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\begin{center}\huge MATH 1550 - EXAM 1 FALL 2018\\
SOLUTION
%\vspace*{2in}
%\large Name: \makebox[3in]{\hrulefill}\\
\end{center}
\begin{flushleft}\vspace{0.5in}
Friday 14 September 2018 \\
Instructor: Tom Cuchta
\end{flushleft}
\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}
\vspace*{10pt}
\end{coverpages}
\begin{questions}
\question[6] Consider the following data representing the pungencies of tabasco peppers (in thousands of Scoville units):\\
\begin{tabular}{llllllll}
35 & 51 & 44 & 42 & 37 & 38 & 36 & 39 \\
44 & 43 & 40 & 40 & 32 & 39 & 41 & 38 \\
\end{tabular}\\
Construct a frequency distribution for the data using $4$ classes. \\
\textit{Solution} (this problem resembles multiple problems in \textbf{HW1}): First compute
\[ \text{range} = \text{max} - \text{min} = 51-32 = 19\]
and
\[ \text{class width} = \dfrac{\text{range}}{\text{\# of classes}} = \dfrac{19}{4} = 4.75 \stackrel{\text{round up}}{\longrightarrow} 5,\]
we get the following distribution:
\[ \begin{array}{|l|l|}
\text{Class} & \text{Frequency} \\
\hline
32--36 & 3 \\
37--41 & 8 \\
42--46 & 4 \\
47--51 & 1
\end{array} \]
\question[10] Consider the following data representing the cholesterol levels of a sample of 9 employees: \\
\begin{tabular}{lllllllllllllll}
154 & 240 & 171 & 188 & 235 & 203 & 184 & 181 & 203
\end{tabular}
\begin{parts}
\part[5] Find the median. \\
\textit{Solution} (this problem resembles \textbf{Quiz 1 and \#13 in HW2}): First order the data to get \\
\begin{center}\begin{tabular}{lllllllllllllll}
154 & 171 &181& 184& 188& 203& 203& 235& 240
\end{tabular}\end{center}
From this ordering, the middle data point is $188$, so that is the median.
\part[5] Find the mode. \\
\textit{Solution} (this problem resembles \textbf{Quiz 1 and \#13 in HW2}): The most common data point is $203$, which is the mode.
\end{parts}
%\question[6] Find the weighted mean score of the following scores from a hypothetical student in MATH 1550 at the end of the course:\\
%\begin{tabular}{lll}
% & \textbf{Score} (out of 100 points) & \textbf{Percent of final grade} \\
%Homework & 72 & 15 \% \\
%Quiz & 45 & 10 \% \\
%Exam 1 & 80 & 15 \% \\
%Exam 2 & 70 & 15 \% \\
%Exam 3 & 68 & 15 \% \\
%Final Exam & 86 & 30 \% \\
%\end{tabular} \\
\question[12] Consider the following data representing the total number of votes for the office of President of the United States (in hundreds of thousands) in West Virginia for general elections from 2000 to 2016:\\
\begin{tabular}{lllll}
631.9 & 750.3 & 701.3 & 655.9 & 678.1
\end{tabular}
\begin{parts}
\part[6] Find the mean. \\
\textit{Solution} (this problem resembles \textbf{Quiz 1 and \#13 in HW2}): Calculate
\[ \text{mean} = \dfrac{631.9+750.3+701.3+655.9+678.1}{5} = \dfrac{3417.5}{5} = 683.5.\]
\part[6] Find the population standard deviation. \\
\textit{Solution} (this problem resembles \textbf{Quiz 1 and \#13 in HW2}): Compute
\[ \begin{array}{ll}
\sigma &= \sqrt{\dfrac{(631.9-683.5)^2+(750.3-683.5)^2+(701.3-683.5)^2+(655.9-683.5)^2+(678.1-683.5)^2}{5}}\\
&= \sqrt{\dfrac{8232.56}{5}} \\
&= \sqrt{1646.512} \\
&=40.577
\end{array}\]
\end{parts}
\newpage
\question[12]
\begin{parts}
\part[6] The mean monthly utility bill for a sample of households in a city is $\$68$, with a standard deviation of $\$7$. Between what two values does about $95\%$ of the data lie? (Assume the data has a bell-shaped distribution) \\
\textit{Solution} (this problem resembles \textbf{\#29 in HW2}): We are told that the mean is $\mu=68$ and the standard deviation is $\sigma=7$. The empirical rule tells us that $95\%$ of the data lies within two standard deviations of the mean. So compute
\[ \mu + 2\sigma = 68 + 2(7) = 68+14 = 82\]
and
\[ \mu - 2\sigma = 68 - 2(7) = 68-14 = 54.\]
Therefore $95\%$ of the data lies between the values $\$54$ and $\$82$.
\part[6] You are conducting a survey on the number of people per household in your region. From a sample with $n=48$, the mean number of pets per household is $2$ pets and the standard deviation is $1$ pet. Using Chebychev's theorem, determine at least how many of the households have $0$ to $4$ pets. \\
\textit{Solution} (this problem resembles \textbf{\#36 in HW2}): First figure out how many standard deviations ($s=1$) away from the mean $\overline{x}=2$ the values $0$ and $4$ are. Since
\[ 2 -2(1) = 0 \]
and
\[ 2 + 2(1) = 4,\]
we see that ``$0$ to $4$ pets" indicates data values $k=2$ standard deviations from the mean. Therefore Chebyshev's theorem tells us that $1-\dfrac{1}{k^2}=1-\dfrac{1}{4}=\dfrac{3}{4}$ of the data are in the range of ``$0$ to $4$ households". From our sample, we can say that at least $\dfrac{3}{4}$ of the $n=48$ houses sampled yields $\left( \dfrac{3}{4} \right) (48) = 36$ households are in this category.
\end{parts}
\question[20] Consider the following data points: \\
\begin{tabular}{llllllllll}
56 & 63 & 57 & 60 & 80 & 54 & 63 \\
\end{tabular}
\begin{parts}
\part[6] Find the quartiles $Q_1$, $Q_2$, and $Q_3$. \\
\textit{Solution} (this problem resembles \textbf{\#11 and \#17 in HW3}): \\ First order the data to get
\begin{tabular}{lllllll}
54& 56 &57 &60 &63& 63& 80
\end{tabular}
First find the median to get $Q_2=60$. Find the median of the data points to the left of $60$ to get $Q_1=56$ and find the median of the data points to the right of $60$ to get $Q_3=63$.
\part[6] Draw a box whisker plot for this data. \\
\textit{Solution}: \begin{tikzpicture}
\begin{axis}
[
ytick={1,2,3},
yticklabels={},
xtick={54,56,60,63,80},
]
\addplot+[
boxplot prepared={
median=60,
upper quartile=63,
lower quartile=56,
upper whisker=80,
lower whisker=54
},
] coordinates {};
\end{axis}
\end{tikzpicture}
\part[4] Find the IQR. \\
\textit{Solution:} \[ IQR = Q_3-Q_1 = 63-56 = 7 \]
\part[4] Identify outliers. \\
\textit{Solution}: First compute
\[ 1.5IQR = 1.5(7) = 10.5, \]
\[ Q_1 - 1.5IQR = 56-10.5 = 45.5, \]
and
\[ Q_3+1.5IQR = 63+10.5 = 73.5.\]
Since the data point 80 is larger than $Q_3+1.5IQR$, we consider it an outlier.
\end{parts}
%\question[5] The distribution of a set of data is approximately bell-shaped. The mean of the data is $18.7$ with a standard deviation of $3.2$. Find the $z$-score of the data point $x=19.3$. \\
\question[18] A probability experiment is carried out by flipping three coins and then rolling a $3$-sided die.
\begin{parts}
\part[6] How many outcomes are in the sample space? \\
\textit{Solution} (this problem resembles \textbf{\#51 in HW3}): 24
\part[6] How many outcomes are in the event of flipping two heads then a tail and then rolling a $3$? What is the probability of this event? \\
\textit{Solution}: There is one outcome for this event. Therefore the probability is $\dfrac{1}{24}$. \\
(\textit{note: there was some confusion on if the coin flips were ordered or not, so I also accepted that there could be $3$ outcomes for this event which would yield probability $\dfrac{3}{24} = \dfrac{1}{8}$})
\part[6] How many outcomes are in the event of the first coin flip being heads and rolling a number bigger than $1$? What is the probability of this event? \\
\textit{Solution}: There are 8 outcomes in this event. Therefore the probability of the event is $\dfrac{8}{24}=\dfrac{1}{3}$.
\end{parts}
\question[11] In a survey, 510 adults were asked whether they drive a pickup truck and whether they drive a Ford. The results showed that three in ten adults surveyed drive a Ford. Of the adults surveyed that drive Fords, two in nine drive a pickup truck. Find the probability that a randomly selected adult drives a Ford and drives a pickup truck.\\
\textit{Solution} (this problem resembles \textbf{Quiz 3 and \#20 and \# 23 in HW4}): We are told that $P\Big(\text{Ford}\Big)=\dfrac{3}{10}$ and $P\Big(\text{pickup truck}|\text{Ford}\Big)= \dfrac{2}{9}$. Therefore using the multiplication rule
\[ P \Big( \text{Ford} \text{ and } \text{pickup truck} \Big)=P(\text{Ford} \Big)P\Big(\text{pickup truck}|\text{Ford} \Big) = \dfrac{3}{10} \cdot \dfrac{2}{9} = \dfrac{6}{90} \approx 0.0666\ldots. \]
\question[11] Of the cartons produced by a company, $5\%$ have a puncture, $8\%$ have a smashed corner, and $0.4\%$ have both a puncture and a smashed corner. Find the probability that a randomly selected carton has a puncture or has a smashed corner.\\
\textit{Solution} (this problem resembles \textbf{\#16 in HW4}): We are told that $P\Big(\text{punctured}\Big)=0.05$, $P\Big(\text{smashed} \Big)=0.08$, and that \\ $P\Big(\text{punctured} \text{ and } \text{smashed} \Big) = 0.004$. Therefore using the addition rule,
\[\begin{array}{ll}
P\Big(\text{punctured or smashed} \Big) &= P\Big( \text{punctured} \Big) + P \Big( \text{smashed} \Big) - P \Big ( \text{punctured and smashed} \Big) \\
&= 0.05 + 0.08 - 0.004 \\
&= 0.126.
\end{array} \]
\end{questions}
\end{document}