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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 16:53:13 1997}
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\section{Exercises\label{Exercises}}

\begin{enumerate}
\item[1.]  \label{q1}Consider a normal random variable with mean $50$ and
standard deviation $10,$ and a random sample of size $80$ from which we are
to compute the values of $\bar{X}$, the sample mean. What is the probability
of getting a value of $\bar{X}$ as low as $46$?\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a1}

\item[2.]  \label{q2}Suppose a working widget deteriorates very little with
age. That is, a widget that has been running for some time will have nearly
the same failure probability during the following hour as it had during its
first hour of operation. Then, the failure times have an exponential
distribution $P(T\leq t)$ of the form $1-e^{-\frac{x}{\mu }}$. Given that
the widget has a mean life of 5 years, what is the probability that the
widget will have a lifetime exceeding 7.5 years? If the widget is guaranteed
for 2 years, what percentage of such widgets can be expected to need
replacement while under warranty?\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{%
0.2214in}{0.205in}{0in}}{}{}{a2}

\item[3.]  \label{q3}A widget has a mean life of 5 years with a standard
deviation of 2 years. Assuming a normal distribution, what is the
probability that the widget will have a lifetime exceeding 7.5 years? If the
widget is guaranteed for 2 years, what percentage of such widgets can be
expected to need replacement while under warranty?\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a3}

\item[4.]  \label{q4}The mean of a continuous distribution with probability
density function $f(u)$ is the integral $\int_{-\infty }^{\infty
}uf(u)du=\mu $ of the product of the variable and the probability density
function. The variance is the integral $\int_{-\infty }^{\infty }\left(
u-\mu \right) ^{2}f(u)du$. Find the mean and variance for each of the
continuous distributions discussed in this chapter.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a4}

\item[5.]  \label{q5}The mean of a discrete distribution with probability
density function $f(u)$ is the sum $\sum_{-\infty }^{\infty }uf(u)=\mu ,$
and the variance is $\sum_{-\infty }^{\infty }\left( u-\mu \right)
^{2}f(u)=\sigma ^{2}$.

\begin{itemize}
\item  Find the mean and variance for the discrete distributions discussed
in this chapter.

\item  If the probability density function for a distribution is $%
f(n)=\left( \frac{1}{2}\right) ^{n},$ what is the mean of the distribution?
What is the variance?\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{%
0.205in}{0in}}{}{}{a5}
\end{itemize}

\item[6.]  \label{q6}A die is cast until a 4 appears. What is the
probability that it must be cast more than five times?

\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a6}

\item[7.]  \label{q7}A telephone switchboard handles 600 calls on average
during a single rush hour. The board can make a maximum of 20 connections
per minute. Use the Poisson distribution to evaluate the probability that
the board will be overtaxed during any given minute of a rush hour.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a7}

\item[8.]  \label{q8}Find the probability that $x^{2}\leq 4$ for a normal
distribution with mean 1 and standard deviation 1.

\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a8}
\end{enumerate}

\section{Solutions\label{Solutions}}

\begin{enumerate}
\item[1.]  \label{a1}To solve this problem, you need to know that the
distribution of the mean of a sample of size $n$ from a normal distribution
of mean $\mu $ and standard deviation $\sigma $ is normal with mean $\mu $
and standard deviation $\frac{\sigma }{\sqrt{n}}$. Thus, the probability is
\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{%
}{q1} 
\[
\Pr \left( \bar{X}\leq 46\right) =\func{NormalDist}\left( 46;50,\frac{1}{2}%
\sqrt{5}\right) =\,1.7331\times 10^{-4} 
\]

\item[2.]  \label{a2}$\func{ExponentialDist}\left( 7.5;5\right)
=\,.78=P(X\leq 7.5)$, so the probability that $X$ is greater than $7.5$ is $%
1-.78=\,.22$.

$\func{ExponentialDist}\left( 2;5\right) =\,.33=P(X\leq 2)$, so the answer
to the second question is ``about 33 percent.''\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{}{q1}

\item[3.]  \label{a3}$\func{NormalDist}\left( 7.5;5,2\right) =\,.894=P(X\leq
7.5)$, so the probability that $X$ is greater than $7.5$ is $%
1-\,.894\,=\,.106$, or $10.6$ percent.\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{}{q3}

$\func{NormalDist}\left( 2;5,2\right) =.067=P(X\leq 2)$, so the answer to
the second question is ``about $7$ percent.''

\item[4.]  \label{a4}For the normal distribution, \textsf{Evaluate} gives 
\begin{eqnarray*}
\int_{-\infty }^{\infty }\limfunc{NormalDen}\left( u;\mu ,\sigma \right) udu
&=&\,\mu \\
\int_{-\infty }^{\infty }\limfunc{NormalDen}\left( u;\mu ,\sigma \right)
\left( u-\mu \right) ^{2}du &=&\,\sigma ^{2}
\end{eqnarray*}

For the Student's t distribution, with five degrees of freedom, \textsf{%
Evaluate} gives 
\[
\int_{-\infty }^{\infty }u\limfunc{TDen}\left( u;5\right) du=\,0 
\]
for the mean and $\,$%
\[
\int_{-\infty }^{\infty }u^{2}\,\limfunc{TDen}\left( u;5\right) du=\,\frac{25%
}{3}\sqrt{5} 
\]
for the variance. (When the parameters are symbolic, there are problems in
carrying out computations of the integrals that give mean and variance. You
should have no difficulty when you specify numerical parameters.)\hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{}{q4}

\item[5.]  \label{a5}Following is a sample solution.

\begin{itemize}
\item  For the binomial distribution, \textsf{Evaluate} followed by \textsf{%
Simplify} gives the mean: 
\[
\sum_{x=1}^{\infty }x\binom{n}{x}p^{x}\left( 1-p\right) ^{n-x}=np\left(
1-p\right) ^{n-1}\left( -\frac{1}{-1+p}\right) ^{n-1}=\,pn 
\]
\textsf{Evaluate} followed by \textsf{Simplify} and then \textsf{Factor}
gives the variance: 
\[
\sum_{x=0}^{n}\left( x-pn\right) ^{2}\binom{n}{x}p^{x}\left( 1-p\right)
^{n-x}=\,\,pn-p^{2}n=\,\left( 1-p\right) pn 
\]
(The intermediate expression for the variance is complicated and does not
appear here. Also, you need to make the simplifications $\left( -1\right)
^{2n}=1$ and $\left( -1\right) ^{2n+1}=-1$. Note that the symbol $\binom{n}{x%
}$ is a binomial fraction, rather than a matrix. To enter a binomial
fraction, from the \textsf{Insert} menu choose \textsf{Binomial}, and choose 
\textsf{None} for \textsf{Line}.)

\item  If the probability density function for a distribution is $%
f(n)=\left( \frac{1}{2}\right) ^{n}$ for $n\geq 0$, then the mean of the
distribution is $\sum_{n=1}^{\infty }n\left( \frac{1}{2}\right) ^{n}=\,2,$
and the variance is $\sum_{n=1}^{\infty }\left( n-2\right) ^{2}\left( \frac{1%
}{2}\right) ^{n}=\,2$.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{%
0.1807in}{0.2188in}{0in}}{}{}{q5}
\end{itemize}

\item[6.]  \label{a6}The probability of getting a $4$ on a single cast is $%
\frac{1}{6}$, so the probability of getting a different result is $\frac{5}{6%
}$. The probability of casting the die five times without getting a $4$ is $%
\left( \frac{5}{6}\right) ^{5}=\,.40188.$\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{}{q6}

\item[7.]  \label{a7}With $600$ calls on average during rush hour, the
average number of calls per minute is $10$. The probability that the number
of connections in a given minute is less than or equal to $20$ is the sum $%
\sum_{k=0}^{20}\limfunc{PoissonDen}(k,10)=\sum_{k=0}^{20}\frac{10^{k}e^{-10}%
}{k!}=\,.99841$. Thus, the probability that the board will be overtaxed is $%
1-.99841=\,.00159.$\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1807in%
}{0.2188in}{0in}}{}{}{q7}

\item[8.]  \label{a8}$x^{2}\leq 4$ when $-2\leq x\leq 2$. So \hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1807in}{0.2188in}{0in}}{}{}{q8} 
\begin{eqnarray*}
\Pr \left( x^{2}\leq 4\right) &=&\Pr (x\leq 2)-\Pr (x\leq -2) \\
&=&\func{NormalDist}(2;1,1)-\func{NormalDist}(-2;1,1) \\
&=&.84
\end{eqnarray*}
\end{enumerate}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Menu Items, Lists, and Matrices}{}{}{DM11-1.tex#Statistics}

\item  \hyperref{Measures of Central Tendency}{}{}{DM11-2.tex#Statistics,
mean}

\item  \hyperref{Measures of Dispersion}{}{}{DM11-3.tex#Mean deviation}

\item  \hyperref{Distributions and Densities}{}{}{DM11-4.tex#Distribution}

\item  \hyperref{Families of Continuous Distributions}{}{}{%
DM11-5.tex#Continuous distributions}

\item  \hyperref{Families of Discrete Distributions}{}{}{DM11-6.tex#Discrete
distributions}

\item  \hyperref{Random Numbers}{}{}{DM11-7.tex#Statistics, random numbers}

\item  \hyperref{Curve Fitting}{}{}{DM11-8.tex#Curve fitting}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid-file
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36.875pt;original-height 22.5625pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Data}{}{}{%
DMIndex.tex#Data}, \hyperref{Distribution}{}{}{DMIndex.tex#Distribution}, 
\hyperref{Matrices}{}{}{DMIndex.tex#Matrices}, \hyperref{Reshape}{}{}{%
DMIndex.tex#Reshape}, \hyperref{Solve}{}{}{DMIndex.tex#Solve}, \hyperref{%
Statistics}{}{}{DMIndex.tex#Statistics}}

\end{document}
