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\begin{document}
\mbox{}\vspace*{-1.5cm}

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

\begin{center}
{\bf Mathematics 214: Introduction to Statistics }
\\
{\bf Review for Exam 3, April 1999}
\end{center}

The following table is to be used for the questions 1) -- 4). 
\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}
\item 
Find $P(X_1\geq1)$. 

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 0.55}$     &
{\bf b)} & ${\Ds 0.35}$ &
{\bf c)} &  ${\Ds 0.45}$  & 
{\bf d)} & ${\Ds 0.25}$ & 
{\bf e)} & ${\Ds 0.3}$
\end{tabular}

\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}

\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}

\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}

\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}

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

\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}

\newpage
\item 
Let $X$ and $Y$ be random variables with variances $V(X)=1$ and $V(Y)=9$ respectively. 
Assume 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}

\item 
Let $X$ and $Y$ be random variables with 
\[
   E(X)=3,\ E(Y)=6,\ V(X)=9,\ V(Y)=1,\ E(XY)=20.
\]
Then the covariance Cov$(X,Y)$ of $X$ and $Y$ is 

\begin{tabular}{cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}cp{2.3cm}}
{\bf a)} & ${\Ds 2}$     &
{\bf b)} & ${\Ds -2}$ &
{\bf c)} &  ${\Ds \frac{2}{3}}$  & 
{\bf d)} & ${\Ds 11}$ & 
{\bf e)} & ${\Ds -\frac{2}{3}}$
\end{tabular}

\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}

\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}

\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}

\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}

\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}

\item
Let $X,\,Y$ be independent random variables with $E(X)=3,\,E(Y)=4$ and $V(X)=9,\,V(Y)=25$.
Then the covariance  Cov$(X,Y)$ is 

\begin{tabular}{cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}}
{\bf a)} & ${\Ds 3}$     &
{\bf b)} & ${\Ds -3}$ &
{\bf c)} &  ${\Ds 0}$  & 
{\bf d)} & ${\Ds 27}$ & 
{\bf e)} & ${\Ds 15}$
\end{tabular}

\item
For an aptitude test for a certain job history shows scores to be normally distributed with a variance of 225.
20 applicants are to take the test.
Find the approximate probability that the sample variance of the test scores will be less 
than 138. (You may use a calculator.) 

\begin{tabular}{cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}}
{\bf a)} & ${\Ds 0.1}$     &
{\bf b)} & ${\Ds 0.9}$ &
{\bf c)} &  ${\Ds 0.975}$  & 
{\bf d)} & ${\Ds 0.025}$ & 
{\bf e)} & ${\Ds 0.05}$
\end{tabular}

\item
One-hour carbon monoxide concentrations in air samples from a large city average 10 ppm, with a standard deviation 
of 9 ppm. 
Find the probability that the average concentration in 100 samples selected randomly will exceed 12 ppm. 

\begin{tabular}{cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}cp{2.5cm}}
{\bf a)} & ${\Ds 0.4129}$     &
{\bf b)} & ${\Ds 0.0132}$ &
{\bf c)} &  ${\Ds 0.4868}$  & 
{\bf d)} & ${\Ds 0.0871}$ & 
{\bf e)} & ${\Ds 0.2514}$
\end{tabular}
\end{problems}

\vspace*{1.5cm}
\begin{center} 
\begin{minipage}[t]{6cm}
{\bf 1.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 2.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 3.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 4.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 5.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 6.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 7.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 8.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e)
\end{minipage}\hspace{2.5cm} \begin{minipage}[t]{6cm}
{\bf \mbox{}\phantom{1}9.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e)  \\[1ex]
{\bf 10.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 11.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 12.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 13.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 14.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 15.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
{\bf 16.}\quad (a) \quad (b) \quad (c) \quad (d) \quad (e) \\[1ex]
\end{minipage}
\end{center}
\end{document}