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@ -81,118 +81,7 @@
\subsection{Theorie Wiederholung} \subsection{Theorie Wiederholung}
\begin{frame} \begin{frame}
\frametitle{Stetige Zufallsvariablen I} \frametitle{sasdf}
\vspace*{-10mm}
\begin{lightgrayhighlightbox}
Erinnerung: Diskrete Zufallsvariablen
\begin{align*}
\text{\normalfont Verteilung: }& P_X(x) = P(X = x) \\
\text{\normalfont Verteilungsfunktion: }& F_X(x) = P(X \le x) =
\sum_{n: x_n \le y} P_X(x)
\end{align*}
\vspace{-10mm}
\end{lightgrayhighlightbox}
\begin{columns}[t]
\pause\column{\kitthreecolumns}
\centering
\begin{itemize}
\item Verteilungsfunktion $F_X(x)$ einer stetiger ZV
\begin{gather*}
F_X(x) = P(X \le x)
\end{gather*}
\end{itemize}
\pause\column{\kitthreecolumns}
\centering
\begin{itemize}
\item Wahrscheinlichkeitsdichte $f_X(x)$ einer stetiger ZV
\begin{gather*}
F_X(x) = \int_{-\infty}^{x} f_X(u) du
\end{gather*}
\end{itemize}
\end{columns}
\begin{columns}[t]
\pause \column{\kitthreecolumns}
\centering
\begin{gather*}
\text{Eigenschaften:} \\[3mm]
\lim_{x\rightarrow -\infty} F_X(x) = 0 \\
\lim_{x\rightarrow +\infty} F_X(x) = 1 \\
x_1 \le x_2 \Rightarrow F_X(x_1) \le F_X(x_2)\\
F_X(x+) = \lim_{h\rightarrow 0^+} F_X (x+h) = F_X(x)
\end{gather*}
\pause \column{\kitthreecolumns}
\centering
\begin{gather*}
\text{Eigenschaften:} \\[3mm]
f_X(x) \ge 0 \\
\int_{-\infty}^{\infty} f_X(x) dx = 1
\end{gather*}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Stetige Zufallsvariablen II}
\begin{minipage}{0.6\textwidth}
\begin{itemize}
\item Wichtige Kenngrößen
\begin{align*}
\begin{array}{rlr}
\text{Erwartungswert: } \hspace{5mm} & E(X) =
\displaystyle\int_{-\infty}^{\infty} x f_X(x) dx
& \hspace{5mm} \big( = \mu \big) \\[3mm]
\text{Varianz: } \hspace{5mm} & V(X) = E\mleft(
\mleft( X - E(X) \mright)^2 \mright) \\[3mm]
\text{Standardabweichung: } \hspace{5mm} &
\sqrt{V(X)} & \hspace{5mm} \big( = \sigma \big)
\end{array}
\end{align*}
\end{itemize}
\end{minipage}
\begin{minipage}{0.38\textwidth}
\begin{lightgrayhighlightbox}
Erinnerung
\begin{align*}
\text{\normalfont Erwartungswert: }& E(X) =
\sum_{n=1}^{\infty} x_n P_X(x) \\
\text{\normalfont Varianz: }& V(X) = E\mleft( \mleft(
X - E(X) \mright)^2 \mright)
\end{align*}
\end{lightgrayhighlightbox}
\end{minipage}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}[c]
\column{\kitthreecolumns}
\centering
\begin{greenblock}{Verteilungsfunktion (kontinuierlich)}
\vspace*{-6mm}
\begin{gather*}
F_X(x) = P(X \le x)\\[4mm]
P(a < X \le b) = F_X(b) - F_X(a) \\[8mm]
\lim_{x\rightarrow -\infty} F_X(x) = 0 \\
\lim_{x\rightarrow +\infty} F_X(x) = 1 \\
x_1 \le x_2 \Rightarrow F_X(x_1) \le F_X(x_2)\\
F_X(x+) = \lim_{h\rightarrow 0^+} F_X (x+h) = F_X(x)
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\centering
\begin{greenblock}{Wahrscheinlichkeitsdichte \phantom{()}}
\vspace*{-6mm}
\begin{gather*}
F_X(x) = \int_{-\infty}^{x} f_X(u) du \\[5mm]
f_X(x) \ge 0 \\
\int_{-\infty}^{\infty} f_X(x) dx = 1
\end{gather*}
\end{greenblock}
\end{columns}
\end{frame} \end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@ -331,6 +220,7 @@
\begin{gather*} \begin{gather*}
x_1 \le x_2 \Rightarrow F_X(x_1) \le F_X(x_2) \\ x_1 \le x_2 \Rightarrow F_X(x_1) \le F_X(x_2) \\
F_X(x+) = \lim_{h\rightarrow 0^+} F_X (x+h) = F_X(x) F_X(x+) = \lim_{h\rightarrow 0^+} F_X (x+h) = F_X(x)
\hspace{5mm}\forall x\in \mathbb{R}
\end{gather*} \end{gather*}
\column{\kitonecolumn} \column{\kitonecolumn}
\end{columns} \end{columns}
@ -377,7 +267,7 @@
= P(1 < X \le 2) = F_X(2) - F_X(1) = e^{-a} - e^{-4a} = P(1 < X \le 2) = F_X(2) - F_X(1) = e^{-a} - e^{-4a}
\end{gather*} \end{gather*}
\end{enumerate} \end{enumerate}
% tex-fmt: on % tex-fmt: off
\end{frame} \end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@ -387,193 +277,7 @@
\subsection{Theorie Wiederholung} \subsection{Theorie Wiederholung}
\begin{frame} \begin{frame}
\frametitle{Die Normalverteilung} \frametitle{sasdf}
\begin{columns}
\column{\kitthreecolumns}
\centering
\begin{gather*}
X \sim \mathcal{N}\mleft( \mu, \sigma^2 \mright)
\end{gather*}%
\vspace{0mm}
\begin{align*}
f_X(x) &= \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(\frac{(x -
\mu)^2}{2 \sigma^2} \right) \\[2mm]
F_X(x) &=
\vcenter{\hbox{\scalebox{1.5}[2.6]{\vspace*{3mm}$\displaystyle\int$}}}_{\hspace{-0.5em}-\infty}^{\,x}
\frac{1}{\sqrt{2\pi
\sigma^2}} \exp\left(\frac{(u - \mu)^2}{2 \sigma^2} \right) du
\end{align*}
\column{\kitthreecolumns}
\centering
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
domain=-4:4,
xmin=-4,xmax=4,
width=15cm,
height=5cm,
samples=200,
xlabel={$x$},
ylabel={$f_X(x)$},
xtick={0},
xticklabels={\textcolor{KITblue}{$\mu$}},
ytick={0},
]
\addplot+[mark=none, line width=1pt]
{(1 / sqrt(2*pi)) * exp(-x*x)};
\addplot+ [KITblue, mark=none, line width=1pt]
coordinates {(-0.5, 0.15) (0.5, 0.15)};
\addplot+ [KITblue, mark=none, line width=1pt]
coordinates {(-0.5, 0.12) (-0.5, 0.18)};
\addplot+ [KITblue, mark=none, line width=1pt]
coordinates {(0.5, 0.12) (0.5, 0.18)};
\node[KITblue] at (axis cs: 0, 0.2) {$\sigma$};
% \addplot +[scol2, mark=none, line width=1pt]
% coordinates {(4.8, -1) (4.8, 2)};
% \addplot +[scol2, mark=none, line width=1pt]
% coordinates {(5.2, -1) (5.2, 2)};
% \node at (axis cs: 4.8, 3) {$S(1-\delta)$};
% \node at (axis cs: 5.2, 3) {$S(1+\delta)$};
\end{axis}
\end{tikzpicture}
\end{figure}
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
domain=-4:4,
xmin=-4,xmax=4,
width=15cm,
height=5cm,
samples=200,
xlabel={$x$},
ylabel={$F_X(x)$},
xtick=\empty,
ytick={0, 1},
]
\addplot+[mark=none, line width=1pt]
{1 / (1 + exp(-(1.526*x*(1 + 0.1034*x))))};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
% TODO: Are Z/z notation used in the lecture?
\begin{frame}
\frametitle{Rechnen mithilfe der Standardnormalverteilung}
\vspace*{-15mm}
\begin{itemize}
\item Die Standardnormalverteilung
\end{itemize}
\begin{minipage}{0.48\textwidth}
\centering
\begin{gather*}
Z \sim \mathcal{N} (0,1) \\[4mm]
\Phi(z) := F_Z(z) = P(Z \le z) \\
\Phi(-z) = 1 - \Phi(z)
\end{gather*}
\end{minipage}%
\begin{minipage}{0.48\textwidth}
\centering
\begin{tabular}{|c|c||c|c||c|c|}
\hline
$z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ \\
\hline
\hline
0{,}00 & 0{,}500000 & 0{,}10 & 0{,}539828 & 0{,}20 & 0{,}579260 \\
0{,}02 & 0{,}507978 & 0{,}12 & 0{,}547758 & 0{,}22 & 0{,}587064 \\
0{,}04 & 0{,}515953 & 0{,}14 & 0{,}555670 & 0{,}24 & 0{,}594835 \\
0{,}06 & 0{,}523922 & 0{,}16 & 0{,}563559 & 0{,}26 & 0{,}602568 \\
0{,}08 & 0{,}531881 & 0{,}18 & 0{,}571424 & 0{,}28 & 0{,}610261 \\
\hline
\end{tabular}\\
\end{minipage}
\pause
\begin{itemize}
\item Standardisierte ZV
\begin{gather*}
\begin{array}{cc}
E(X) &= 0 \\
V(X) &= 1
\end{array}
\hspace{45mm}
\text{Standardisierung: } \hspace{5mm}
\widetilde{X} = \frac{X - E(X)}{\sqrt{V(X)}}
= \frac{X - \mu}{\sigma}
\end{gather*}
\end{itemize}
\vspace*{5mm}
\pause
\begin{lightgrayhighlightbox}
Rechenbeispiel
\begin{gather*}
X \sim \mathcal{N}(\mu = 1, \sigma^2 = 0{,}5^2) \\[2mm]
P\left(X \le 1{,}12 \right)
= P\left(\frac{X - 1}{0{,}5} \le \frac{1{,}12 - 1}{0{,}5}\right)
= P\left(\frac{X - 1}{0{,}5} \le
0{,}24\right) = \Phi\left(0{,}24\right) = 0{,}594835
\end{gather*}
\end{lightgrayhighlightbox}
\end{frame}
% TODO: Are Z/z notation used in the lecture?
\begin{frame}
\frametitle{Zusammenfassung}
\vspace*{-15mm}
\begin{columns}[t]
\column{\kitthreecolumns}
\centering
\begin{greenblock}{Standardnormalverteilung}
\vspace*{-10mm}
\begin{gather*}
Z \sim \mathcal{N} (0,1) \\[4mm]
\Phi(z) := F_Z(z) = P(Z \le z) \\
\Phi(-z) = 1 - \Phi(z)
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\centering
\begin{greenblock}{Standardisierung}
\vspace*{-10mm}
\begin{gather*}
\widetilde{X} = \frac{X - E(X)}{\sqrt{V(X)}}
= \frac{X - \mu}{\sigma}
\end{gather*}
\end{greenblock}
\end{columns}
\vspace{5mm}
\begin{table}
\centering
% \cdots
\begin{tabular}{|c|c||c|c||c|c||c|c|}
\hline
$z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ \\
\hline
\hline
1{,}40 & 0{,}919243 & 2{,}80 & 0{,}997445 & 3{,}00 & 0{,}998650 & 4{,}20 & 0{,}999987 \\
1{,}42 & 0{,}922196 & 2{,}82 & 0{,}997599 & 3{,}02 & 0{,}998736 & 4{,}22 & 0{,}999988 \\
1{,}44 & 0{,}925066 & 2{,}84 & 0{,}997744 & 3{,}04 & 0{,}998817 & 4{,}24 & 0{,}999989 \\
1{,}46 & 0{,}927855 & 2{,}86 & 0{,}997882 & 3{,}06 & 0{,}998893 & 4{,}26 & 0{,}999990 \\
1{,}48 & 0{,}930563 & 2{,}88 & 0{,}998012 & 3{,}08 & 0{,}998965 & 4{,}28 & 0{,}999991 \\
\hline
\end{tabular}
% \cdots
\end{table}
\end{frame} \end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@ -625,7 +329,7 @@
\column{\kitthreecolumns} \column{\kitthreecolumns}
\centering \centering
\pause \begin{gather*} \pause \begin{gather*}
X \sim \mathcal{N} \mleft( \mu = 0{,}5, \sigma^2 = 0{,}07^2 \mright) X \sim \mathcal{N} \mleft( \mu = 0{,}5, \sigma = 0{,}07^2 \mright)
\end{gather*} \end{gather*}
\begin{align*} \begin{align*}
P(E_\text{a}) &= P \Big( \big( X < S(1-\delta) \big) P(E_\text{a}) &= P \Big( \big( X < S(1-\delta) \big)
@ -633,8 +337,8 @@
&= P(X < S(1 - \delta)) + P(X > S(1 + \delta)) \\[2mm] &= P(X < S(1 - \delta)) + P(X > S(1 + \delta)) \\[2mm]
&= P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right) &= P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+ P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm] + P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
&\approx \Phi(-2{,}86) + \left(1 - \Phi(2{,}86)\right) \\ &\approx \Phi(-2.86) + \left(1 - \Phi(2.86)\right) \\
&= 2 - 2\Phi(2{,}86) \approx 0{,}424\text{\%} &= 2 - 2\Phi(2.86) \approx 0{,}424\text{\%}
\end{align*} \end{align*}
\column{\kitthreecolumns} \column{\kitthreecolumns}
\centering \centering
@ -655,8 +359,8 @@
\addplot+[mark=none, line width=1pt] \addplot+[mark=none, line width=1pt]
{1 / sqrt(2*3.1415*0.07*0.07) * exp(-(x - 5)*(x-5)/(2*0.07*0.07))}; {1 / sqrt(2*3.1415*0.07*0.07) * exp(-(x - 5)*(x-5)/(2*0.07*0.07))};
\addplot +[KITblue, mark=none, line width=1pt] coordinates {(4.8, -1) (4.8, 2)}; \addplot +[scol2, mark=none, line width=1pt] coordinates {(4.8, -1) (4.8, 2)};
\addplot +[KITblue, mark=none, line width=1pt] coordinates {(5.2, -1) (5.2, 2)}; \addplot +[scol2, mark=none, line width=1pt] coordinates {(5.2, -1) (5.2, 2)};
\node at (axis cs: 4.8, 3) {$S(1-\delta)$}; \node at (axis cs: 4.8, 3) {$S(1-\delta)$};
\node at (axis cs: 5.2, 3) {$S(1+\delta)$}; \node at (axis cs: 5.2, 3) {$S(1+\delta)$};
@ -680,7 +384,7 @@
dass nur noch halb so viele Ladegeräte wie in a) aussortiert dass nur noch halb so viele Ladegeräte wie in a) aussortiert
werden. Auf welchen Wert müsste er dazu $\sigma$ senken? werden. Auf welchen Wert müsste er dazu $\sigma$ senken?
\pause\begin{gather*} \pause\begin{gather*}
P(E_\text{b}) = \frac{1}{2} P(E_\text{a}) \approx 0{,}212\text{\%} \\ P(E_\text{b}) = \frac{1}{2} P(E_\text{a}) \approx 0.212\text{\%} \\
\end{gather*} \end{gather*}
\vspace*{-18mm} \vspace*{-18mm}
\begin{columns} \begin{columns}
@ -698,10 +402,10 @@
\pause\column{\kitthreecolumns} \pause\column{\kitthreecolumns}
\centering \centering
\begin{gather*} \begin{gather*}
2 - 2\Phi\left(\frac{0{,}2}{\sigma'}\right) = 2{,}12\cdot 10^{-3} \\[2mm] 2 - 2\Phi\left(\frac{0.2}{\sigma'}\right) = 2{,}12\cdot 10^{-3} \\
\Rightarrow \Phi\left(\frac{0{,}2}{\sigma'}\right) \approx 0{,}9989 \\[2mm] \Rightarrow \Phi\left(\frac{0.2}{\sigma'}\right) \approx 0.9989 \\
\Rightarrow \sigma' \approx \frac{0{,}2}{\Phi^{-1}(0{,}9989)} \Rightarrow \sigma' \approx \frac{0{,}2}{\Phi^{-1}(0{,}9989)}
\approx \frac{0{,}2}{3{,}08} \approx 0{,}65 \approx \frac{0{,}2}{3{,}08} \approx 0.65
\end{gather*} \end{gather*}
\end{columns} \end{columns}
\pause \vspace*{-5mm}\item Durch einen Produktionsfehler verschiebt sich der \pause \vspace*{-5mm}\item Durch einen Produktionsfehler verschiebt sich der
@ -711,7 +415,7 @@
P(E_\text{c}) &\overset{\text{a)}}{=} P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right) P(E_\text{c}) &\overset{\text{a)}}{=} P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+ P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm] + P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
&\approx \Phi(-4{,}29) + (1 - \Phi(1{,}43)) \\ &\approx \Phi(-4{,}29) + (1 - \Phi(1{,}43)) \\
& = 2 - \Phi(4{,}29) - \Phi(1{,}43) \approx 7{,}78 \text{\%} & = 2 - \Phi(4{,}29) - \Phi(1{,}43) \approx 7.78 \text{\%}
\end{align*} \end{align*}
\end{enumerate} \end{enumerate}
% tex-fmt: on % tex-fmt: on