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\begin{document}

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

Directions: {\bf \Large Work only on this sheet} (on both sides, if
needed); do not turn in any supplementary sheets of paper. There is
actually plenty of room for your answers, as long as you organize
yourself BEFORE starting writing.

{\bf 1.} Consider once again the Jack and Jill example, Sec.  3.14.1.
Write you answers as a decimal expression, e.g. $\sqrt{0.5^3-0.2} \cdot
1.8$.

\begin{itemize}

\item [(a)] (20) Find $F_{X_3}(1.3)$.

\item [(b)] (15) Find $F_{X_3,Y_3}(1.3,0.1)$.  

\end{itemize}

{\bf 2.}  Consider the Ethernet example, Sec. 5.5.4.  Answer the
following with either decimal expressions (see above) or integrals.  The
latter must be definite integrals that calculus students could evaluate
to actual numbers.

\begin{itemize}

\item [(a)] (10) Find Var(X).

\item [(b)] (10) Find $P(X < 0.32)$.

\item [(c)] (15) Find $P(X = Y)$.

\item [(d)] (15) ({\bf BONUS PROBLEM:} Points plus Extra Credit) Suppose
transmission time T for a message is also random, exponentially
distributed with mean 0.1.  Find $P(X < Y \textrm{ and there is no
collision})$.

\end{itemize}

{\bf 3.} (15) Fill in the following R code to find $P(Y > 552)$ in Sec.
4.4.7.2:

\begin{lstlisting}
xpois(blank2, blank3)
\end{lstlisting}

The "x" here is blank1.

\onecolumn

{\bf Solutions:}

{\bf 1.a}

$$
F_{X_3}(1.3) = P(X_3 \leq 1.3) = P(X = 0 \textrm{ or } X = 1)  =
(0.5)^3 + 3 (0.5)^3 
$$

{\bf 1.b} 

\begin{eqnarray}
F_{X_3,Y_3}(1.3,0.1) &=& P(X_3 \leq 1.3, ~~ Y_3 \leq 0.1) \\ 
&=& P[(X = 0 \textrm{ or } X = 1) \textrm{ and }
(Y = 0)] \\
&=& P(X = 0 \textrm{ or } X = 1) \cdot P(Y = 0) \\
&=& \left [ (0.5)^3 + 3 (0.5)^3 \right ] \cdot 0.5
\end{eqnarray}

{\bf 2.a} Since $\lambda = 0.01$, then $Var(X) = (1/0.01)^2$ (p.95).

{\bf 2.b} $P(X < 0.32) = \int_{0}^{0.32} 5 e^{-5 u} ~ du$.

{\bf 2.c} 0

{\bf 2.d} Here T will be the transmission time for X.  We need to find
$P(X+T < Y)$.  Following the computation in Sec. 5.5.6, we have

$$
f_{X+T} (v) = e^{-5v} - e^{-10v} 
$$

So, similar to the reasoning in (5.87), we have

$$
P(X+T < Y) = \int_{0}^{\infty} 5e^{-5w}
\int_{0}^{w} [e^{-5v} - e^{-10v}] ~ dv
~ dw
$$

{\bf 3.} Use Sec. 4.4.6.1.

\begin{lstlisting}
ppois(4,5.52)
\end{lstlisting}

\end{document}