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\rightline{November 15, 2005}
\vskip 8 pt
\centerline{\bf Math 102 / Core 143 Sections BX --- Exam II}

\vspace*{0.5cm}
\noindent
Show all work clearly for partial credit; an unevaluated
expression is worth more than the numerical answer without explanation.

\begin{enumerate}

\item[1] {\it(30 points)} Parapsychologists, who purport to make
scientific studies of extra-sensory perception (ESP) including
precognition (``foreknowing''), use special decks of cards.  Let us
suppose we have such a deck of 20 cards, four each of the following:
\newline
\centerline{\includegraphics{e2f05.png}}
All of the following draws are made at random, after the deck is
shuffled.
\begin{itemize}
\item[(a)] If two cards are drawn from the deck \underbar{without}
replacement, what is the probability that they have the same symbol?
(Note: It doesn't matter what symbol is on the first card.  What is
the probability that the second one matches?)
\item[(b)] If two cards are drawn \underbar{with} replacement, what
is the probability that the first is a square and the second is a
circle?
\item[(c)] If two cards are drawn with replacement, what
is the probability that either the first is a square or the second is a
circle, or both?
\item[(d)] If seven cards are drawn with replacement, what is the
probability that at least one is a square?
\item[(e)] Checking a subject for precognition, a parapsychologist asks
the subject to pick one of the five symbols and then picks a card from
the deck and notes whether the card's symbol matches the subject's
choice.  If this experiment is repeated 8 times (replacing the card in
the deck each time), what is the probability that card and choice will
agree at least 7 times, just by chance?
\end{itemize}

\item[2.] {\it(20 points)} In the game of Varl, you have 3 chances in
10 to win; when you win, you get \$2, and when you lose, you lose \$1.
\begin{itemize}
\item[(a)] Consider a box (values on tickets, number of tickets with
each value) so that a draw from the box correctly models a single
play of Varl.  What are the average and SD of this box?  (It may help
with partial credit if you sketch the box.)
\item[(b)] If you play the game only once, what is the probability that
your winnings will be within one SD of the box average?  (Note: Is the
box distribution normal?)
\item[(c)] If you play the game 400 times, what is the probability that
your average winnings will be within one SD of the box average?
\item[(d)] Same question as (c), but with SE in place of SD.
\end{itemize}

\newpage

\item[3.] {\it(10 points)} Suppose a die is rolled (I)~60 times and
(II)~600 times. Give short explanations of your answers to the
following: In which case, (I), (II) or neither, is each of the
following more likely?
\begin{itemize}
\item[(a)] The number of 4's is within 5 rolls
of exactly ${1\over6}$ of the total number of rolls.
\item[(b)] The number of 4's is within 5\% of exactly ${1\over6}$
of the total number of rolls.
\item[(c)] The number of 4's is smaller than the number of 3's.
\end{itemize}

\item[4.] {\it(30 points)} A survey of 1600 randomly chosen
workers in New York State reveals that 30\% of the sample have
access to a cafeteria at work, and that the average cost of a
lunch (whether brought from home or bought at work) is \$6, with
an SD of \$3.
\begin{itemize}
\item[(a)] Find a 95\% confidence interval for the average cost
of a NYS worker's lunch.
\item[(b)] Suppose a similar survey were conducted in Montana, which
has only 1/16 of the population of New York, and the average and SD
of the sample were the same (\$6 and \$3 respectively).  How would the
95\% confidence interval change?
\item[(c)] How many workers in New York would have to be surveyed to
make the confidence interval only 1/3 as wide? (Again, assume average
and SD of the sample are \$6 and \$3.)
\item[(d)] Suppose the survey were conducted monthly for a period of
10 years.  In about how many of the 120 surveys would the correct
average cost of lunch for all NYS workers (at that time) be in a
95\% confidence interval computed from that survey's data?
\item[(e)] What is an 80\% confidence interval for the percentage of
NYS workers who have access to a cafeteria at work?
\end{itemize}

\item[5.] {\it(10 points)} Related to the article ``What is the chance
of your being guilty'' by John Kay, how is understanding of contingent
[i.e., conditional] probability of crucial importance to the case of
Sally Clark (accused of murder after two of her children succumbed to cot death)?

\end{enumerate}

\hrule

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


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

\begin{enumerate}

\item[1.] \begin{itemize}
\item[(a)] Three cards of the remaining 19 match the first one, so 3/19.
\item[(b)] $(4/20)(4/20) = 1/25$.
\item[(c)] Using (b): $(4/20)+(4/20)-1/25 = 9/25$.
\item[(d)] Subtract from one the probability that none is a square:
$1 - (16/20)^7 = 1-(4/5)^7 \approx .79$.
\item[(e)] The probability of chance agreement on any one trial is $1/5$,
so the probability on exactly 7 or exactly 8 trials is
$C(8,7)(1/5)^7(4/5)^1 + C(8,8)(1/5)^8 = (8(4)+1)/5^8 = 33/390,625
=0.00008448$.
\end{itemize}

\item[2.] \begin{itemize}
\item[(a)] The box contains 3 tickets of 2's and 7 of $-1$'s, so
the average is $(3(2)+7(-1))/10 = 1/10$ and the SD, by the shortcut
formula, is $(2-(-1))\sqrt{(.3)(.7)} = 3\sqrt{.21} \approx 1.37$.
\item[(b)] If the distribution were normal, we would say this would
happen about 68\% of the time.  But because 2 is not within one SD
of the average and $-1$ \underbar{is} within this interval, we can
say that it will happen 70\% of the time.
\item[(c)] The SE of the average is the SD of the box divided by
$\sqrt{400}$, and the averages \underbar{are} distributed (nearly)
normally, the probability of the sample average being within the
SD of the box average is the probability of a $z$-value between
$-20$ and 20, which is virtually 100\%
\item[(d)] This probability is the probability of a $z$-value
between $-1$ and 1, or about 68\%.
\end{itemize}

\item[3.] \begin{itemize}
\item[(a)] (I): There is more variability in the count as the
number of rolls gets larger.
\item[(b)] (II): There is more variability in the percentage
when there are fewer rolls.
\item[(c)] The answer I expected, which received full credit, is
neither, since the number of 4's has the same expected value as
the number of 3's, no matter how many times the die is rolled.
But the correct answer is (II), because in that case the probability
$p$ that there are exactly as many 4's as 3's is lower, and so the
probability that the number of 4's is larger is higher, because
it  is equal to the probability that the number of 3's is larger,
and each of these is half of the probability $1-p$
that the number of 4's is not equal to the number of 3's.
\end{itemize}

\item[4.] \begin{itemize}
\item[(a)] $\$6 \pm 2(\$3/\sqrt{1600}) = \$(6\pm.15)$
\item[(b)] It would not change, because 1600 workers is still a
negligible fraction of Montana's work force, so the correction
factor is still essentially 1.
\item[(c)] Because of the square root in the denominator, to make
the denominator 3 times as large, we need 9 times as many workers,
or 14,400.
\item[(d)] About 95\% of the surveys, or 114, will yield an interval
including the correct value.
\item[(e)] The bootstrapped value for the SE of percentage is
$(1-0)\sqrt{(.3)(.7)}/\sqrt{1600} \approx 1.1\%$.  Using the normal
table for an area of 80\%, we get a $z$-value of about 1.3; so the
desired confidence interval is $30\%\pm 1.3(1.1\%) \approx 30\%\pm1.5\%$.
\end{itemize}

\item[5.] Kay's point is that it is not correct, in computing the probability
of two events both happening, to multiply the probabilities of the two events,
if the events are not independent. And two cot deaths in the same household
are probably not independent; the connection may be a murdering mother, but
it may also be an inherited respiratory weakness or household contamination.
Here are the relevant sentences: ``And Bayes' theorem shows that probability
and statistics cannot decide
in cases like Mrs Clark's. Unexplained infant death is rare but when it occurs
it more often results from natural causes than from murder. A single incident
should therefore, in the absence of other evidence, be treated as natural. But
what of a series? The simple theory of probability, as applied by Johnnie
Cochran and Professor Sir Roy Meadow, an expert witness in the Clark case,
multiplies the probabilities of a series of events together on the assumption
that they are unrelated. But it is unlikely that these effects are unrelated.
Murderous intent may persist in the mind of evildoers. Genetic and environmental
influences that cause death may affect siblings. We simply do not know the
relative significance of these factors, so we must look to other evidence - such
as the blood sample that, after Ms Clark had spent two years in prison,
indicated that her son Harry had probably died of a respiratory infection.''

\end{enumerate}
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