%% This document created by Scientific Notebook (R) Version 3.0 \documentclass[12pt,thmsa]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{sw20jart} %TCIDATA{TCIstyle=article/art4.lat,jart,sw20jart} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2606} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %TCIDATA{Created=Mon Aug 19 14:52:24 1996} %TCIDATA{LastRevised=Monday, October 04, 2004 11:35:13} %TCIDATA{} %TCIDATA{CSTFile=Exam.cst} %TCIDATA{PageSetup=72,72,36,36,0} \input{tcilatex} \begin{document} \section{Exam} \subsection{Comment} Skrivning i kontinuerliga f\"{o}rdelningar \subsection{Text} \section{Continous probability distributions} Namn: %TCIMACRO{\HTML{}}% %BeginExpansion %% %EndExpansion \quad Till m\aa nga av flervalsfr\aa gorna kan mer \"{a}n ett svar vara m\"{o}% jligt. Du f\"{o}rv\"{a}ntas ge det i sammanhanget b\"{a}sta m\"{o}jliga svaret. F\"{o}r att en dugga skall bli godk\"{a}nd m\aa ste du ha minst 3 godk\"{a}% nda p\aa\ teoridelen \textbf{och} minst 3 godk\"{a}nda p\aa\ problemdelen. Hj\"{a}lpmedel: Eget formelblad om h\"{o}gst en A4 sida skrivet med hj\"{a}% lp av Scientific NoteBook. Du har m\"{o}jlighet att redovisa en dugga p\aa\ ett avsnitt endast en g\aa % ng. \bigskip \subsection{Setup} Choices: Break, Radio, Permute Question space: 1 Print Choices: (a), (b), (c), (d), (e), (f) Submit: Click to grade Title: Continous probability distributions $\limfunc{nplaces}(x,n)=1.0\left\lfloor 10^{n}x+0.5\right\rfloor /10^{n}$ Seed: 15 \section{Question} \subsection{Text} {\Large Teorifr\aa gor} \rule{20cm}{0.1cm} \section{Variant} \subsection{Statement} Picture a relative frequency histogram that measures, on the horizontal axis, various (equal--sizes) lifetime ranges of appliance. The height of any histogram column then represents \subsection{Choices} \begin{itemize} \item the relative frequency density \item the proportion of all appliance units the lifetime of which falls within the range that constitutes the base of the column \item the probability that any randomly chosen unit of this appliance has a lifetime within the range that constitutes the base of the column \end{itemize} \begin{description} \item[Fixed] all of the above$\correctchoice{}$ \end{description} \section{Variant} \subsection{Statement} Picture a probability density function that measures, on the horizontal axis, various possible average total costs (ATC) of producing a given product. The probability associated with ATC=10.937 USD \subsection{Choices} \begin{itemize} \item equals zero$\correctchoice{}$ \item is measured by the height of the function above the stated ATC value \item is measured by the height of the function above the stated ATC value relative to the total area under the function \item cannot possbly be determined without seeing what the function looks like \end{itemize} \section{Variant} \subsection{Statement} Which one of the following is not a continous probability distribution \subsection{Choices} \begin{itemize} \item The Poisson probability distribution$\correctchoice{}$ \item The exponential probability distribution \item The normal probability distribution \item The uniform probability distribution \end{itemize} \section{Question} \section{Variant} \subsection{Statement} The normal probability distribution function is characterized by \subsection{Choices} \begin{itemize} \item an equality of the mean, median and mode \item tails extending indefinitely in both directions from the center \item a positive probability of finding a value of the random variable within any given range from minus infinity to plus infinity \end{itemize} \begin{description} \item[Fixed] all of the above$\correctchoice{}$ \end{description} \section{Variant} \subsection{Statement} A normal curve will appear the more peaked \subsection{Choices} \begin{itemize} \item the smaller is the value of $\sigma _{X}\correctchoice{}$ \item the larger is the value of $\sigma _{X}$ \item the larger is the value of $\mu _{X}$ \item the more closely mean, median and mode coincide \end{itemize} \section{Variant} \subsection{Statement} A normal probability density function with a mean of zero and a standard deviation of one \subsection{Choices} \begin{itemize} \item is called a standard normal curve$\correctchoice{}$ \item is called a standard normal deviate \item is a logically impossibility (because mean equals median in a normal curve) \item is a logically impossibility (because a normal curve has points of inflectio at $\mu _{X}\pm \sigma _{X}$) \end{itemize} \section{Question} \section{Variant} \subsection{Statement} If the standard normal deviate of an observation equals $+1.5$, while the mean and standard deviation of a normal random variable equal $12$ and $3$, respectively, the observation itself \subsection{Choices} \begin{itemize} \item must equal $16.5\correctchoice{}$ \item must equal $7.5$ \item must equal $21$ \item cannot possibly be determined without additional information \end{itemize} \section{Variant} \subsection{Statement} The normal distribution is a continous probability distribution \subsection{Choices} \begin{itemize} \item nevertheless, it can be used to approximate certain discrete probability distributions, such as the binomial, hypergeometric or Poisson distributions.$\correctchoice{}$ \item therefore, it cannot be used to approximate discrete probability distributions \item however, when $\mu \geq 20$, it becomes a discrete probability distributions \end{itemize} \begin{description} \item[Fixed] all the above statements about it are pure nonsense \end{description} \section{Variant} \subsection{Statement} The continuity correction factor turns any discrete $x$ value into a small range of $x$ \subsection{Choices} \begin{itemize} \item by adding half the distance between discrete values to any given $x$ value$\correctchoice{}$ \item by deducting $\mu $ from $x$ and dividing the result by $\sigma $ \item by deducting $\mu /2$ from every $x$ \item by using the interval from $x$ to $x+1$ \end{itemize} \section{Question} \section{Variant} \subsection{Statement} Whenever the number of occurences of an event is determined by a Poisson process, the likelihood of encountering specified intervals of time or space between consecutive occurrences can be described \subsection{Choices} \begin{itemize} \item the exponential probability distribution$\correctchoice{}$ \item the Poisson probability distribution \item the uniform probability distribution \item the standard normal deviate \end{itemize} \section{Variant} \subsection{Statement} Any quantitative variable the numerical value of which is determined by chance is called \subsection{Choices} \begin{itemize} \item a random variable$\correctchoice{}$ \item a standard normal deviate \item an expected value \end{itemize} \begin{description} \item[Fixed] none of the above \end{description} \section{Variant} \subsection{Statement} Any regular probability distribution (that shows probabilities for individual values of a random variable) can easily be converted into \subsection{Choices} \begin{itemize} \item a cumulative probability distribution$\correctchoice{}$ \item a binomial probability distribution \item a hypergeometric probability distribution \end{itemize} \begin{description} \item[Fixed] none of the above \end{description} \section{Question} \section{Variant} \subsection{Statement} According to a common rule of thumb, a normal probability distribution can nicely approximate a Poisson distribution whenever the value of $\mu $ in the Poisson distribution \subsection{Choices} \begin{itemize} \item exceeds $20\correctchoice{}$ \item exceeds $5$ \item equals $10$ \item is smaller than $10$ \end{itemize} \section{Variant} \subsection{Statement} According to a common rule of thumb, a normal probability distribution can nicely approximate a binomial distribution whenever \subsection{Choices} \begin{itemize} \item $n$ is large and $\pi $ is close to $0.5\correctchoice{}$ \item $n$ is large and $\pi $ is close to $1$ \item $n$ is small and $\pi $ is close to $0$ \item $n$ is small and $\pi $ is close to $1$ \end{itemize} \section{Variant} \subsection{Statement} The uniform probability distribution \subsection{Choices} \begin{itemize} \item is equally likely to take on any of the values within a given range$% \correctchoice{}$ \item is single--peaked above the random variable's mean, median and mode \item is characterized by tails extending indefinitely in both directions from the central mean \item is correctly described by two of the choices above \end{itemize} \section{Question} \subsection{Setup} Choices: No Break \subsection{Text} {\Large Problem} \rule{20cm}{0.1cm} \section{Variant} \subsection{Setup} $\mu :=\func{rand}\left( 25,50\right) $ $\sigma :=\func{rand}\left( 5,15\right) $ $p:=\func{rand}\left( 1,9\right) /10$ $\func{svar}:=\limfunc{nplaces}(\func{NormalInv}\left( p;\mu ,\sigma \right) ,2)$ Condition: $\left( p\neq 0.5\right) $ \subsection{Statement} The time required by a bank teller to cash a check has a mean of $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}% %BeginExpansion \mu % %EndExpansion $ seconds and a standard deviation of $% %TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluatenum}}% %BeginExpansion \sigma % %EndExpansion $ seconds. Find the times representing the $% %TCIMACRO{\FORMULA{100p}{100p}{evaluate}}% %BeginExpansion 100p% %EndExpansion $:th percentile under the assumption of normally distributed variables. Answer with 2 decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $\mu :=\func{rand}\left( 15,25\right) $ $\sigma :=\func{rand}\left( 10,20\right) /10$ $p_{1}:=0.3+\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $p_{2}:=p_{1}+\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $a:=\left\lfloor 10\left( \mu +\sigma \func{NormalInv}\left( p_{1};0,1\right) \right) +0.5\right\rfloor /10$ $b:=\left\lfloor 10\left( \mu +\sigma \func{NormalInv}\left( p_{2};0,1\right) \right) +0.5\right\rfloor /10$ $\func{svar}:=\limfunc{nplaces}(\func{NormalDist}\left( \frac{b-\mu }{\sigma };0,1\right) -\func{NormalDist}\left( \frac{a-\mu }{\sigma };0,1\right) ,3)$ \subsection{Statement} The time required to install a new aircraft engine is a normally distributed random variable with a mean of $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}% %BeginExpansion \mu % %EndExpansion $ hours and a standard deviation of $% %TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluatenum}}% %BeginExpansion \sigma % %EndExpansion $ hour. What is the probability that the next installation takes between $% %TCIMACRO{\FORMULA{a}{a}{evaluatenum}}% %BeginExpansion a% %EndExpansion $ and $% %TCIMACRO{\FORMULA{b}{b}{evaluatenum}}% %BeginExpansion b% %EndExpansion $ hours? Answer with 3 decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $\mu :=\func{rand}\left( 100000,150000\right) $ $\sigma :=\func{rand}\left( 5000,15000\right) $ $a:=\mu +\func{rand}\left( -\left\lceil 2\sigma \right\rceil ,\left\lfloor 2\sigma \right\rfloor \right) $ $\func{svar}:=\limfunc{nplaces}(1-\func{NormalDist}\left( \frac{a-\mu }{% \sigma };0,1\right) ,3)$ \subsection{Statement} The weekly number of checks cleared by a bank is a normally distributed random variable with a mean of $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluate}}% %BeginExpansion \mu % %EndExpansion $ checks cleared and a standard deviation of $% %TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluate}}% %BeginExpansion \sigma % %EndExpansion $ checks. In what proportion of weeks will the bank have to clear more than $% %TCIMACRO{\FORMULA{a}{a}{evaluate}}% %BeginExpansion a% %EndExpansion $ checks? \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Question} \subsection{Setup} Choices: No Break \section{Variant} \subsection{Setup} $\mu :=\func{rand}\left( 75,85\right) $ $\sigma :=\func{rand}\left( 5,10\right) $ $N:=\func{rand}\left( 300,3000\right) $ $p_{1}:=\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $p_{2}:=0.5+\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $v_{1}:=\left\lceil \mu +\sigma \func{NormalInv}\left( p_{1};0,1\right) \right\rceil $ $v_{2}:=\left\lceil \mu +\sigma \func{NormalInv}\left( p_{2};0,1\right) \right\rceil $ $n:=\left\lceil Np_{1}\right\rceil $ $\func{svar}:=\left\lfloor n\frac{1-\func{NormalDist}\left( \frac{v_{2}-\mu }{\sigma };0,1\right) }{\func{NormalDist}\left( \frac{v_{1}-\mu }{\sigma }% ;0,1\right) }+0.5\right\rfloor $ Condition: $\left( v_{1}+v_{2}\neq 2\mu \right) $ \subsection{Statement} A quality inspector tests the strength of window panes. Their breaking pressure is a normally distributed random variable with a mean of $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluate}}% %BeginExpansion \mu % %EndExpansion $ mph wind velocity and a standard deviation of $% %TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluate}}% %BeginExpansion \sigma % %EndExpansion $ mph. Some $% %TCIMACRO{\FORMULA{n}{n}{evaluate}}% %BeginExpansion n% %EndExpansion $ panes broke under simulated wind velocities of $% %TCIMACRO{\FORMULA{v_{1}}{v_{1}}{evaluate}}% %BeginExpansion v_{1}% %EndExpansion $ mph and less. How many withstood the hurricane velocity of $% %TCIMACRO{\FORMULA{v_{2}}{v_{2}}{evaluate}}% %BeginExpansion v_{2}% %EndExpansion $ mph or more? Answer with a whole number. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $n:=\func{rand}\left( 45000,55000\right) $ $\mu :=\func{rand}\left( 165,180\right) /10$ $\sigma :=\func{rand}\left( 700,900\right) /1000$ $p_{1}:=0.6+\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $p_{2}:=0.4+\func{rand}\left( \left\{ 0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.24,0.25,0.26,0.27,0.28,0.29,0.3\right\} \right) $ $b:=\left\lfloor 10\left( \mu +\sigma \func{NormalInv}\left( p_{1};0,1\right) \right) +0.5\right\rfloor /10$ $a:=\left\lfloor 10\left( \mu +\sigma \func{NormalInv}\left( 1-p_{2};0,1\right) \right) +0.5\right\rfloor /10$ $\func{svar}:=\left\lfloor 100\left( 1-\func{NormalDist}\left( \frac{a-\mu }{% b-\mu }\func{NormalInv}\left( p_{1};0,1\right) ;0,1\right) \right) +0.5\right\rfloor $ Condition:$\left( \left( a\neq b\right) \wedge \left( a\neq \mu \right) \wedge \left( b\neq \mu \right) \right) $ \subsection{Statement} A study of $% %TCIMACRO{\FORMULA{n}{n}{evaluate}}% %BeginExpansion n% %EndExpansion $ coal miners shows that the daily output per worker equals $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluatenum}}% %BeginExpansion \mu % %EndExpansion $ tons on the average. Some $% %TCIMACRO{\FORMULA{100p_{1}}{100.0p_{1}}{evaluatenum}}% %BeginExpansion 100.0p_{1}% %EndExpansion $ percent produce $% %TCIMACRO{\FORMULA{b}{b}{evaluatenum}}% %BeginExpansion b% %EndExpansion $ tons or fewer. What percentage produces more than $% %TCIMACRO{\FORMULA{a}{a}{evaluatenum}}% %BeginExpansion a% %EndExpansion $ tons? Assume normal probability distribution. Answer with no decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} Choices: Break $\mu :=\func{rand}\left( 48,56\right) $ $\sigma :=\func{rand}\left( 5,15\right) $ $a:=\mu +\func{rand}\left( 1,\left\lceil 0.1\mu \right\rceil \right) $ \subsection{Statement} A personnel manager has discovered that the hours of sick leave taken by employees during a year are a normal distributed random variable with a mean of $% %TCIMACRO{\FORMULA{\mu }{\mu }{evaluate}}% %BeginExpansion \mu % %EndExpansion $ hours and a standard deviation of $% %TCIMACRO{\FORMULA{\sigma }{\sigma }{evaluate}}% %BeginExpansion \sigma % %EndExpansion $ hours. She considers the hours taken by A ($% %TCIMACRO{\FORMULA{a}{a}{evaluate}}% %BeginExpansion a% %EndExpansion $) highly unusual. Do you agree? \subsection{Choices} \begin{itemize} \item No, because $% %TCIMACRO{% %\FORMULA{\left\lfloor 100\left( 1-\func{NormalDist}\left( \frac{a-\mu }{\sigma }\right) \right) +0.5\right\rfloor }{\left\lfloor 100.\,\allowbreak 5-100.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) \right\rfloor \allowbreak }{evaluatenum}}% %BeginExpansion \left\lfloor 100.\,\allowbreak 5-100.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) \right\rfloor \allowbreak % %EndExpansion $ percent of the personnel can be expected to have this sick leave$% \correctchoice{}$ \item No, because $% %TCIMACRO{% %\FORMULA{\left\lfloor 100\func{NormalDist}\left( \frac{a-\mu }{\sigma }\right) +0.5\right\rfloor }{\left\lfloor 100.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) +.\,\allowbreak 5\right\rfloor \allowbreak }{evaluatenum}}% %BeginExpansion \left\lfloor 100.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) +.\,\allowbreak 5\right\rfloor \allowbreak % %EndExpansion $ percent of the personnel can be expected to have this sick leave \item Yes, because only $% %TCIMACRO{% %\FORMULA{\left\lfloor 100\left( 1-\func{NormalDist}\left( \frac{a-\mu }{\sigma }\right) \right) ^{2}+0.5\right\rfloor }{\left\lfloor 100.0\left( 1.0-1.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) \right) ^{2}+.\,\allowbreak 5\right\rfloor \allowbreak }{evaluatenum}}% %BeginExpansion \left\lfloor 100.0\left( 1.0-1.0\func{statevalf}_{\func{cdf},\func{normald}}\left( \frac{a-\mu }{\sigma }\right) \right) ^{2}+.\,\allowbreak 5\right\rfloor \allowbreak % %EndExpansion $ percent of the personnel can be expected to have this sick leave \item Yes, because only $% %TCIMACRO{% %\FORMULA{\left\lfloor 10\sqrt{\func{NormalDist}\left( \frac{a-\mu }{\sigma }\right) }+0.5\right\rfloor }{\left\lfloor 10.0\func{statevalf}_{\func{cdf},\func{normald}}^{\frac{1}{2}}\left( \frac{a-\mu }{\sigma }\right) +.\,\allowbreak 5\right\rfloor \allowbreak }{evaluatenum}}% %BeginExpansion \left\lfloor 10.0\func{statevalf}_{\func{cdf},\func{normald}}^{\frac{1}{2}}\left( \frac{a-\mu }{\sigma }\right) +.\,\allowbreak 5\right\rfloor \allowbreak % %EndExpansion $ percent of the personnel can be expected to have this sick leaveQuestion \end{itemize} \section{Question} \subsection{Setup} Choices: No Break \section{Variant} \subsection{Setup} $n:=\func{rand}\left( 20,45\right) $ $p:=\func{rand}\left( 10,20\right) /100$ $k:=\func{rand}\left( \max \left( \left\lceil np-2\sqrt{np\left( 1-p\right) }% \right\rceil ,1\right) ,\max \left( \left\lfloor np+2\sqrt{np\left( 1-p\right) }\right\rfloor ,1\right) \right) $ $\func{svar}:=\limfunc{nplaces}(\func{NormalDist}\left( \frac{k+0.5-np}{% \sqrt{np\left( 1-p\right) }};0,1\right) -\func{NormalDist}\left( \frac{% -0.5-np}{\sqrt{np\left( 1-p\right) }};0,1\right) ,3)$ \subsection{Statement} Experience shows that $% %TCIMACRO{\FORMULA{100p}{100p}{evaluate}}% %BeginExpansion 100p% %EndExpansion $ percent of the people entering a store make a purchase. What is the probability that that at most $% %TCIMACRO{\FORMULA{k}{k}{evaluate}}% %BeginExpansion k% %EndExpansion $ of the next $% %TCIMACRO{\FORMULA{n}{n}{evaluate}}% %BeginExpansion n% %EndExpansion $ customers do purchase (use approximation with correction)? Answer with 3 decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $n:=\func{rand}\left( 90,150\right) $ $p:=\func{rand}\left( 2,12\right) /100$ $k:=\func{rand}\left( \left\lceil n\left( 1-p\right) \right\rceil +2,\left\lfloor n\left( 1-p\right) -0.5+2\sqrt{np\left( 1-p\right) }% \right\rfloor -2\right) $ $\func{svar}:=\limfunc{nplaces}(1-\func{NormalDist}\left( \frac{% k+0.5-n\left( 1-p\right) }{\sqrt{np\left( 1-p\right) }};0,1\right) ,3)$ \subsection{Statement} The absentee rate in a class of $% %TCIMACRO{\FORMULA{n}{n}{evaluate}}% %BeginExpansion n% %EndExpansion $ statistics students is $% %TCIMACRO{\FORMULA{100p}{100p}{evaluate}}% %BeginExpansion 100p% %EndExpansion $ percent. With the help of an appropriate approximation, compute the probability that in the next class more than $% %TCIMACRO{\FORMULA{k}{k}{evaluate}}% %BeginExpansion k% %EndExpansion $ will attend. Use approximation with correction. Answer with 3 decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $n:=\func{rand}\left( 90,150\right) $ $p:=\func{rand}\left( 2,12\right) /100$ $k_{1}:=\func{rand}\left( \left\lceil n\left( 1-p\right) -2\sqrt{np\left( 1-p\right) }\right\rceil +2,\left\lfloor n\left( 1-p\right) \right\rfloor \right) $ $k_{2}:=k_{1}+\func{rand}\left( 1,\max \left( \left\lceil 2\sqrt{np\left( 1-p\right) }\right\rceil -2,2\right) \right) $ $\func{svar}:=\limfunc{nplaces}(\func{NormalDist}\left( \frac{% k_{2}+0.5-n\left( 1-p\right) }{\sqrt{np\left( 1-p\right) }};0,1\right) -% \func{NormalDist}\left( \frac{k_{1}-0.5-n\left( 1-p\right) }{\sqrt{np\left( 1-p\right) }};0,1\right) ,3)$ \subsection{Statement} The absentee rate in a class of $% %TCIMACRO{\FORMULA{n}{n}{evaluate}}% %BeginExpansion n% %EndExpansion $ statistics students is $% %TCIMACRO{\FORMULA{100p}{100p}{evaluate}}% %BeginExpansion 100p% %EndExpansion $ percent. With the help of an appropriate approximation, compute the probability that in the next class between $% %TCIMACRO{\FORMULA{k_{1}}{k_{1}}{evaluate}}% %BeginExpansion k_{1}% %EndExpansion $ and $% %TCIMACRO{\FORMULA{k_{2}}{k_{2}}{evaluate}}% %BeginExpansion k_{2}% %EndExpansion $ will attend. (Use approximation with correction) \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Question} \subsection{Setup} Choices: No Break \section{Variant} \subsection{Setup} $a:=\func{rand}\left( 1,60\right) $ $b:=\func{rand}\left( a,60\right) $ $\func{svar}:=\limfunc{nplaces}(\frac{b+a}{2},1)$ Condition: $\left( \left( a\neq b\right) \wedge \left( 2a\neq b\right) \right) $ \subsection{Statement} The arrival of a bus is equally likely at between $% %TCIMACRO{\FORMULA{a}{a}{evaluate}}% %BeginExpansion a% %EndExpansion $ and $% %TCIMACRO{\FORMULA{b}{b}{evaluate}}% %BeginExpansion b% %EndExpansion $ minutes from now. Determine your expected waiting time. Answer with 1 decimal. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $b:=\func{rand}\left( 5,60\right) $ $k:=\func{rand}\left( 1,b\right) $ $\func{svar}:=\limfunc{nplaces}(1-\frac{k}{b},3)$ Condition: $\left( 2k\neq b\right) $ \subsection{Statement} The arrival of a bus is equally likely at any time during the next $% %TCIMACRO{\FORMULA{b}{b}{evaluate}}% %BeginExpansion b% %EndExpansion $ minutes. Determine the probability that you have to wait more than $% %TCIMACRO{\FORMULA{k}{k}{evaluate}}% %BeginExpansion k% %EndExpansion $ minutes. Answer with 3 decimals. \subsection{Choices} InputField(MATH) GradeProc: givecredit($\limfunc{response}=\func{svar}$) \subsection{Answer} R\"{a}tt svar \"{a}r $% %TCIMACRO{\FORMULA{\func{svar}}{\func{svar}}{evaluatenum}}% %BeginExpansion \func{svar}% %EndExpansion $ \section{Variant} \subsection{Setup} $b:=\func{rand}\left( 5,60\right) $ $k_{1}:=\func{rand}\left( 1,b\right) $ $k_{2}:=k_{1}+\func{rand}\left( 1,b\right) $ $\func{svar}:=\limfunc{nplaces}(\frac{k_{2}-k_{1}}{b},3)$ Condition: $\left( \left( k_{2}