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\begin{center}\huge MATH 1550 - EXAM 2 - FALL 2018\\
SOLUTION
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%\large Name: \makebox[3in]{\hrulefill}\\
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\begin{flushleft}\vspace{0.5in}
19 October 2018 \\
Instructor: Tom Cuchta
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\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}


\begin{minipage}{0.5\textwidth}
\begin{tabular}{|l|}
\hline 
Binomial: $P(x)=\dfrac{n!}{(n-x)!x!} p^x q^{n-x}$ \\
\hline
Geometric: $P(x)=pq^{x-1}$ \\
\hline 
Poisson: $P(x)=\dfrac{\mu^x e^{-\mu}}{x!}$ \\
\hline
$E=z_c\dfrac{\sigma}{\sqrt{n}}; \bar{x}-E < \mu < \bar{x}+E$\\
\hline
$E=t_c \dfrac{s}{\sqrt{n}}; \bar{x}-E < \mu < \bar{x}+E$\\
\hline 
$E=z_c\sqrt{\dfrac{\hat{p}\hat{q}}{n}}; \hat{p}-E < p < \hat{p}+E$ \\
\hline
$\dfrac{(n-1)s^2}{\chi_R^2} < \sigma^2 < \dfrac{(n-1)s^2}{\chi_L^2}$ \\
\hline
$\sqrt{\dfrac{(n-1)s^2}{\chi_R^2}} < \sigma < \sqrt{\dfrac{(n-1)s^2}{\chi_L^2}}$ \\
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{tabular}{|l|}
\hline 
Finding $\chi_R^2$ and $\chi_L^2$: \\
use $\dfrac{1+c}{2}$ for $\chi_L^2$ \\
use $\dfrac{1-c}{2}$ for $\chi_R^2$ \\
\hline 
$\mu_{\bar{x}}=\mu$ and $\sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}$\\
\hline 
$z = \dfrac{x-\mu}{\sigma}$ \\
\hline
$z=\dfrac{\bar{x}-\mu}{\sigma_{\bar{x}}}; \sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}$ \\
\hline
$t=\dfrac{\bar{x}-\mu}{s_{\bar{x}}}; s_{\bar{x}}=\dfrac{s}{\sqrt{n}}$\\
\hline
$z=\dfrac{\hat{p}-p}{\sigma_{\hat{p}}}; \sigma_{\hat{p}}=\sqrt{\dfrac{pq}{n}}$ \\
\hline
$\chi^2 = \dfrac{(n-1)s^2}{\sigma^2}$ \\
\hline
\end{tabular}
\end{minipage}
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\begin{questions}
\question[16] 
\begin{parts}
\part[8] An auto parts seller finds that 1 in every 100 parts sold is defective. Find the probability that the first defective part is the third part sold. \\
\textit{Solution}: This is a geometric distribution with $p=\dfrac{1}{100}$ and $q=\dfrac{99}{100}$. The distribution is given by $P(x)=\dfrac{1}{100} \left( \dfrac{99}{100} \right)^{x-1}$. Therefore the requested probability is
\[ P(3) = \dfrac{1}{100} \left( \dfrac{99}{100} \right)^{3-2} = \dfrac{1}{100} \left( \dfrac{99}{100} \right)^{1} = 0.0099.\]
\part[8] The mean number of heart transplants performed per day in the U.S. in a recent year was about six. Find the probability that the number of heart transplants performed on any given day is no more than 1. \\
\textit{Solution}: This is a Poisson distribution with $\mu=6$. The distribution is given by $P(x)=\dfrac{6^x e^{-6}}{x!}$. The requested probability is
\[ P(x \leq 1) = P(0) + P(1) = \dfrac{6^0 e^{-6}}{0!} + \dfrac{6e^{-6}}{1!} = e^{-6}+6e^{-6} \approx 0.01735\]
\end{parts}

\question[15] The monthly utility bills in a city are normally distributed, with a mean of $\$100$ and a standard deviation of $\$12$. Find the probability that a randomly selected utility bill is between $\$90$ and $\$120$. \\
\textit{Solution}: Here $\mu=100$ and $\sigma=12$. We want to find $P(90 \leq x \leq 120)$. Subtracting $\mu$ and dividing by $\sigma$ yields
\[ \dfrac{90-100}{12} \leq \underbrace{\dfrac{x-\mu}{\sigma}}_{=z} \leq \dfrac{120-100}{12},\]
i.e.
\[ -0.83 \leq z \leq 1.66.\]
To use the normal table we can compute the desired probability by the following calculation:
\[ P \left( -0.83 \leq z \leq 1.66 \right) = P\left( z \leq 1.66 \right) - P \left( -0.83 \right) \approx 0.9515 - 0.2033 = 0.7482 .\]

\question[13] The time spent (in days) waiting for a kidney transplant for people ages 35-49 can be approximated by a normal distribution with mean $1674$ and standard deviation $212.5$. What waiting time represented the 80th percentile? \\
\textit{Solution}: Here $\mu=1674$ and $\sigma=212.5$. The 80th percentile is $P_{80}$ corresponds to the critical value $z=0.84$. Therefore we see that 
\[ x = z\sigma+\mu = (0.84)(212.5)+1674 = 1852.5 \text{ days}.\]
\question[14] During a certain week, the mean price of gasoline in California was $\$4.117$ per gallon. A random sample of 38 gas stations is selected from this population. What is the probability that the mean price from the sample was between $\$4.128$ and $\$4.143$ that week? Assume $\sigma=\$0.049$.  \\
\textit{Solution}: Here $\mu_{\bar{x}}=\mu=4.117$ and $\sigma_{\bar{x}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.049}{\sqrt{38}} = 0.007948$. Now convert the inequality $4.128 \leq \bar{x} \leq 4.143$ into a $z$ score:
\[ \dfrac{4.128 - 4.117}{0.007948} \leq z \leq \dfrac{4.143 - 4.117}{0.007948},\]
i.e. 
\[ 1.38 \leq z \leq 3.27.\]
Therefore the requested probability is
\[ P( 4.128 \leq \bar{x} \leq 4.143 ) = P ( 1.38 \leq z \leq 3.27 ) = P(z \leq 3.27) - P(z \leq 1.38) = 0.0833.\]

\question[14] In a random sample of five people, the mean driving distance to work was 22.2 miles and the sample standard deviation was $5.8$ miles. Construct a $95\%$ confidence interval for the population mean.
\begin{parts}
\part[8] Summarize the data by filling out the table below: \\
\begin{center}\begin{tabular}{|l|l|}
\hline
$c$ & $=0.95$ \\
$n$ & =5 \\
$\text{d.f.}$ & = \underline{\textbf{4}} \\
(Critical value) $t_c$ & = \underline{\textbf{2.776}} \\
$\bar{x}$ & =$22.2$ \\
$s$ & =$5.8$ \\
$E$ & =\underline{$\mathbf{t_c \dfrac{s}{\sqrt{n}} = 2.776 \dfrac{5.8}{\sqrt{5}} = 7.2004}$}\\
\hline
\end{tabular}\end{center}

\part[6] (Fill in the \textbf{three} blanks) The confidence interval is: \\
\begin{center} $\underline{\mathbf{\bar{x}-E=14.9996}} < \underline{\mu} < \underline{\mathbf{\bar{x}+E=29.4004}}$\end{center}
\vfill

\end{parts}

\question[14] In a survey of 2303 U.S. adults, 734 believe in UFOs. Construct a $90\%$ confidence interval for the population proportion of U.S. adults who believe in UFOs.
\begin{parts}
\part[8] Summarize the data by filling out the table below: \\
\begin{center}\begin{tabular}{|l|l|}
\hline
$c$ & $=0.9$ \\
(Critical value) $z_c$ & =1.645 \\
$n$ & =\underline{\textbf{2303}}\\
$\underline{\mathbf{\hat{p}}}$ & =0.31 \\
$\hat{q}$ & =$\underline{\mathbf{1-\hat{p}=1-0.31=0.69}}$\\
$E$ & =$\underline{\mathbf{z_c \sqrt{\dfrac{\hat{p}\hat{q}}{n} }=1.645 \sqrt{\dfrac{(0.31)(0.69)}{2303}}=0.0158}}$\\
\hline
\end{tabular}\end{center}

\part[6] (Fill in the \textbf{three} blanks) The confidence interval is: \\
\begin{center} $\mathbf{\underline{\hat{p}-E=0.31-0.0158=0.2942}} < \underline{\mathbf{p}}  < \mathbf{\underline{\hat{p}+E=0.31+0.0158=0.3258}}$\end{center}
\end{parts}

\newpage
\question[14] A magazine includes a report on the prices of subcompact digital cameras. The article states that 11 randomly selected subcompact digital cameras have a sample standard deviation of $\$109$. Use an $80\%$ level of confidence to construct a confidence interval for the population standard deviation.
\begin{parts}
\part[8] Summarize the data by filling out the table below: \\
\begin{center}\begin{tabular}{|l|l|}
\hline
$c$ & =$0.8$ \\
$n$ & =11 \\
d.f. & =\underline{\textbf{10}} \\
\underline{\textbf{s}} & =109 \\
$\dfrac{1+c}{2}$ & =0.9 \\
$\dfrac{1-c}{2}$ & = 0.1 \\
$\chi_R^2$ & =\underline{\textbf{15.987}}\\
$\chi_L^2$ & =\underline{\textbf{4.865}} \\
\hline
\end{tabular}\end{center}
\part[6] (Fill in the \textbf{three} blanks) The confidence interval is: \\
\begin{center} $\mathbf{\underline{\sqrt{\dfrac{(n-1)s^2}{\chi_R^2}}=\sqrt{\dfrac{10(109^2)}{15.987}}=86.207}} <  \underline{\mathbf{\sigma}} < \mathbf{\underline{\sqrt{\dfrac{(n-1)s^2}{\chi_L^2}}=\sqrt{\dfrac{10(109^2)}{4.865}}=156.237}}$\end{center}
\end{parts}
\vfill

\end{questions}

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