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     \begin{list}{\bf{\arabic{prob}.}\hfill}{\usecounter{prob}
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\begin{document}
\mbox{}\vspace*{-1.5cm}

{\bf Name:} \underline{\mbox{\hspace*{7cm}}}
\hfill{\bf Instructor:} Heide Gluesing-Luerssen\hspace*{.7cm}

\begin{center}
{\bf Mathematics 214: Introduction to Statistics }
\\
{\bf Spring Semester 1998}
\\
{\bf Exam 3}
\\
{\bf April 17, 1998}
\end{center}
%\vspace*{.4cm}
\begin{tabular}{p{18cm}}\hline
This
\rule[0cm]{0cm}{.8cm}Examination
contains 16 questions 6 points worth each.
You start with 4 points,  and the highest possible score is 100.
Fill in your answers on this cover sheet by placing an {\sf X} through one
letter for each problem except the first two problems.
Write the answers for 1 and 2 in the corresponding empty boxes of the list
below.
Books and notes are not allowed.
\\[1.7ex]
\hline
\end{tabular}
\vspace*{.7cm}

\mbox{}\hspace*{1.5cm}\begin{tabular}[t]{|p{.6cm}||p{.5cm}|p{.5cm}|p{.5cm}|p{.5c
m}|p{.5cm}||p{.8cm}|}\hline
 \ 1      &\multicolumn{5}{|c||}{} &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 2      &\multicolumn{5}{|c||}{}  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 3      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 4     &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 5      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 6      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 7      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 8      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
\end{tabular} \qquad\quad
\begin{tabular}[t]{|p{.6cm}||p{.5cm}|p{.5cm}|p{.5cm}|p{.5cm}|p{.5cm}||p{.8cm}|}\
hline
 \ \ 9      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 10      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 11      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 12     &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 13      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 14      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 15      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \ 16      &\  a &\ b &\ c &\ d &\ e  &   \rule[-.5cm]{0cm}{1.2cm} \\ \hline
 \end{tabular}
\\[1.3ex]
\mbox{}\hspace*{14.3cm} {\bf Total} \fbox{\rule[-.3cm]{0cm}{1cm}\qquad\ }

\vspace*{1.2cm}
{\bf Sign the pledge:}
\\[1ex]
``On my honor, I have neither given nor received unauthorized aid on this
Exam.''

{\bf Signature:} \underline{\mbox{\hspace*{7cm}}}

\vspace*{.3cm}
\begin{center}     {\bf GOOD LUCK}\end{center}

\newpage
\begin{problems}
\item
Find $P(Z\leq 2)$ where $Z$ is a standard normal random variable.
{\bf Write your answer on the cover sheet!}
\vspace*{1cm}

\item
Assume $X$ is a normal random variable with mean 3 and standard deviation
2. Find $P(-1\leq X\leq 5)$.
{\bf Write your answer on the cover sheet!}
\vspace*{1.5cm}

\item
A fair die is tossed 600 times. What is the approximate probability that at
most 90 times a ``six'' comes up?

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.5cm}cp{2.5cm}}
{\bf a)} & ${\Ds \frac{9}{60}}$     &
{\bf b)} & ${\Ds 0.3508}$ &
{\bf c)} &  ${\Ds 0.1492}$  &
{\bf d)} & ${\Ds \frac{9}{10}}$ &
{\bf e)} & ${\Ds 0.8508}$
\end{tabular}
\vspace*{3.5cm}

\item
The grade-point averages of a large population of college students are
approximately normally distributed
with mean equal to 2.4  and standard deviation equal to 0.5.
Suppose that three students are randomly selected from the student body.
What is the probability that all three will possess a grade-point average
in excess of 3.0?

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.5cm}cp{2.5cm}}
{\bf a)} & ${\Ds (0.3849)^3}$     &
{\bf b)} & ${\Ds (0.1151)^3}$ &
{\bf c)} &  ${\Ds (0.8849)^3}$  &
{\bf d)} & ${\Ds 1-(0.1151)^3}$ &
{\bf e)} & ${\Ds 1-(0.8849)^3}$
\end{tabular}
\vspace{2.5cm}

\item
The fracture strengths of a certain type of glass average 14 (in thousands
of pounds per square inch)  and
have a standard deviation of 2.
What is the probability that the average fracture strength for 100 pieces
of this glass  exceeds 14.5?

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 0.0202}$     &
{\bf b)} & ${\Ds 0.4938}$ &
{\bf c)} &  ${\Ds 0.9938}$  &
{\bf d)} & ${\Ds 0.0062}$ &
{\bf e)} & ${\Ds 0.4013}$
\end{tabular}
\end{problems}

\newpage
The following table is to be used for the questions 6) -- 9).
\begin{center}
\begin{tabular}{p{1cm}|p{1cm}|p{1cm}p{1cm}p{1cm}|p{.7cm}}
     &       & \multicolumn{3}{|c|}{$X_1$}&    \\ \hline
    &        &  \ 0   &  \ 1     & \ 2      &    \\ \hline
    & \  -1   &0.2 & 0.15 & 0.1   &    \\
 $X_2$   & \   0    &0.1 & 0.05  & 0.05 &   \\
   &  \  1    & 0.2 & 0.1    &   0    &     \\
  &  \   2    & 0.05 & 0     &   0    &    \\ \hline
   &         &         &        &         &   %\rule[-.3cm]{0cm}{1cm}
 \end{tabular}
\end{center}

\begin{problems}
\setcounter{prob}{5}
\item
Find $P(X_1=2)$.

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 0.05}$     &
{\bf b)} & ${\Ds 0}$ &
{\bf c)} &  ${\Ds 0.15}$  &
{\bf d)} & ${\Ds 0.1}$ &
{\bf e)} & ${\Ds 0.3}$
\end{tabular}
\vspace{2cm}

\item
Find $E(X_2)$.

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 1 }$     &
{\bf b)} & ${\Ds -0.05}$ &
{\bf c)} &  ${\Ds 0.6}$  &
{\bf d)} & ${\Ds -0.1}$ &
{\bf e)} & ${\Ds .15}$
\end{tabular}
\vspace{3cm}

\item
Find $E(X_1 X_2)$.

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 0}$     &
{\bf b)} & ${\Ds 1}$ &
{\bf c)} &  ${\Ds -0.25}$  &
{\bf d)} & ${\Ds -0.15}$ &
{\bf e)} & ${\Ds 0.6}$
\end{tabular}
\vspace{3cm}

\item
Find $P(X_1=2\,|\,X_2=0)$.

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 0.03}$     &
{\bf b)} & ${\Ds 0.05}$ &
{\bf c)} &  ${\Ds \frac{1}{3}}$  &
{\bf d)} & ${\Ds \frac{1}{11}}$ &
{\bf e)} & ${\Ds \frac{1}{4}}$
\end{tabular}
\vspace{3cm}

\item
Let $X$ and $Y$ be random variables with variances $V(X)=1$ and $V(Y)=9$
respectively.
Assume {\sl Cov}$(X,Y)$, the covariance of $X$ and $Y$, is 2.
Then the variance of $2X-Y$ is

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 3}$     &
{\bf b)} & ${\Ds -9}$ &
{\bf c)} &  ${\Ds -5}$  &
{\bf d)} & ${\Ds 21}$ &
{\bf e)} & ${\Ds 5}$
\end{tabular}
\vspace{3cm}

\item
Let $X$ and $Y$ be random variables with variances $V(X)=1$ and $V(Y)=9$
respectively.
Assume {\sl Cov}$(X,Y)$, the covariance of $X$ and $Y$, is 2.
Then the correlation coefficient $\rho$ between $X$ and $Y$ is

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds \frac{2}{3}}$     &
{\bf b)} & ${\Ds \frac{4}{3}}$ &
{\bf c)} &  ${\Ds \frac{2}{9}}$  &
{\bf d)} & ${\Ds \frac{4}{9}}$ &
{\bf e)} & ${\Ds \frac{2}{\sqrt{3}}}$
\end{tabular}
\vspace{3.5cm}

\item
The times to failure of four turbine blades in jet engines, in $10^3$
hours, were as follows:
\[
           3.2,\ 1.8,\  2.6,\  4.4
\]
The sample mean of these data is

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds \frac{3}{4}}$     &
{\bf b)} & ${\Ds 3}$ &
{\bf c)} &  ${\Ds 12}$  &
{\bf d)} & ${\Ds 4}$ &
{\bf e)} & ${\Ds \frac{11}{4}}$
\end{tabular}
\vspace{2.5cm}

\item
The times in minutes between eruptions for three of the eruptions of the
geyser {\em Old Faithful} on
August 1, 1985 were 77,\ 80,\ 80.
The sample mean of these data is 79.  Find the sample variance.

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 3}$     &
{\bf b)} & ${\Ds 2}$ &
{\bf c)} &  ${\Ds 6}$  &
{\bf d)} & ${\Ds 6244}$ &
{\bf e)} & ${\Ds 3123.5}$
\end{tabular}
\vspace{3cm}

\item
The diameter measurements of an electric cable, taken at 100 points along
the cable, yield a sample mean of
2.1 centimeters and a sample standard deviation of 0.3 centimeter.
Construct a 90\% confidence interval for the average diameter of the cable.

\begin{tabular}{cp{3.5cm}cp{3.5cm}cp{3.5cm}}
{\bf a)} & ${\Ds 0.3\pm1.645\frac{2.1}{10}}$     &
{\bf b)} & ${\Ds 2.1\pm1.96\frac{0.09}{10}}$ &
{\bf c)} &  ${\Ds 2.1\pm1.645\frac{0.3}{100}}$  \\[2.5ex]
{\bf d)} & ${\Ds 2.1\pm1.645\frac{0.3}{10}}$ &
{\bf e)} & ${\Ds 2.1\pm1.96\frac{0.3}{10}}$
\end{tabular}
\vspace{3cm}

\item
Suppose that a random sample of only 15 black cherry trees has been taken
to estimate the average
volume of a black cherry tree in the Allegheny National Forest.
Assume that the sample mean for the volumes (in cubic feet) of the 15 trees
is $\bar{x}=30.2$ and the sample
variance is $s^2=268.96\ (=16.4^2)$.
If we assume further that the distribution of the volumes of all the black
cherry trees is normal, then a
confidence interval for the average volume of a tree in this forest with
confidence coefficient 0.98 is:

\begin{tabular}{cp{3.5cm}cp{3.5cm}cp{3.5cm}}
{\bf a)} & ${\Ds 30.2\pm1.345\frac{16.4}{\sqrt{15}}}$     &
{\bf b)} & ${\Ds 30.2\pm2.33\frac{16.4}{\sqrt{15}}}$ &
{\bf c)} &  ${\Ds 30.2\pm29.1413\frac{16.4}{\sqrt{15}}}$  \\[2.5ex]
{\bf d)} & ${\Ds 30.2\pm2.624\frac{16.4}{\sqrt{14}}}$ &
{\bf e)} & ${\Ds 30.2\pm2.624\frac{16.4}{\sqrt{15}}}$
\end{tabular}
\vspace{3cm}

\item
Suppose $n=13$ observations are taken on  normally distributed measurements.
The sample variance is $s^2=5$.
A confidence interval for the variance $\sigma^2$ with confidence
coefficient 0.99 is approximately given
by

\begin{tabular}{cp{3.5cm}cp{3.5cm}cp{3.5cm}}
{\bf a)} & ${\Ds \Big(\frac{60}{21.03},\,\frac{60}{5.23}\Big)}$     &
{\bf b)} & ${\Ds \Big(\frac{60}{28.3},\,\frac{60}{3.07}\Big)}$ &
{\bf c)} &  ${\Ds \Big(-\frac{60}{28.3},\,\frac{60}{28.3}\Big)}$  \\[2.5ex]
{\bf d)} & ${\Ds \Big(\frac{60}{29.82},\,\frac{60}{3.57}\Big)}$ &
{\bf e)} & ${\Ds \Big(-\frac{60}{3.07},\,\frac{60}{28.3}\Big)}$
\end{tabular}
\vspace{2cm}



\end{problems}
\end{document}
