 pt
\hrule
\vskip 12 pt

\noindent
{\bf Some possibly useful formulas:}
$$
\frac{n!}{k!(n-k)!}\,p^k(1-p)^{n-k} \qquad\qquad
\sqrt{\frac{{\rm total\ population}-{\rm sample\ size}}{{\rm total\ population}-1}}
$$
$$
\sigma = ({\rm larger}-{\rm smaller})
   \sqrt{\rm(fraction\ with\ larger)(fraction\ with\ smaller)}
   = \sqrt{{\rm average\ of\ }(x-\overline{x})^2}
$$
$$
EV{\rm\ of\ sum\ (or\ count)} = n\cdot(AV{\rm\ of\ box)} \qquad\qquad
SE{\rm\ of\ sum\ (or\ count)} = \sqrt{n}\cdot(SD{\rm\ of\ box)}
$$
$$
EV{\rm\ of\ avg\ (or\ \%)} = AV{\rm\ of\ box} \qquad\qquad
SE{\rm\ of\ avg\ (or\ \%)} = \frac{SD{\rm\ of\ box}}{\sqrt{n}}
$$

\centerline{\begin{tabular}{|cc|cc|cc|cc|cc|}
\hline\multicolumn{10}{|c|}{Normal table (Area between $-z$ and $z$)}\\\hline
$z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%) & $z$ & Area(\%)\\
\hline
0.0 & 0.0 & 0.9 & 63.19 & 1.8 & 92.81 & 2.7 & 99.31 & 3.6 & 99.968 \\
0.05 & 3.99 & 0.95 & 65.79 & 1.85 & 93.57 & 2.75 & 99.4 & 3.65 & 99.974 \\
0.1 & 7.97 & 1 & 68.27 & 1.9 & 94.26 & 2.8 & 99.49 & 3.7 & 99.978 \\
0.15 & 11.92 & 1.05 & 70.63 & 1.95 & 94.88 & 2.85 & 99.56 & 3.75 & 99.982 \\
0.2 & 15.85 & 1.1 & 72.87 & 2 & 95.45 & 2.9 & 99.63 & 3.8 & 99.986 \\
0.25 & 19.74 & 1.15 & 74.99 & 2.05 & 95.96 & 2.95 & 99.68 & 3.85 & 99.988 \\
0.3 & 23.58 & 1.2 & 76.99 & 2.1 & 96.43 & 3 & 99.73 & 3.9 & 99.99 \\
0.35 & 27.37 & 1.25 & 78.87 & 2.15 & 96.84 & 3.05 & 99.771 & 3.95 & 99.992 \\
0.4 & 31.08 & 1.3 & 80.64 & 2.2 & 97.22 & 3.1 & 99.806 & 4 & 99.9937 \\
0.45 & 34.73 & 1.35 & 82.3 & 2.25 & 97.56 & 3.15 & 99.837 & 4.05 & 99.9949 \\
0.5 & 38.29 & 1.4 & 83.85 & 2.3 & 97.86 & 3.2 & 99.863 & 4.1 & 99.9959 \\
0.55 & 41.77 & 1.45 & 85.29 & 2.35 & 98.12 & 3.25 & 99.885 & 4.15 & 99.9967 \\
0.6 & 45.15 & 1.5 & 86.64 & 2.4 & 98.36 & 3.3 & 99.903 & 4.2 & 99.9973 \\
0.65 & 48.43 & 1.55 & 87.89 & 2.45 & 98.57 & 3.35 & 99.919 & 4.25 & 99.9979 \\
0.7 & 51.61 & 1.6 & 89.04 & 2.5 & 98.76 & 3.4 & 99.933 & 4.3 & 99.9983 \\
0.75 & 54.67 & 1.65 & 90.11 & 2.55 & 98.92 & 3.45 & 99.944 & 4.35 & 99.9986 \\
0.8 & 57.63 & 1.7 & 91.09 & 2.6 & 99.07 & 3.5 & 99.953 & 4.4 & 99.9989 \\
0.85 & 60.47 & 1.75 & 91.99 & 2.65 & 99.2 & 3.55 & 99.961 & 4.45 & 99.9991 \\
\hline
\end{tabular}}

\newpage

\Large

\noindent
Name: \mbox{\rule[-1pt]{3truein}{1pt}}

\vskip 12 pt
\noindent
{\bf Your answers to Exam II}

\begin{itemize}

\item[1.] \begin{itemize}
\item[(a)]
\item[(b)]
\item[(c)]
\item[(d)]
\item[(e)]
\end{itemize}

\item[2.]

\item[3.] \begin{itemize}
\item[(a)]
\item[(b)]
\item[(c)]
\item[(d)]
\item[(e)]
\end{itemize}

\item[4.]

\item[5.]\begin{itemize}
\item[(a)]
\item[(b)]
\item[(c)]
\item[(d)]
\end{itemize}

\item[6.] and 7. on back.

\end{itemize}

\newpage
\noindent
6.
\vskip 4 true in

\noindent
7.

\normalsize

\newpage

\noindent
{\bf Math 102 / Core 143 --- Solutions to Exam II}

\begin{enumerate}

\item[1.]\begin{itemize}
\item[(a)] Because four of the seven numbers 6 through 12 are even, the
probability is $4/7$.
\item[(b)] The probability that the first is even is $1/2$, and the
probability that the second is less than 3 is $2/12 = 1/6$, and the
events are independent, so the probability that both happen is $(1/2)(1/6)
= 1/12$.
\item[(c)] $1/2 + 1/6 - 1/12 = 7/12$.
\item[(d)] This is the complement of the event ``getting no number
less than 3 on 5 rolls'', so: $1 - (5/6)^5 \approx 0.6$.
\item[(e)] $C(8,7)(1/6)^7(5/6)^1 + C(8,8)(1/6)^8 = [8(5)+1]/6^8 \approx
0.000024$.
\end{itemize}

\item[2.] The averages of 100 rolls have a nearly normal distribution,
with an EV of 6.5 (the average of the possible values) and an SE of about
$3.45/\sqrt{100} = .345$, so, using the normal table,
$$
P(6<{\rm avg}<7) = P\left(\frac{6-6.5}{.345}<z<\frac{7-6.5}{.345}\right)
= P(-1.45<z<1.45) \approx .85 \ .
$$

\item[3.] \begin{itemize}
\item[(a)] There is more variability in the average with smaller samples
so the average is more likely to be above 8 with 20 cards: I.
\item[(b)] Because 5 is below the expected value of the average, and the
sample average tends to be closer to the expected value of the average when 
the samples are larger, this is more likely with II.
\item[(c)] Because there is more variability in the sum with a larger
sample, this is more likely with II.
\item[(d)] Despite the fact that it is phrased in terms of sum, this
is really the same as part (a), so the answer is I.
\item[(e)] The ``deck'' is now essentially 1's and 0's, meaning 6's (of
which there are 4) and non-6's (of which there are 32).  With
this box model, we should expect only $1/9 \approx 11\%$ of the cards to
be 6's, so 20\% is higher than expected; and variability in percentage is
larger with small samples, so this is more likely with I.
\end{itemize}

\item[4.] Of the 95\% of the population who do not have the disease,
the test will erroneously say that 10\%, or 9.5\% of the population,
actually have it.  Of the 5\% who have the disease, the test will
say that 85\% of them, or 4.25\% of the population, have it.  Thus,
of the 13.75\% of the population who test positive, only 4.25\%
actually have the disease: $4.25\%/13.75\% \approx 31\%$.  So you only
have about a 31\% chance of actually having the disease even though
you tested positive.

\item[5.] \begin{itemize}
\item[(a)] We know this is usually neither normal nor uniform; it has
a long tail (skew) to the right.
\item[(b)] Averages of large samples, even from a non-normal population,
are normally distributed, by the Central Limit Theorem.
\item[(c)] Uniform distribution (at least roughly).
\item[(d)] Normally distributed (the cumulative effect of many small
influences, usually genetic).  Other answers are acceptable, for example
taking into account that men's and women's average heights differ.
\end{itemize}

\item[6.] Just because of chance variation, some census tracts will have
higher than average cases of cancer, and some will have lower.  In fact,
there will be some that have very high incidences, and of course it is
possible that there is some carcinogen in the environment of such a tract.
But some of the high-incidence tracts may be high-incidence just by
accident --- even some that seem significantly (i.e., meaningfully) high,
say 80 or 100 percent above the average rate.  Deciding whether it is chance
or some undiscovered carcinogen is what overwhelms that office.
What level seems ``significant'' to you is of course your own decision.  But
for example, if you thought 80\% above average is significant, then you 
should be guessing how much variability there is in the cancer rate, to 
decide how often that high a cancer rate appears.

\item[7.] Here is the relevant section of the article:

{\leftskip .3 true in
\rightskip .3 true in
The prosecutor's
fallacy is the assertion that, because the story before the court is highly
improbable, the defendant's innocence is equally improbable. But all accounts
of events in high-profile legal cases are highly improbable. That is why
they are high-profile legal cases. The courts do not hear reports of happy
families and normal behaviour. Their services are required only for bizarre
and unlikely incidents --- such as the saga of Nicole and O.J. Simpson and
the tragedy of the Clark family.

So juries are not asked to decide whether the events before them are out of
the run of everyday experience. They are asked to decide on the most
probable explanation of improbable events.

}

We might rephrase this as follows: The prosecutor's fallacy is to encourage
the jury to assume that the probability of the offense being committed is the
same as the probability that the defendant is somehow associated with the
offense (by the investigation) but did not commit it.  Kay points out
that the offense \underbar{is} improbable (fortunately for civilization), but
the jury should be trying to judge the \underbar{conditional} probability that,
given that the offense has occurred (as it has), the defendant was not the one
who committed it.

\end{enumerate}
\end{document}

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