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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 the 2t/15 example, Sec. 4.3.3.  Suppose this is the
density of light bulb lifetimes L (on the time scale of years).  Note:
In all parts below, give each answer as decimal expression, e.g.
$\sqrt{(0.123)^2+5}$, or as a common fraction reduced to lowest terms.
You may cite equations in that section.

\begin{itemize}

\item [(a)] (20) Find the proportion of bulbs with lifetime less than 
the mean lifetime.

\item [(b)] (20) Find E(1/L).

\item [(c)] (20) If I test many bulbs, on average how long will it take to
find two that have lifetimes longer than 2.5?

\item [(d)] (20) Suppose I've been using bulb A for 2.5 years now in a
certain lamp, and am continuing to use it.  But at this time I put a new
bulb, B, in a second lamp.  I am curious as to which bulb is more 
likely to burn out within the next six months.  Find the two
probabilities.

\end{itemize}

{\bf 2.} (20) The expected value of a chi-square random variable with k
degrees of freedom turns out to be k.  Derive this fact in a
step-by-step manner, citing mailing tubes, and NOT using material past
p.94.

{\bf Solutions:}

{\bf 1.a} 

\begin{equation}
P(L < 2.8) = \int_{1}^{2.8} 2t/15 ~ dt = (2.8^2 - 1)/15
\end{equation}

{\bf 1.b}

\begin{equation}
E(1/L) = \int_{1}^{4} \frac{1}{t} \cdot 2t/15 ~ dt = \frac{2}{5}
\end{equation}

{\bf 1.c}

Use (3.110) with k = 2 and p = 0.65.

{\bf 1.d}  First find the probability of NOT burning out in the next six
months.  For bulb A, use (6.3), yielding

\begin{equation}
\frac{P(L > 3.1)}{P(L > 2.5)} =
\frac
{\int_{3.1}^{4} 2t/15 ~ dt} 
{0.65} =
\frac
{(16-3.1^2)/15}
{0.65}
\end{equation}

{\bf 2.} Let Y be as in (4.55).  Then

\begin{eqnarray}
E(Y) &=& E(Z_1^2+...+Z_k^2) \\ 
&=& k E(Z_1^2) ~~~~ (((3.13), \textrm{ident. distrib.)} \\ 
&=& k [Var(Z_1) + (EZ_1)^2] ~~~~ ((3,29)) \\
&=& k
\end{eqnarray}
\end{document}