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\rightline{April 13, 2010}

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

\vskip 12 pt
\noindent
An unsimplified answer like $12\sqrt{3.51}+6/7$ is usually worth more
than 23.3, because it is easier to understand where it came from.

\begin{enumerate}

\item[1.] \textit{(25 points)} A regular icosahedron (a 20-sided solid)
has its sides numbered 1 through 20.
\begin{itemize}
\item[(a)] What is the conditional probability, given that a roll of this
20-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
statements are true)?
\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 6 are less
than 3?
\end{itemize}

\item[2.] \textit{(10 points)} Which term describes each of the following
probability distributions, related to thousands of rolls of the 20-sided die
in Question~1?  The choices are: ``(approximately) normal'',
``(approximately) uniform'' and
``other''; they may be used more than once or not at all.
\begin{itemize}
\item[(a)] The numbers rolled
\item[(b)] The averages of every 100 rolls of the die
\item[(c)] The products of every 100 rolls of the die
\end{itemize}

\item[3.] \textit{(15 points)} The aces, deuces, treys, and 4's through 9's
are selected from a straight ($=$ bridge $=$ poker) deck, making a 36-card
deck in which the average card value is 5.
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 or neither, would the following be
more likely?
\begin{itemize}
\item[(a)] The average value of the cards chosen is at least 4.
\item[(b)] The average value of the cards chosen is at least 8.
\item[(c)] The sum of the cards chosen is at least 8 more than
5 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}

\newpage

\item[4.] \textit{(25 points)} A person stands in a shopping mall.  She
stops 400 people and asks
each how many times per month they go out for hamburgers, and which of
two national hamburger chains they prefer.  For the 400, the average
number of burger runs per month is 20, with an SD of 15, and 250 say
they prefer MacTavish's hamburgers to Burger Empire's.
\begin{itemize}
\item[(a)] What is a 95\% confidence interval for
the average number of visits to a hamburger restaurant for a person
in that city?
\item[(b)] How many people would she have had to question to reduce
the width of her confidence interval by a factor of 3 (assuming a
similar average and SD in her sample)?
\item[(c)] What is an 80\% confidence interval for the fraction of the
population who prefer MacTavish's burgers to Burger Empire's?
\end{itemize}

\item[5.] \textit{(15 points)} Suppose we have a large population whose incomes
we know; they do not follow the normal curve (as with most income
distributions, they are strongly skewed to the right), but they average
\$35,000, with an SD of \$24,000.  From this population
we randomly choose 900 people and average their incomes.  We repeat
this 400 times.
\begin{itemize}
\item[(a)] Should we expect the distribution of these 400 averages
to be close to a normal curve? Why or why not?
\item[(b)] What should we expect the average of the 400 averages to be?
\item[(c)] How much variability should we expect in the 400 averages?
(Answer in the form ``\blank\ percent of them are likely to be in the
range \blank\ to \blank~.'')
\item[(d)] If instead of 400 times we did this only 100 times, how
would the answers to (a), (b) and (c) change, if at all?
\item[(e)] If instead of using samples of 900, we used samples of 1600
(but again 400 samples), how would the answers to (a), (b) and (c),
if at all?
\end{itemize}

\item[6.] \textit{(10 points)} Related to the articles ``Why Doctors
Hate Science" and "The Myth of Early Detection", both by Sharon Begley:
After these columns were written, the FDA changed its recommendations on
how frequently women should get mammograms. In view of these columns,
do you think they recommended more or less frequently, and why?

\end{enumerate}

\newpage

%\vskip 12 pt
%\hrule
%\vskip 12 pt

\vspace{12pt}
Some Possibly Useful Formulas:
$$
\frac{n!}{k!(n-k)!}\,p^k(1-p)^{n-k}
\qquad\qquad
$$
$$
EV{\rm\ of\ sum\ (count)} = n\cdot(AV{\rm\ of\ box)} \qquad\qquad
SE{\rm\ of\ sum\ (count)} = \sqrt{n}\cdot(SD{\rm\ of\ box)}
$$
$$
EV{\rm\ of\ avg\ (\%)} = AV{\rm\ of\ box} \qquad\qquad
SE{\rm\ of\ avg\ (\%)} = \frac{SD{\rm\ of\ box}}{\sqrt{n}}
$$
$$
SD = ({\rm larger}-{\rm smaller})
   \sqrt{\rm(fraction\ with\ larger)(fraction\ with\ smaller)}
$$


\vskip 24 pt

\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
\strut
\newpage

{
\Large
\baselineskip 24 pt

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

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

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

\end{itemize}

Answer Question~6 on the back of this page.

\newpage

\begin{itemize}
\item[6.]
\end{itemize}

}

\newpage

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

\begin{enumerate}

\item[1.] \begin{itemize}
\item[(a)] Because 8 of the 15 numbers 6 through 20 are even, the
probability is $8/15$.
\item[(b)] The probability that the first is even is $1/2$, and the
probability that the second is less than 3 is $2/20 = 1/10$, and the
events are independent, so the probability that both happen is $(1/2)(1/10)
= 1/20$.
\item[(c)] $1/2 + 1/10 - 1/20 = 11/20$.
\item[(d)] This is the complement of the event ``getting no number
less than 3 on 5 rolls'', so: $1 - (9/10)^5 \approx 0.41$.
\item[(e)] $C(8,6)(1/10)^6(9/10)^2 + C(8,7)(1/10)^7(9/10)^1 + C(8,8)(1/10)^8
= [(8\cdot7/2\cdot1)9^2+ (8)9+1]/10^8 = 0.00002341$.
\end{itemize}

\item[2.] (a) (approximately) uniform.\qquad (b) (approximately) normal.\qquad
(c) other.

\item[3.] \begin{itemize}
\item[(a)] There is more variability in the average with smaller samples,
so the average is more likely to be at least 4 with 40 cards: II.
\item[(b)] Because 8 is above 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 I.
\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 (b), 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.] \begin{itemize}
\item[(a)] Roughly $20\pm 2(15/\sqrt{400}) = 20\pm 1.5$ visits per month.
\item[(b)] To reduce the size of the interval to $20\pm 0.5$ visits per month,
she would have had to interview 9 times as many people, or 3600.
\item[(c)] Using the normal table for an area of 80\%:
$$
\frac{250}{400} \pm
1.3\left(\frac{(1-0)\sqrt{\frac{250}{400}\cdot\frac{150}{400}}}{\sqrt{400}}\right)
= 62.5\% \pm 3.1\%\ .
$$
\end{itemize}

\item[5.] \begin{itemize}
\item[(a)] Yes: By the Central Limit Theorem, no matter what
distribution you start with, the average of many draws from the
distribution will have a distribution that is close to a normal
curve.
\item[(b)] The $EV$ for the average, i.e., the average income
of the population, \$35,000.
\item[(c)] The $SE$ for the average is the $SD$ of the
population divided by the square root of the number in the sample:
\$24,000$/30 =$ \$800. So we should expect about 68\% of the
averages to be between \$34,200 and \$35,800.  (We could also
say that we expect 95\% are between \$33,400 and \$36,600.)
\item[(d)] None of the answers would change: The 400 did not
enter into these calculations; we were just looking at 400
numbers from the distribution of averages.
\item[(e)] The 900 \underbar{did} enter into the computation in
(c): The new $SE$ for the average is \newline
\$24,000$/40 =$ \$600, so about 68\% of the averages are
between \$34,400 and \$35,600.
\end{itemize}

\item[6.] The FDA advised fewer screenings, at longer intervals, because
early detection often meant unnecessary and possibly dangerous treatment,
like removal of tumors that were not threatening.


\end{enumerate}
\end{document}

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