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{\bf STAT 324 Sample midterm problems }\\


\begin{enumerate}
\item 1/7 of men and 1/9 of women are color-blind. $60\%$ of
population are females.\\
(a) A person is chosen at random and that person is color-blind. What is the probability that the person is male?\\
{\em Solution.} Let $C$ be the event the person is color blind. Let
$M$ and $F$ be the events the person is male and female
respectively. Then we need to compute
$$P(M|C) = \frac{P(M \cap C)}{P(C)} = \frac{P(C|M)P(M)}{P(C|M)P(M) + P(C|F)P(F)}
=\frac{\frac{1}{7} \cdot \frac{4}{10}}{\frac{1}{7} \cdot
\frac{4}{10} + \frac{1}{9}\cdot \frac{6}{10}}.$$

(b) A person is chosen at random and that person is male. What is
the probability that the person is color-blind?\\
{\em Solution.} $P(C|M) = \frac{1}{7} \cdot \frac{4}{10}$.

\item The life time $X$ of an electronic component follows the distribution $f(x) = e^{-x}$ for $x
> 0$ and $f(x)=0$ otherwise. Three of these components operate
independently in a piece of equipment. The equipment fails if at
least two of the components fail. Find the probability that the
equipment will operate for at least 2 hours without failure.\\
{\em Solution.} The probability of a single component failure is
$$p=P(X<2) =\int_0^{2} e^{-x} \;dx = 1-e^{-2}.$$

The equipment will fail if at least two out of 3 components fail. So
the probability of the equipment failure is
$${3 \choose 2} p^2(1-p) + {3 \choose 3}p^3 = 3p^2(1-p) + p^3.$$
The probability of the equipment operating for at least 2 hours is
then $$1 -3p^2(1-p) - p^3.$$

\item The achievement scores for a college entrance examination are
normally distributed with mean 70 and variance 10. What fraction of
the scores lies between 75 and 80? You may use the following {\tt R}
output.
\begin{verbatim}
> pnorm(c(1:4),1,2)
[1] 0.50 0.69 0.84 0.93
\end{verbatim}
{\em Solution.} Let $X$ be the achievement score and let
$Z=\frac{X-70}{10} \sim N(0,1)$. Need to find $$p=P(
\frac{75-70}{10}< Z < \frac{80-70}{10}) = P( 0.5< Z < 1).$$ However,
the {\tt R} output is given in terms of $W \sim N(1,2^2)$. The
relationship between $W$ and $Z$ is $W= 2Z+1$. So \bq p&=&P(2\cdot
0.5 +1 < W < 2 \cdot 1
+1)=P(2 <W<3)\\
&=& {\tt pnorm}(3,1,2) - {\tt pnorm}(2,1,2) = 0.84-0.69=0.15. \eq



\item The density function of he random variable $X$
is $f(x)= \frac{x}{2}$ for $0 < x <2$ and $f(x)=0$ otherwise.\\
(a) Find the cumulative distribution function of $X$ and plot its graph.\\
{\em Solution.} $F_X(x) = \int_0^x \frac{x}{2} \;dx = \frac{x^2}{4}$
for $0 <x<2$, $F_X(x) = 0$ for $x \leq 0$ while $F_X(x) = 1$ for $x
\geq 2$.

(b) Derive the probability density function of $Y=2X+1$ using the
cumulative distribution function of $X$.\\
{\em Solution.} $F_Y(y) = P(Y\leq y) = P(3X + 1 \leq y) =
F_X(\frac{y-1}{3}) = \frac{1}{4}\Big(\frac{y-1}{3}\Big)^2$ for
$1<y<5$. $f_Y(y) = \frac{dF_Y(y)}{dy}= \frac{1}{6}\frac{y-1}{3}$ for
$1<y<5$.




\item Let $X$ and $Y$ have joint probability density
$$f(x,y) = cx, \mbox{ for } 0<x<1, 0<y<1 \mbox{ and } 1 < x+y <2$$
and $f(x,y)=0$ otherwise.\\
(a) Find constant $c$.\\
{\em Solution.} Draw the domain where the joint density is defined.
$$1=\int_0^1 \int_{1-y}^1 cx \;dx \;dy = \frac{c}{2} \int_0^1(1-(1-y)^2) \;dy
=\frac{c}{2}(y +
\frac{1}{3}(1-y)^3)\Big|_0^1=\frac{c}{2}(1-\frac{1}{3}).$$ So
$c=3$.\\
(b) Find the marginal distributions of $X$ and $Y$.\\
{\em Solution.} $f_X(x) = \int_{1-x}^1 3x \;dy = 3x^2 $ for $0<
x<1$. $f_Y(y) = \int_{1-y}^1 3x \;dx = \frac{3}{2}(1-(1-y)^2)$ for
$0<y<1$.\\
(c) Find the conditional probability $P(X < 1 |X+Y > 1)$.\\
{\em Solution.} This is trivially $$\frac{P(X < 1, X+Y >1)}{P(X+Y
>1)}=1$$

\item Let $X_1, \cdots, X_n$ be a random sample from Bernoulli distribution with parameter $p$.
(a) What is $\mathbb{E}(S^2/p^2)$? $S^2$ is the sample variance.
Explain your results.\\
{\em Solution.} The sample variance is an unbiased estimator of the
population variance. Hence $\mathbb{E}(S^2/p^2) =\mathbb{E}(S^2)/p^2
= {\bf Var}(X_i)/p^2.$ The variance for a Bernoulli random variable
can be esily computed as ${\bf Var} (X_i) = p(1-p)$. So
$\mathbb{E}(S^2/p^2)=(1-p)/p$. \\
(b) Find an unbiased estimator for $p^2$. Explain your results.\\
{\em Solution.} There are many solutions to this problem. We know
$S^2$ and $\bar X$ will be unbiased estimators of population
variance $p(1-p)$ and mean $p$ respecively. So $ {\bf E}(\bar X)
-{\bf E}(S^2) = p -p(1-p) = p^2$. Hence, $\bar X - S^2$ is an
unbiased estimator for $p^2$.


\item Let $X_1 ,X_2$ be a random sample from $N(0, 1/\theta)$. Note that the sample size is 2 and the density function for $X_i$ is $f(x_i) = \frac{\sqrt{\theta}}{\sqrt{2\pi}}\exp(-\theta x_i^2/2)$.
 Find the likelihood function and use it to obtain the maximum
likelihood estimator of $\theta$.\\
{\em Solution.} The likelihood function is $L(\theta)
=\theta/(2\pi)\exp(-\theta(x_1^2 +x_2^2)/2)$. Get log-likelihood
function $\ln L(\theta) = const + \ln \theta - \theta
(x_1^2+x_2^2)/2$. Differentiate with respect to $\theta$ to get the
maximum:
$$\frac{d \ln L(\theta)}{d\theta} = \frac{1}{\theta} - \frac{1}{2}(x_1^2+x_2^2)=0.$$
Solving the equation, we get $\hat \theta = 2/(x_1^2 + x_2^2)$.

\end{enumerate}
\end{document}