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\rightline{9 November 2010}

\centerline{\bf Math 102 / Core 143 CX --- 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{(17 points)} A regular dodecahedron
(a 12-sided solid) has its sides numbered 1 through 12.
\begin{itemize}
\item[(a)] What is the conditional probability that, given
a roll of this 12-sided die is at least 6, 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 5 rolls,
exactly 3 are less than 3?
\end{itemize}

\item[2.] \textit{(9 points)} Which term describes each
of the following probability distributions, related to
rolls of the 12-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{(12 points)} The 3's through 9's are
selected from a straight ($=$ bridge $=$ poker) deck
of cards, making a 28-card deck in which the average
card value is 6.  We could (I) pick 100 cards from the
deck (replacing the card and shuffling between picks)
or (II) pick 10 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 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.
\end{itemize}

\item[4.] \textit{(12 points)} If we flip a coin an
even number of times, the most likely outcome is that
the number of heads is exactly half the number of
flips. \begin{itemize}
\item[(a)] What is the probability of that outcome if
the number of flips is 2?
\item[(b)] What is the probability of that outcome if
the number of flips is 4?
\item[(c)] Your answer in (b) should be smaller than
that in (a).  Does this contradict the Law of Averages
in the case when an experiment is repeated only a few
times?
\end{itemize}

\newpage

\item[5.] \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 25, with an SD of 12,
and 280 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 burger runs 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[6.] \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~.'')
\end{itemize}

\item[7.] \textit{(10 points)} Related to the article
``Often wrong, but never in doubt,'' by Richard H.
Thaler: How did you do in the "quiz" posed by the
author? Do you feel you are "miscalibrated"? (Remember
that, in reading your answers, I am not looking for a
``correct'' response of some sort, just evidence that
you have read the article and thought about your
answer.)

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

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

\item[4.]\begin{itemize}
\item[(a)]
\item[(b)]
\item[(c)]
\end{itemize}

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

\item[6.]\begin{itemize}
\item[(a)]
\item[(b)]
\item[(c)]
\end{itemize}

\end{itemize}

Answer Question~7 on the back of this page.

\newpage

\begin{itemize}
\item[7.]
\end{itemize}

}

\newpage

\noindent
{\bf Solutions to Exam II}

\begin{enumerate}

\item[1.] \begin{itemize}
\item[(a)] Because 4 of the 7 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:
$1 - (5/6)^5 \approx 0.598$.
\item[(e)] $C(5,2)(1/6)^3(5/6)^2 = 10\cdot 25/6^5 
= 125/3888 \approx 3.2\%$.
\end{itemize}

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

\item[3.] \begin{itemize}
\item[(a)] There is less variability in the average with larger samples,
so the average is more likely to be closer to the average of 6, and hence at
least 4, with more picks: I.
\item[(b)] Because 8 is above the expected value of the average, 6, to get
to 8 or above is to get \underbar{away} from the expected value, which is
more likely with smaller samples: II
\item[(c)] Because there is more variability in the \underbar{sum} with a
larger sample, this is more likely with I.
\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 II.
\end{itemize}

\item[4.] \begin{itemize}
\item[(a)] $C(2,1)(1/2)^1(1/2)^1 = 2(1/2)(1/2) = 1/2$.
\item[(b)] $C(4,2)(1/2)^2(1/2^2) = 6(1/4)(1/4) = 3/8$.
\item[(c)] No, it doesn't. The Law of Averages says that the fraction
of heads will usually be \underbar{closer} to the expected fraction, 1/2,
with more flips (because the $SE$ for that fraction gets smaller as $n$
gets larger); but ``closer'' doesn't mean exactly at that fraction.  (Math
Nerd Alert: What follows is too technical for an intro class, which
is why I didn't talk about it.) The probability of
getting exactly half heads in $2n$ flips (let's call it $Q(n)$) is
$$
Q(n) = C(2n,n)(1/2)^n(1/2)^n
   = \frac{(2n)!}{(n!)^2\cdot 2^{2n}} \ ,
$$
and the probability of getting exactly half heads in $2(n+1)$ flips is
$$
Q(n+1) = \frac{(2n+2)!}{((n+1)!)^2\cdot 2^{2n+2}}\ ,
$$
so
$$
\frac{Q(n+1)}{Q(n)} = \frac{(2n+2)(2n+1)}{(n+1)^2\cdot 2^2}
   = \frac{(2n+2)(2n+1)}{(2n+2)(2n+2)} = \frac{2n+1}{2n+2} < 1
$$
so $Q(n+1)<Q(n)$.  Thus, the probability of exactly half heads
always goes down as $2n$ goes up.

\end{itemize}

\item[5.] \begin{itemize}
\item[(a)] The CI is $25\pm 2(12/\sqrt{400}) = 25\pm 1.2$ visits per month.
\item[(b)] To reduce the size of the interval to $25\pm 0.4$ 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\% gives $z = 1.3$:
$$
\frac{280}{400} \pm
1.3\left(\frac{(1-0)\sqrt{\frac{280}{400}\cdot\frac{120}{400}}}{\sqrt{400}}\right)
\approx 70\% \pm 3\%\ .
$$
\end{itemize}

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

\item[7.] Of course I don't know how you did on the quiz, or whether
you feel you are ``miscalibrated.''  My own response to the quiz was
pretty close to right, because I knew something about the relative sizes
of the earth and the moon.  I think at my advanced age I have gotten
better ``calibrated,'' i.e., I have a better sense than I once did of
how likely an event is to occur.  But I have gained that sense only by
being proven wrong so many times.


\end{enumerate}
\end{document}

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