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\begin{document}

\lstset{xleftmargin=5mm,framexleftmargin=10mm,basicstyle=\footnotesize}

Name: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Directions: {\bf \Large Work only on this sheet} (on both sides, if
needed).  MAKE SURE TO COPY YOUR ANSWERS TO A SEPARATE SHEET FOR SENDING
ME AN ELECTRONIC COPY LATER.

{\bf Important note:}  Remember that in problems calling for R code, you
are allowed to use any built-in R function, e.g. {\bf choose()}, {\bf
sum()}, etc.  

{\bf 1.}  Consider the simulation on p.62.

\begin{itemize}

\item [(a)] (15) Suppose line 7 were replaced by

\begin{lstlisting}
mean(dvals^2) - (mean(dvals))^2
\end{lstlisting}

(Recall that the {\bf return()} is not ncessary.)  What would replace
``ED'' in the comment in line 6?  Your answer should reflect the likely
goal of the programmer, and you must use official notation from our
course (which is fairly standard in probability books); ED is an
example, as is P(D = 3) and so on.

\item [(b)] (10) Suppose line 7 were replaced instead by

\begin{lstlisting}
mean(dvals > 18)
\end{lstlisting}

What would replace ``ED'' in line 6 in this case?

\end{itemize}

{\bf 2.}  This problem involves the parking space example, Sec.
3.12.3.2.

\begin{itemize}

\item [(a)] (15) Find P(D = 12).

\item [(b)] (10) Find Var(N).

\item [(c)] (10) Good news!  I found a parking place just one space away
from the destination.  Find the probability that I am parked in the same
block as the destination.

\end{itemize}

{\bf 3.} (10) We have two vectors, {\bf x} and {\bf y}.  The elements of
the latter are either "a", "b" or "c".  We want to create a new vector
with the following property:  For any element in {\bf y} that has the
value "a", the new vector's corresponding element will be 100, with the
new value being 200 in the case of "b".  In the case of "c", the element
in the new vector will be the corresponding element of {\bf x}.  Write a
single line of code that creates and prints out this new vector.  You
are not allowed to use loops.  

Example:

\begin{lstlisting}
> x
[1]  5 12 13  3  4  5
> y
[1] "a" "c" "b" "b" "a" "a"
>  # single line of code here, 
>  # maybe with semicolons
[1] 100  12 200 200 100 100
\end{lstlisting}

(My grading script will set global variables {\bf x} and {\bf y}.  Your
code should NOT do this.)

{\bf 4.} (30) The function {\bf rpmf()} below generates {\bf n} random
values from a distribution with probability mass function {\bf pmf} and
support {\bf supp}.  The term {\it support} for a distribution is just a
fancy name for the set of values a random variable with that
distribution can take on.  The function is is then used to find the
approximate probability that in 3 consecutive stops in the bus ridership
example (Sec. 2.11), a total of 2 passengers board.  Fill in the blanks,
without using loops.

\begin{lstlisting}
rpmf <- function(n,pmf,supp) blank (a)

bvals <- rpmf(3000,c(0.5,0.4,0.1),0:2)
m <- matrix( blank (b) )
# the following call sets sums[i]
# to the sum of row i of m
sums <- apply(m,1,sum)
blank (c)
\end{lstlisting}

\onecolumn

\newpage

{\bf Solutions:}

{\bf 1.a} Var(D)

{\bf 1.b} P(D $>$ 18)

{\bf 2.a}

$$
(1-0.15)^{22} ~ 0.15
$$

{\bf 2.b}

$$
\frac{1-0.15}{0.15^2}
$$

{\bf 2.c}

\begin{eqnarray}
P(N = 12 ~|~ N = 10 \textrm{ or } N = 12) &=& 
\frac{P(N = 12)}{PN = 10 \textrm{ or } N = 12)} \\
&=& \frac
{(1-0.15)^{11} ~ 0.15}
{(1-0.15)^{9} ~ 0.15 + (1-0.15)^{11} ~ 0.15}
\end{eqnarray}

{\bf 3.}

\begin{lstlisting}
ifelse(y == "b",200,ifelse(y == "a",100,x))
\end{lstlisting}

{\bf 4.}

\begin{lstlisting}
rpmf <- function(n,pmf,supp) sample(supp,n,prob=pmf,replace=T)

bvals <- rpmf(3000,c(0.5,0.4,0.1),0:2)
m <- matrix(bvals,ncol=3)
# the following call sets sums[i]
# to the sum of row i of m
sums <- apply(m,1,sum)
mean(sums == 2)
\end{lstlisting}

\end{document}