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\input{tcilatex}
\begin{document}


\section{Exam}

\subsubsection{Comment}

I detta exempel testar vi studentens kunskaper i ber\"{a}kning av olika
typer av moment.

\subsubsection{Text}

\section{Moment}

\subsection{Comment}

seed:=12345

\subsection{Setup}

Errors: report

Choices: No Break

Title: Probability

Submit:Click to Grade

$\limfunc{nplaces}(x,n)=1.0\left\lfloor 10^{n}x+0.5\right\rfloor /10^{n}$

\section{Part}

\section{Text}

\section{Problemdel}

\section{Question}

\subsubsection{Setup}

Select: 1

\subsection{Comment}

V\"{a}ntev\"{a}rde och varians

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{\sum_{k=1}^{n}x_{k}}{n},\frac{%
\sum_{k=1}^{n}x_{k}^{2}}{n}-\left( \frac{\sum_{k=1}^{n}x_{k}}{n}\right)
^{2}\right) $

\subsubsection{Statement}

Givet en diskret likformig f\"{o}rdelning p\aa\ $\Omega =\left\{ x_{k}\mid
k=1,2,\ldots ,n\right\} $.

\subsubsection{Substatement}

Ge formeln f\"{o}r v\"{a}ntev\"{a}rdet:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Ge formeln f\"{o}r variansen som en summa av tv\aa\ termer:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ x_{k}\mid k=1,2,\ldots ,n\right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\sum_{k=1}^{n}x_{k}P\left( X=x_{k}\right) =\frac{%
\sum_{k=1}^{n}x_{k}}{n} \\
E\left( X^{2}\right) & =\sum_{k=1}^{n}x_{k}^{2}P\left( X=x_{k}\right) =\frac{%
\sum_{k=1}^{n}x_{k}^{2}}{n} \\
V\left( X\right) & =E\left( X^{2}\right) -E^{2}\left( X\right) =\frac{%
\sum_{k=1}^{n}x_{k}^{2}}{n}-\left( \frac{\sum_{k=1}^{n}x_{k}}{n}\right) ^{2}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( p,p\left( 1-p\right) \right) $

\subsubsection{Statement}

Givet en Bernoullif\"{o}rdelning med sannolikheten $p=P\left( X=1\right) $ f%
\"{o}r att lyckas.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ 0,1\right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =0\times \left( 1-p\right) +1\times p=p \\
E\left( X^{2}\right) & =0^{2}\times \left( 1-p\right) +1^{2}\times p=p \\
V\left( X\right) & =p-p^{2}=p\left( 1-p\right)
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( np,np\left( 1-p\right) \right) $

\subsubsection{Statement}

{Givet} en binomialf\"{o}rdelning med parametrarna $n$ och $p$.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

En binomialf\"{o}rdelning \"{a}r en summa av oberoende Bernoullif\"{o}%
rdelningar och vi har 
\begin{equation*}
X=\sum_{k=1}^{n}X_{k}\in Bin\left( n,p\right) \quad \text{d\"{a}r }X_{k}\in
Ber\left( p\right)
\end{equation*}%
h\"{a}rav f\"{o}ljer 
\begin{align*}
E\left( X\right) & =\sum_{k=1}^{n}E\left( X_{k}\right) =np \\
V\left( X\right) & =\sum_{k=1}^{n}V\left( X_{k}\right) =np\left( 1-p\right)
\end{align*}%
ty oberoende.

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{1}{p},\frac{1-p}{p^{2}}\right) $

\subsubsection{Statement}

Givet en geometrisk f\"{o}rdelning med parametern $p$ f\"{o}r att lyckas.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ 1,2,3,\ldots \right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\sum_{k=1}^{\infty }kp\left( 1-p\right) ^{k-1}=p\frac{d}{%
dp}\sum_{k=0}^{\infty }\left( -\frac{d}{dp}\left( 1-p\right) ^{k}\right) \\
& =-p\frac{d}{dp}\sum_{k=0}^{\infty }\left( 1-p\right) ^{k}=-p\frac{d}{dp}%
\frac{1}{1-\left( 1-p\right) } \\
& =-p\frac{d}{dp}\frac{1}{p} \\
& =\frac{1}{p} \\
E\left( X^{2}\right) & =p\sum_{k=1}^{\infty }k^{2}\left( 1-p\right)
^{k-1}=p\sum_{k=1}^{\infty }\left( k+1\right) k\left( 1-p\right)
^{k-1}-p\sum_{k=1}^{\infty }k\left( 1-p\right) ^{k-1} \\
& =p\frac{d^{2}}{dp^{2}}\sum_{k=0}^{\infty }\left( 1-p\right) ^{k}-\frac{1}{p%
} \\
& =\frac{2}{p^{2}}-\frac{1}{p} \\
V\left( X\right) & =\frac{2}{p^{2}}-\frac{1}{p}-\frac{1}{p^{2}}=\frac{1}{%
p^{2}}-\frac{1}{p} \\
& =\frac{1-p}{p^{2}}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{n}{p},\frac{n\left( 1-p\right) }{p^{2}}\right) 
$

\subsubsection{Statement}

Givet en negativ binomialf\"{o}rdelning med parametrarna $p$ och $n$.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

En negativ binomialf\"{o}rdelning $X$ \"{a}r en summa av oberoende
geometriska f\"{o}rdelningar $X_{1},\cdots ,X_{n}$ varav 
\begin{align*}
X& =\sum_{k=1}^{n}X_{k} \\
E\left( X\right) & =\sum_{k=1}^{n}E\left( X_{k}\right) =\frac{n}{p} \\
V\left( X\right) & =\sum_{k=1}^{n}V\left( X_{k}\right) =\frac{n\left(
1-p\right) }{p^{2}}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \lambda ,\lambda \right) $

\subsubsection{Statement}

{Givet} en Poissonf\"{o}rdelning med parametern $\lambda $.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ 0,1,2,3,\ldots \right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\sum_{k=0}^{\infty }k\frac{\lambda ^{k}}{k!}e^{-\lambda
}=e^{-\lambda }\lambda \sum_{k=1}^{\infty }\frac{\lambda ^{k-1}}{\left(
k-1\right) !}=e^{-\lambda }\lambda \underset{e^{\lambda }}{\underbrace{%
\sum_{k=0}^{\infty }\frac{\lambda ^{k}}{k!}}}=\lambda \\
E\left( X^{2}\right) & =\sum_{k=0}^{\infty }k^{2}\frac{\lambda ^{k}}{k!}%
e^{-\lambda }=\sum_{k=0}^{\infty }k\left( k-1\right) \frac{\lambda ^{k}}{k!}%
e^{-\lambda }+\sum_{k=0}^{\infty }k\frac{\lambda ^{k}}{k!}e^{-\lambda } \\
& =e^{-\lambda }\lambda ^{2}\sum_{k=0}^{\infty }\frac{\lambda ^{k}}{k!}%
+\lambda =\lambda ^{2}+\lambda \\
V\left( X\right) & =\lambda ^{2}+\lambda -\lambda ^{2}=\lambda
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{a+b}{2},\frac{\left( b-a\right) ^{2}}{12}%
\right) $

\subsubsection{Statement}

{Givet} en kontinuerlig likformig f\"{o}rdelning med parametrarna $a$ och $b$%
.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ x\mid a<x<b\right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\int_{a}^{b}x\frac{1}{b-a}\,dx=\frac{a+b}{2} \\
E\left( X^{2}\right) & =\int_{a}^{b}x^{2}\frac{1}{b-a}\,dx=\frac{1}{3}\left(
a^{2}+ab+b^{2}\right) \\
V\left( X\right) & =\frac{1}{3}\left( a^{2}+ab+b^{2}\right) -\left( \frac{a+b%
}{2}\right) ^{2}=\frac{\left( b-a\right) ^{2}}{12}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{1}{\lambda },\frac{1}{\lambda ^{2}}\right) $

\subsubsection{Statement}

{Givet} en exponentialf\"{o}rdelning med parametern $\lambda $.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ x\mid 0<x<\infty \right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\int_{0}^{\infty }xe^{-\lambda x}\,dx=\left[ -\frac{1}{%
\lambda }xe^{-\lambda x}\right] _{0}^{\infty }+\frac{1}{\lambda }%
\int_{0}^{\infty }e^{-\lambda x}\,dx=\frac{1}{\lambda } \\
E\left( X^{2}\right) & =\int_{0}^{\infty }x^{2}e^{-\lambda x}\,dx=\left[ -%
\frac{1}{\lambda }x^{2}e^{-\lambda x}\right] _{0}^{\infty }+\frac{2}{\lambda 
}\int_{0}^{\infty }xe^{-\lambda x}\,dx=\frac{2}{\lambda ^{2}} \\
V\left( X\right) & =\frac{2}{\lambda ^{2}}-\frac{1}{\lambda ^{2}}=\frac{1}{%
\lambda ^{2}}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \mu ,\sigma ^{2}\right) $

\subsubsection{Statement}

{Givet} en normalf\"{o}rdelning med parametrarna $\mu $ och $\sigma $.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ x\mid -\infty <x<\infty \right\} $ erh\aa lls f\"{o}r $%
\mu =0$ och $\sigma =1$ att%
\begin{align*}
E\left( X\right) & =\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty }xe^{-%
\frac{x^{2}}{2}}\,dx=\frac{1}{\sqrt{2\pi }}\left[ -e^{-\frac{x^{2}}{2}}%
\right] _{-\infty }^{\infty }=0 \\
E\left( X^{2}\right) & =\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty
}x^{2}e^{-\frac{x^{2}}{2}}\,dx=\frac{1}{\sqrt{2\pi }}\left[ -xe^{-\frac{x^{2}%
}{2}}\right] _{-\infty }^{\infty }+\frac{1}{\sqrt{2\pi }}\int_{-\infty
}^{\infty }e^{-\frac{x^{2}}{2}}\,dx=1 \\
V\left( X\right) & =1-0^{2}=1
\end{align*}%
Det allm\"{a}nna fallet f\"{o}ljer ur sambandet $Y=\mu +X\sigma $ d\"{a}r $%
X\in N\left( 0,1\right) $

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{\alpha }{\lambda },\frac{\alpha }{\lambda ^{2}}%
\right) $

\subsubsection{Statement}

{Givet} en gammaf\"{o}rdelning med parametrarna $\alpha $ och $\lambda $.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Med $\Omega =\left\{ x\mid 0\leq x<\infty \right\} $ erh\aa lls%
\begin{align*}
E\left( X\right) & =\int_{0}^{\infty }x\frac{\lambda ^{\alpha }}{\Gamma
\left( \alpha \right) }x^{\alpha -1}e^{-\lambda x}\,dx=\frac{\alpha }{%
\lambda }\int_{0}^{\infty }\frac{\lambda ^{\alpha +1}}{\Gamma \left( \alpha
+1\right) }x^{\alpha }e^{-\lambda x}\,dx=\frac{\alpha }{\lambda } \\
E\left( X^{2}\right) & =\int_{0}^{\infty }x^{2}\frac{\lambda ^{\alpha }}{%
\Gamma \left( \alpha \right) }x^{\alpha -1}e^{-\lambda x}\,dx=\frac{\alpha
\left( \alpha +1\right) }{\lambda ^{2}}\int_{0}^{\infty }\frac{\lambda
^{\alpha +2}}{\Gamma \left( \alpha +2\right) }x^{\alpha +1}e^{-\lambda
x}\,dx=\frac{\alpha \left( \alpha +1\right) }{\lambda ^{2}} \\
V\left( X\right) & =\frac{\alpha \left( \alpha +1\right) }{\lambda ^{2}}%
-\left( \frac{\alpha }{\lambda }\right) ^{2}=\frac{\alpha }{\lambda ^{2}}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( n,2n\right) $

\subsubsection{Statement}

{Givet en }$\chi ^{2}$-f\"{o}rdelning med parametern $n$.

\subsubsection{Substatement}

Best\"{a}m dess v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m dess varians:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Det g\"{a}ller att en $\chi ^{2}$--f\"{o}rdelning med parametern $n$ \"{a}r
densamma som en gammaf\"{o}rdelning $Gam\left( \frac{n}{2},0.5\right) $ och d%
\"{a}rf\"{o}r erh\aa lls%
\begin{align*}
E\left( X\right) & =\frac{n/2}{0.5}=n \\
V\left( X\right) & =\frac{n/2}{\left( 0.5\right) ^{2}}=2n
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p:=\left( \frac{1}{8},\frac{3}{8},\frac{3}{8},\frac{1}{8}\right) $

$m:=p\left( 0,1,2,3\right) ^{T}$

$v:=p\left( 0^{2},1^{2},2^{2},3^{2}\right) ^{T}-m_{1}^{2}$

$s:=\limfunc{nplaces}\left( \sqrt{v_{1}},3\right) $

\paragraph{$\limfunc{svar}:=1$}

\subsubsection{Statement}

\AA terkom till denna!!!En stokastisk variabel $X$ har f\"{o}rdelningen 
\begin{equation*}
P\left( X=k\right) =\left\{ 
\begin{array}{ll}
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluatenum} }%
%BeginExpansion
p_{1}
%EndExpansion
& k=0 \\ 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluatenum} }%
%BeginExpansion
p_{2}
%EndExpansion
& k=1 \\ 
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{evaluatenum} }%
%BeginExpansion
p_{3}
%EndExpansion
& k=2 \\ 
%TCIMACRO{\FORMULA{p_{4}}{p_{4}}{evaluatenum} }%
%BeginExpansion
p_{4}
%EndExpansion
& k=3%
\end{array}%
\right.
\end{equation*}%
d\"{a}r $\Omega _{X}=\{0,1,2,3\}$. Best\"{a}m f\"{o}rdelningens v\"{a}ntev%
\"{a}rde, varians och standardavvikelse samt 
\begin{equation*}
E\left( \sum_{i=1}^{n}X_{i}\right)
\end{equation*}%
och 
\begin{equation*}
V\left( \sum_{i=1}^{n}X_{i}\right)
\end{equation*}%
n\"{a}r alla $X_{i}$ har f\"{o}rdelningen ovan och $X_{1},\ldots ,X_{n}$ 
\"{a}r oberoende.$%
%TCIMACRO{\FORMULA{m}{m}{evaluatenum}}%
%BeginExpansion
m%
%EndExpansion
%TCIMACRO{\FORMULA{s}{s}{evaluatenum}}%
%BeginExpansion
s%
%EndExpansion
%TCIMACRO{\FORMULA{mn}{mn}{evaluatenum}}%
%BeginExpansion
mn%
%EndExpansion
%TCIMACRO{\FORMULA{vn}{nv}{evaluatenum}}%
%BeginExpansion
nv%
%EndExpansion
$

\subsubsection{Substatement}

Best\"{a}m f\"{o}rdelningens v\"{a}ntev\"{a}rde:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=m$)

\subsubsection{Substatement}

Best\"{a}m f\"{o}rdelningens standardavvikelse (tre decimaler):

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=s$)

\subsubsection{Substatement}

Best\"{a}m $E\left( \sum_{i=1}^{n}X_{i}\right) $ n\"{a}r alla $X_{i}$ har f%
\"{o}rdelningen ovan och $X_{1},\ldots ,X_{n}$ \"{a}r oberoende:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=mn$)

\subsubsection{Substatement}

Best\"{a}m $V\left( \sum_{i=1}^{n}X_{i}\right) $ n\"{a}r alla $X_{i}$ har f%
\"{o}rdelningen ovan och $X_{1},\ldots ,X_{n}$ \"{a}r oberoende:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=vn$)

\paragraph{Solution}

F\"{o}rst ber\"{a}knar vi f\"{o}rsta och andra momentet 
\begin{align*}
E\left( X\right) & =\sum_{i\in \Omega _{X}}kP\left( X=k\right) =0\times 
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluatenum}}%
%BeginExpansion
p_{1}%
%EndExpansion
+1\times 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluatenum}}%
%BeginExpansion
p_{2}%
%EndExpansion
+2\times 
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{evaluatenum}}%
%BeginExpansion
p_{3}%
%EndExpansion
+3\times 
%TCIMACRO{\FORMULA{p_{4}}{p_{4}}{evaluatenum}}%
%BeginExpansion
p_{4}%
%EndExpansion
=%
%TCIMACRO{%
%\FORMULA{\sum_{k=0}^{3}kp_{k+1}}{p_{2}+2.0p_{3}+3.0p_{4}}{evaluatenum} }%
%BeginExpansion
p_{2}+2.0p_{3}+3.0p_{4}
%EndExpansion
\\
E\left( X^{2}\right) & =\sum_{i\in \Omega _{X}}k^{2}P\left( X=k\right)
=0^{2}\times 
%TCIMACRO{\FORMULA{p_{1}}{p_{1}}{evaluatenum}}%
%BeginExpansion
p_{1}%
%EndExpansion
+1^{2}\times 
%TCIMACRO{\FORMULA{p_{2}}{p_{2}}{evaluatenum}}%
%BeginExpansion
p_{2}%
%EndExpansion
+2^{2}\times 
%TCIMACRO{\FORMULA{p_{3}}{p_{3}}{evaluatenum}}%
%BeginExpansion
p_{3}%
%EndExpansion
+3^{2}\times 
%TCIMACRO{\FORMULA{p_{4}}{p_{4}}{evaluatenum}}%
%BeginExpansion
p_{4}%
%EndExpansion
=%
%TCIMACRO{%
%\FORMULA{\sum_{k=0}^{3}k^{2}p_{k+1}}{p_{2}+4.0p_{3}+9.0p_{4}}{evaluatenum}}%
%BeginExpansion
p_{2}+4.0p_{3}+9.0p_{4}%
%EndExpansion
\end{align*}

varav 
\begin{align*}
V\left( X\right) & =E\left( X^{2}\right) -E^{2}\left( X\right) =%
%TCIMACRO{%
%\FORMULA{\sum_{k=0}^{3}k^{2}p_{k+1}}{p_{2}+4.0p_{3}+9.0p_{4}}{evaluatenum}}%
%BeginExpansion
p_{2}+4.0p_{3}+9.0p_{4}%
%EndExpansion
-%
%TCIMACRO{\FORMULA{m}{m}{evaluatenum}}%
%BeginExpansion
m%
%EndExpansion
^{2}=%
%TCIMACRO{\FORMULA{v}{v}{evaluatenum} }%
%BeginExpansion
v
%EndExpansion
\\
D\left( X\right) & =\sqrt{%
%TCIMACRO{\FORMULA{v}{v}{evaluatenum}}%
%BeginExpansion
v%
%EndExpansion
}=%
%TCIMACRO{\FORMULA{\sqrt{v_{1}}}{\sqrt{v_{1}}}{evaluatenum}}%
%BeginExpansion
\sqrt{v_{1}}%
%EndExpansion
\end{align*}

samt 
\begin{align*}
E\left( \sum_{i=1}^{n}X_{i}\right) & =\sum_{i=1}^{n}E\left( X_{i}\right) =%
%TCIMACRO{\FORMULA{m}{m}{evaluatenum}}%
%BeginExpansion
m%
%EndExpansion
n\text{,} \\
V\left( \sum_{i=1}^{n}X_{i}\right) & =\left\{ \text{ty oberoende}\right\}
=\sum_{i=1}^{n}V\left( X_{i}\right) =%
%TCIMACRO{\FORMULA{v}{v}{evaluatenum}}%
%BeginExpansion
v%
%EndExpansion
n\text{.}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 1,0.4,\limfunc{nplaces}\left( 0.35456,3\right)
\right) $

\subsubsection{Statement}

Den stokastiska variabeln $X$ har t\"{a}thetsfunktionen 
\begin{equation*}
f\left( x\right) =\left\{ 
\begin{array}{ll}
k\sqrt{x} & \text{om }0\leq x\leq 1\text{,} \\ 
0 & \text{annars.}%
\end{array}%
\right.
\end{equation*}

\subsubsection{Substatement}

Best\"{a}m konstanten $k$:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m v\"{a}ntev\"{a}rde f\"{o}r $X$:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\subsubsection{Substatement}

Best\"{a}m standardavvikelsen f\"{o}r $X$:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\paragraph{Solution}

F\"{o}r momenten erh\aa lls 
\begin{align*}
E\left( X\right) & =\int_{0}^{1}x\sqrt{x}\,dx=0.4 \\
E\left( X^{2}\right) & =\int_{0}^{1}x^{2}\sqrt{x}\,dx=0.28571
\end{align*}%
varvid 
\begin{equation*}
D\left( X\right) =\sqrt{0.28571-0.4^{2}}=0.35456
\end{equation*}

\section{Question}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 62.6,2.88\right) $

\subsubsection{Statement}

Dygnsmedeltemperaturen i juli i Stockholm varierar fr\aa n \aa r till \aa r
och kan anses vara en stokastisk variabel med v\"{a}ntev\"{a}rdet $17.0$
grader och standardavvikelsen $1.6$ grader. F\"{o}r att hj\"{a}lpa
amerikanska turister ville man r\"{a}kna om ovanst\aa ende v\"{a}rden i
Celsiusgrader till Fahrenheitgrader. Om 
\begin{align*}
X& =\text{temperaturen i Celsius,} \\
Y& =\text{temperaturen i Fahrenheit,}
\end{align*}%
s\aa\ g\"{a}ller sambandet 
\begin{equation*}
Y=\frac{9}{5}X+32\text{.}
\end{equation*}

\subsubsection{Substatement}

Best\"{a}m $E\left( Y\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m $D\left( Y\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Det g\"{a}ller att 
\begin{align*}
E\left( Y\right) & =E\left( \frac{9}{5}X+32\right) =\frac{9}{5}E\left(
X\right) +32=\frac{9}{5}17+32=62.6 \\
V\left( Y\right) & =V\left( \frac{9}{5}X+32\right) =\left( \frac{9}{5}%
\right) ^{2}V\left( X\right) =\left( \frac{9}{5}\right) ^{2}1.6^{2} \\
D\left( Y\right) & =\frac{9}{5}1.6=2.88
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 4,11.22\right) $

\subsubsection{Statement}

De stokastiska variablerna $X_{1},X_{2}$ och $X_{3}$ \"{a}r oberoende och
alla har v\"{a}ntev\"{a}rdet $2$ respektive standardavvikelsen $3$. S\"{a}tt 
\begin{equation*}
Y=3X_{1}-2X_{2}+X_{3}
\end{equation*}

\subsubsection{Substatement}

Best\"{a}m $E\left( Y\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m $D\left( Y\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Det g\"{a}ller att 
\begin{align*}
E\left( Y\right) & =E\left( 3X_{1}-2X_{2}+X_{3}\right) =3E\left(
X_{1}\right) -2E\left( X_{2}\right) +E\left( X_{3}\right) \\
& =3\times 2-2\times 2+2=4 \\
V\left( Y\right) & =V\left( 3X_{1}-2X_{2}+X_{3}\right) \\
\left\{ \text{ty oberoende}\right\} & =9V\left( X_{1}\right) +4V\left(
X_{2}\right) +V\left( X_{3}\right) =9\times 9+4\times 9+9 \\
& =126 \\
D\left( Y\right) & =\sqrt{126}=11.22
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}(\func{rand}(700,900)/1000,2)$

$\limfunc{svar}:=\left( 20,20,0.0025,0.00025\right) $

\subsubsection{Statement}

Tr\"{a}plankors l\"{a}ngder varierar slumpm\"{a}ssigt till f\"{o}ljd av fel
i uppm\"{a}tningen f\"{o}re tills\aa gning. Uppm\"{a}tningarna som sker med
tumstock kan ses som utfall av oberoende likaf\"{o}rdelade stokastiska
variabler med v\"{a}ntev\"{a}rde $2$ meter och standardavvikelse $0.005$
meter.

Vid ett tillf\"{a}lle beh\"{o}ver man $10$ plankor som skall l\"{a}ggas
efter varandra s\aa\ att den sammanlagda l\"{a}ngden blir $20$ meter. Man
har att v\"{a}lja mellan f\"{o}ljande tv\aa\ metoder:

\begin{enumerate}
\item M\"{a}t upp en planka, $X$, och s\aa ga d\"{a}refter $10$ lika l\aa %
nga vilket ger den sammanlagda l\"{a}ngden 
\begin{equation*}
Y_{1}=10X
\end{equation*}

\item M\"{a}t upp varje planka f\"{o}r sig vilket ger den sammanlagda l\"{a}%
ngden 
\begin{equation*}
Y_{2}=\sum_{k=1}^{10}X_{k}
\end{equation*}
\end{enumerate}

Vilken metod skall l\"{a}mpligen anv\"{a}ndas? Motivera ditt svar med att
best\"{a}mma

\subsubsection{Substatement}

$E\left( Y_{1}\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

$E\left( Y_{2}\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\subsubsection{Substatement}

$V\left( Y_{1}\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\subsubsection{Substatement'}

$V\left( Y_{2}\right) $:

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{4}$)

\paragraph{Solution}

Man finner 
\begin{align*}
E\left( Y_{1}\right) & =10E\left( X\right) =20 \\
E\left( Y_{2}\right) & =\sum_{k=1}^{10}E\left( X_{k}\right) =20 \\
V\left( Y_{1}\right) & =100V\left( X\right) =0.0025 \\
V\left( Y_{2}\right) & =\sum_{k=1}^{10}V\left( X_{k}\right) \quad \text{%
p\thinspace g\thinspace a oberoendet} \\
& =10V\left( X_{1}\right) =0.00025
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$p_{1}:=\limfunc{nplaces}(\func{rand}(700,900)/1000,2)$

$m:=\left( 3.5,3.5\right) $

$k:=\frac{49}{4}$

$s:=\left( 1.71,1.71\right) $

$c:=\left( 0,0\right) $

\subsubsection{Statement}

Best\"{a}m $E\left( X_{1}\right) $, $E\left( X_{2}\right) $, $E\left(
X_{1}X_{2}\right) $, $D\left( X_{1}\right) $ och $D\left( X_{2}\right) $
samt \"{a}ven $C\left( X_{1},X_{2}\right) $ och $\rho \left(
X_{1},X_{2}\right) $ vid kast med tv\aa\ t\"{a}rningar d\"{a}r%
\begin{equation*}
X_{i}=\text{antal prickar t\"{a}rning }i\text{.}
\end{equation*}

\subsubsection{Substatement}

Best\"{a}m $E\left( X_{1}\right) $ (svara med tv\aa\ decimaler).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=m_{1}$)

\subsubsection{Substatement}

Best\"{a}m $E\left( X_{1}X_{2}\right) $ (svara p\aa\ formen $\frac{a}{b}$).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=k$)

\subsubsection{Substatement}

Best\"{a}m $D\left( X_{2}\right) $ (svara med tv\aa\ decimaler).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=s_{2}$)

\subsubsection{Substatement}

Best\"{a}m $C\left( X_{1},X_{2}\right) $.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=c_{1}$)

\paragraph{Solution}

Det g\"{a}ller att 
\begin{align*}
E\left( X_{i}\right) & =1\times \frac{1}{6}+2\times \frac{1}{6}+\cdots
+6\times \frac{1}{6}=3.5\text{,} \\
V\left( X_{i}\right) & =2\frac{11}{12}\text{,} \\
D\left( X_{i}\right) & \approx 1.71\text{,} \\
E\left( X_{1}X_{2}\right) & =\sum_{i=1}^{6}\sum_{j=1}^{6}ij\frac{1}{36}=%
\frac{6\times 7}{2}\frac{6\times 7}{2}\frac{1}{36}=\left( \frac{7}{2}\right)
^{2}\text{,} \\
C\left( X_{1},X_{2}\right) & =\left( \frac{7}{2}\right) ^{2}-\frac{7}{2}%
\frac{7}{2}=0\text{,} \\
\rho \left( X_{1},X_{2}\right) & =0\text{.}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 1.8,3.8,1.25,1.1\right) $

\subsubsection{Statement}

Definiera de stokastiska variablerna 
\begin{align*}
X& =\text{antal felaktiga moduler i ett program} \\
Y& =\text{antal dagar som g\aa r \aa t f\"{o}r att avlusa programmet}
\end{align*}%
och antag att sannolikhetsfunktionen f\"{o}r $\left( X,Y\right) $ \"{a}r 
\begin{equation*}
\begin{tabular}{llllllll}
&  &  &  & $y$ &  &  &  \\ \cline{2-8}\cline{3-8}
& \multicolumn{1}{|l}{$p\left( x,y\right) $} & \multicolumn{1}{r}{$0$} & 
\multicolumn{1}{r}{$1$} & \multicolumn{1}{r}{$2$} & \multicolumn{1}{r}{$3$}
& \multicolumn{1}{r|}{$4$} & $5$ \\ \cline{2-2}\cline{2-8}\cline{3-8}
& \multicolumn{1}{|c}{$0$} & \multicolumn{1}{|r}{$0.2$} & \multicolumn{1}{r}{%
$0$} & \multicolumn{1}{r}{$0$} & \multicolumn{1}{r}{$0$} & 
\multicolumn{1}{r}{$0$} & $0$ \\ 
& \multicolumn{1}{|c}{$1$} & \multicolumn{1}{|r}{$0.08$} & 
\multicolumn{1}{r}{$0.06$} & \multicolumn{1}{r}{$0.04$} & \multicolumn{1}{r}{%
$0.02$} & \multicolumn{1}{r}{$0$} & $0$ \\ 
$x$ & \multicolumn{1}{|c}{$2$} & \multicolumn{1}{|r}{$0.03$} & 
\multicolumn{1}{r}{$0.09$} & \multicolumn{1}{r}{$0.09$} & \multicolumn{1}{r}{%
$0.06$} & \multicolumn{1}{r}{$0.03$} & $0$ \\ 
& \multicolumn{1}{|c}{$3$} & \multicolumn{1}{|r}{$0.02$} & 
\multicolumn{1}{r}{$0.04$} & \multicolumn{1}{r}{$0.06$} & \multicolumn{1}{r}{%
$0.04$} & \multicolumn{1}{r}{$0.02$} & $0.02$ \\ 
& \multicolumn{1}{|c}{$4$} & \multicolumn{1}{|r}{$0.01$} & 
\multicolumn{1}{r}{$0.01$} & \multicolumn{1}{r}{$0.02$} & \multicolumn{1}{r}{%
$0.03$} & \multicolumn{1}{r}{$0.02$} & $0.01$ \\ \cline{2-2}
\end{tabular}%
\end{equation*}

Best\"{a}m $E\left( X_{1}\right) $, $E\left( X_{2}\right) $, $E\left(
X_{1}X_{2}\right) $, $D\left( X_{1}\right) $ och $D\left( X_{2}\right) $
samt $C\left( X_{1},X_{2}\right) $ och $\rho \left( X_{1},X_{2}\right) $.

\subsubsection{Substatement}

Best\"{a}m $E\left( X_{1}\right) $.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m $E\left( X_{1}X_{2}\right) $.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\subsubsection{Substatement}

Best\"{a}m $D\left( X_{1}\right) $.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{3}$)

\subsubsection{Substatement}

Best\"{a}m $C\left( X_{1},X_{2}\right) $.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{4}$)

\paragraph{Solution}

F\"{o}r att hitta sannolikhetsfunktionen f\"{o}r \ftex variabeln $X$
konstaterar vi att enligt Kolmogorovs axiom g\"{a}ller 
\begin{equation*}
P\left( X=x\right) =\sum_{y=0}^{5}P\left( X=x,Y=y\right)
\end{equation*}%
vilket ger 
\begin{equation*}
\begin{tabular}{c|rrrrr}
$x$ & $0$ & $1$ & $2$ & $3$ & $4$ \\ \hline
$p\left( x\right) $ & $0.2$ & $0.2$ & $0.3$ & $0.2$ & $0.1$%
\end{tabular}%
\end{equation*}%
varur $E\left( X\right) =1.8$ och $D\left( X\right) =1.25$. P\aa\ samma s%
\"{a}tt erh\aa lls att $E\left( Y\right) =1.5$ och $D\left( Y\right) =1.42$.
Eftersom 
\begin{equation*}
E\left( XY\right) =\sum_{x=0}^{4}\sum_{y=0}^{5}xyP\left( X=x,Y=y\right) =3.8
\end{equation*}%
erh\aa lls $C\left( X,Y\right) =1.1$ och $\rho \left( X,Y\right) =0.62$.

\subsection{Variant}

\subsubsection{Setup}

Choices: Permute

\subsubsection{Statement}

Definiera en stokastisk variabel $X$ med egenskapen 
\begin{equation*}
E\left( X\right) =E\left( X^{3}\right) =0
\end{equation*}%
och s\"{a}tt $Y=X^{2}-1$. Best\"{a}m $C\left( X,Y\right) $. \"{A}r
variablerna $X$ och $Y$ oberoende?

\paragraph{Choices}

\begin{itemize}
\item $\neq 0$, ja

\item $\neq 0$, nej

\item $=0$, ja

\item $=0$, nej\correctchoice{}
\end{itemize}

\paragraph{Solution}

F\"{o}r kovariansen erh\aa lls 
\begin{align*}
C\left( X,Y\right) & =E\left[ \left( X-E\left( X\right) \right) \left(
Y-E\left( Y\right) \right) \right]  \\
& =E\left( X\left( X^{2}-1\right) \right)  \\
& =E\left( X^{3}-X\right)  \\
& =0
\end{align*}%
men sj\"{a}lvklart \"{a}r $X$ och $Y$ ej oberoende!

\section{Question}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 2.0067,0.1155\right) $

\subsubsection{Statement}

Den stokastiska variabeln $X$ \"{a}r $R\left( -0.05,0.05\right) $. Best\"{a}%
m $E\left( Y\right) $ och $D\left( Y\right) $ dels "exakt" enligt definition
och dels med Gauss approximationsformler n\"{a}r $Y=\frac{1}{0.5+X}$.

\subsubsection{Substatement}

Ange det "exakta" v\"{a}rdet p\aa\ $E\left( Y\right) $ (fyra decimaler).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Ange det approximativa v\"{a}rdet p\aa\ $D\left( Y\right) $ (fyra decimaler).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Det g\"{a}ller 
\begin{align*}
E\left( Y\right) & =\int_{-0.05}^{0.05}\frac{1}{0.5+x}\frac{1}{0.05-\left(
-0.05\right) }\,dx=10\left[ \ln \left( 0.5+x\right) \right]
_{-0.05}^{0.05}=2.006707 \\
E\left( Y^{2}\right) & =\int_{-0.05}^{0.05}\left( \frac{1}{0.5+x}\right) ^{2}%
\frac{1}{0.05-\left( -0.05\right) }\,dx=10\left[ -\frac{1}{0.5+x}\right]
_{-0.05}^{0.05}=4.040404 \\
D\left( Y\right) & =4.040404-2.006707^{2}=0.01353
\end{align*}%
Gauss approximationsformler ger%
\begin{equation*}
E\left( Y\right) \approx \frac{1}{0.5+E\left( X\right) }=2
\end{equation*}%
samt%
\begin{align*}
V\left( Y\right) & \approx \frac{\left( 0.05+0.05\right) ^{2}}{12}\left[ -%
\frac{1}{\left( 0.5+0\right) ^{2}}\right] ^{2}=\frac{0.01}{12}\times
16=0.013333 \\
D\left( Y\right) & \approx 0.11547
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( \frac{\pi d^{3}}{6},\frac{\pi ^{2}d^{4}\sigma ^{2}}{4%
}\right) $

\subsubsection{Statement}

Volymen $V$ av ett klot uppskattas med $\frac{\pi X^{3}}{6}$ d\"{a}r $X$ 
\"{a}r klotets diameter. F\"{o}r variabeln $X$ g\"{a}ller att 
\begin{equation*}
E\left( X\right) =d\text{ och }V\left( X\right) =\sigma ^{2}\text{.}
\end{equation*}

\subsubsection{Substatement}

Best\"{a}m v\"{a}ntev\"{a}rdet f\"{o}r klotets volym (svara p\aa\ formen $%
\frac{a}{b}$).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

Best\"{a}m variansen f\"{o}r klotets volym (svara p\aa\ formen $\frac{a}{b}$%
).

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

Det g\"{a}ller att 
\begin{equation*}
V=\frac{\pi X^{3}}{6}
\end{equation*}%
och med hj\"{a}lp av Gauss approximationsformler erh\aa lls 
\begin{align*}
E\left( V\right) & \approx \frac{\pi d^{3}}{6} \\
V\left( X\right) & \approx \left( \frac{3\pi }{6}E^{2}\left( X\right)
\right) ^{2}V\left( X\right) =\left( \frac{\pi d^{2}}{2}\right) ^{2}\sigma
^{2}
\end{align*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=\left( 0.5,0.99\right) $

\subsubsection{Statement}

Antalet mobiltelefoner av m\"{a}rket TD98 som Ericsson producerar per vecka
vid en av sina fabriker kan betraktas som en stokastiska variabel $X$ vars v%
\"{a}ntev\"{a}rde \"{a}r $500$ och varians \"{a}r $100$. Best\"{a}m en \"{o}%
vre gr\"{a}ns f\"{o}r sannolikheten att det en vecka produceras mer \"{a}n $%
1000$ telefoner samt en undre gr\"{a}ns f\"{o}r sannolikheten att
produktionen ligger mellan $400$ och $600$ telefoner per vecka.

\subsubsection{Substatement}

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{1}$)

\subsubsection{Substatement}

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}_{2}$)

\paragraph{Solution}

F\"{o}r att finna den f\"{o}rsta sannolikheten anv\"{a}nder vi samma teknik
som i Tjebysjovs olikhet%
\begin{equation*}
E\left( X\right) =\sum_{x=0}^{\infty }xP\left( X=x\right) \geq
\sum_{x=1001}^{\infty }xP\left( X=x\right) \geq \sum_{x=1001}^{\infty
}1000P\left( X=x\right) =1000P\left( X>1000\right) 
\end{equation*}%
Den f\"{o}rsta sannolikheten kan nu skrivas 
\begin{equation*}
P\left( X>1000\right) \leq \frac{E\left( X\right) }{1000}=\frac{500}{1000}%
=0.5
\end{equation*}%
Den andra sannolikheten erh\aa lls med Tjebysjovs olikhet till 
\begin{equation*}
P\left( \left\vert X-500\right\vert >100\right) \leq \frac{100}{100^{2}}=0.01
\end{equation*}%
varf\"{o}r 
\begin{equation*}
P\left( \left\vert X-500\right\vert \leq 100\right) \geq 0.99
\end{equation*}

\subsection{Variant}

\subsubsection{Setup}

$\limfunc{svar}:=10000$

\subsubsection{Statement}

P\aa\ en sp\aa nskiva av l\"{a}ngd $b$ cm g\"{o}r man $n$ m\"{a}tningar av
dess l\"{a}ngd och erh\aa ller d\"{a}rvid ett stickprov $X_{1},X_{2},\ldots
,X_{n}$. F\"{o}r detta stickprov g\"{a}ller den statistiska modellen%
\begin{equation*}
X_{i}=b+\epsilon _{i}\text{ d\"{a}r }E\left( \epsilon _{i}\right) =0\text{
och }V\left( \epsilon _{i}\right) =1\text{ cm.}
\end{equation*}%
Hur stort m\aa ste $n$ vara f\"{o}r att vi skall kunna skatta $b$ med en
noggrannhet om $0.1$ cm med en sannolikhet som \"{a}r $0.99$ eller st\"{o}%
rre.

\paragraph{Choices}

InputField(MATH)

GradeProc: givecredit($\limfunc{response}=\limfunc{svar}$)

\paragraph{Solution}

Vi vill att skillnaden mellan skattningen och det sanna v\"{a}rdet skall
vara mindre \"{a}n eller lika med $0.1$ cm dvs%
\begin{equation*}
P\left( \left\vert \bar{X}-b\right\vert \leq 0.1\right) \geq 0.99\text{.}
\end{equation*}%
Tjebysjov:s olikhet s\"{a}ger att%
\begin{equation*}
P\left( \left\vert \bar{X}-b\right\vert \geq \epsilon \right) \leq \frac{%
\sigma _{\bar{X}}^{2}}{\epsilon ^{2}}
\end{equation*}%
vilket i v\aa rt fall kan skrivas%
\begin{equation*}
P\left( \left\vert \bar{X}-b\right\vert <0.1\right) \geq 1-\frac{\frac{1}{n}%
}{0.1^{2}}\geq 0.99
\end{equation*}%
Ur denna olikhet l\"{o}ser vi $n$ till att vara st\"{o}rre \"{a}n $10000$.

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