Add solution for exercise 1

.latexmkrc

@@ -1,4 +1,3 @@
 $pdflatex="pdflatex -shell-escape -interaction=nonstopmode -synctex=1 %O %S";
 $out_dir = 'build';
 $pdf_mode = 1;
-

Makefile

@@ -1,19 +1,25 @@
 PRESENTATIONS := $(patsubst src/%/presentation.tex,build/presentation_%.pdf,$(wildcard src/*/presentation.tex))
 HANDOUTS := $(patsubst build/presentation_%.pdf,build/presentation_%_handout.pdf,$(PRESENTATIONS))
 
+RC_PDFLATEX := $(shell grep '$$pdflatex' .latexmkrc \
+	       | sed -e 's/.*"\(.*\)".*/\1/' -e 's/%S//' -e 's/%O//')
+
 .PHONY: all
 all: $(PRESENTATIONS) $(HANDOUTS)
 
 build/presentation_%.pdf: src/%/presentation.tex build/prepared
-	TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk $<
-	mv build/presentation.pdf $@
+	TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
+		latexmk -outdir=build/$* $<
+	cp build/$*/presentation.pdf $@
 
 build/presentation_%_handout.pdf: src/%/presentation.tex build/prepared
-	TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk -pdflatex='pdflatex %O "\def\ishandout{1}\input{%S}"' $<
-	mv build/presentation.pdf $@
+	TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
+		latexmk -outdir=build/$*_handout \
+		-pdflatex='$(RC_PDFLATEX) %O "\def\ishandout{1}\input{%S}"' $<
+	cp build/$*_handout/presentation.pdf $@
 
 build/prepared:
-	mkdir -p build
+	mkdir build
 	touch build/prepared
 
 .PHONY: clean

src/2025-12-19/presentation.tex

@@ -81,7 +81,118 @@
 \subsection{Theorie Wiederholung}
 
 \begin{frame}
-    \frametitle{sasdf}
+    \frametitle{Stetige Zufallsvariablen I}
+
+    \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 stetigen 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 stetigen 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: Diskrete Zufallsvariablen
+            \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}[t]
+        \column{\kitthreecolumns}
+        \centering
+        \begin{greenblock}{Verteilungsfunktion (stetige ZV)}
+            \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}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -159,9 +270,9 @@
                 \end{align*}
                 \pause\begin{gather*}
                     \int_{-\infty}^{\infty} f_X(x) dx
-                    = \int_{-\infty}^{\infty} C\cdot x e^{-ax^2} dx
-                    = \frac{C}{-2a} \int_{-\infty}^{\infty} (-2ax) e^{-ax^2} dx  \\
-                    = \frac{C}{-2a} \int_{-\infty}^{\infty} (e^{-ax^2})' dx
+                    = \int_{0}^{\infty} C\cdot x e^{-ax^2} dx
+                    = \frac{C}{-2a} \int_{0}^{\infty} (-2ax) e^{-ax^2} dx  \\
+                    = \frac{C}{-2a} \int_{0}^{\infty} (e^{-ax^2})' dx
                     = \frac{C}{-2a} \mleft[ e^{-ax^2} \mright]_0^{\infty} \overset{!}{=} 1 \hspace{10mm} \Rightarrow C = 2a
                 \end{gather*}
                 \centering
@@ -220,7 +331,6 @@
                 \begin{gather*}
                     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)
-                    \hspace{5mm}\forall x\in \mathbb{R}
                 \end{gather*}
                 \column{\kitonecolumn}
             \end{columns}
@@ -267,7 +377,7 @@
                 = P(1 < X \le 2) = F_X(2) - F_X(1) = e^{-a} - e^{-4a}
             \end{gather*}
     \end{enumerate}
-    % tex-fmt: off
+    % tex-fmt: on
 \end{frame}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -277,7 +387,203 @@
 \subsection{Theorie Wiederholung}
 
 \begin{frame}
-    \frametitle{sasdf}
+    \frametitle{Die Normalverteilung}
+
+    \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}
+
+\begin{frame}
+    \frametitle{Rechnen mithilfe der Standardnormalverteilung}
+
+    \vspace*{-15mm}
+
+    \begin{itemize}
+        \item Die Standardnormalverteilung
+    \end{itemize}
+
+    \begin{minipage}{0.48\textwidth}
+        \centering
+        \begin{gather*}
+            X \sim \mathcal{N} (0,1) \\[4mm]
+            \Phi(x) := F_X(x) = P(X \le x) \\
+            \Phi(-x) = 1 - \Phi(x)
+        \end{gather*}
+    \end{minipage}%
+    \begin{minipage}{0.48\textwidth}
+        \centering
+        \begin{tabular}{|c|c||c|c||c|c|}
+            \hline
+            $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ \\
+            \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*{3mm}
+
+    \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\big(\underbrace{\widetilde{X}}_{\sim
+            \mathcal{N}(0,1)} \le 0{,}24\big)
+            = \Phi\left(0{,}24\right) = 0{,}594835
+        \end{gather*}
+    \end{lightgrayhighlightbox}
+\end{frame}
+
+\begin{frame}
+    \frametitle{Zusammenfassung}
+
+    \vspace*{-15mm}
+
+    \begin{columns}[t]
+        \column{\kitthreecolumns}
+        \centering
+        \begin{greenblock}{Standardnormalverteilung}
+            \vspace*{-10mm}
+            \begin{gather*}
+                X \sim \mathcal{N} (0,1) \\[4mm]
+                \Phi(x) := F_X(x) = P(X \le x) \\
+                \Phi(-x) = 1 - \Phi(x)
+            \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
+            $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$
+            & $\Phi(x)$ \\
+            \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}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -329,16 +635,16 @@
             \column{\kitthreecolumns}
             \centering
             \pause \begin{gather*}
-                X \sim \mathcal{N} \mleft( \mu = 0{,}5, \sigma = 0{,}07^2 \mright)
+                X \sim \mathcal{N} \mleft( \mu = 0{,}5, \sigma^2 = 0{,}07^2 \mright)
             \end{gather*}
             \begin{align*}
                 P(E_\text{a}) &= P \Big( \big( X < S(1-\delta) \big)
                         \cup \big( X > S(1 + \delta) \big) \Big) \\
                      &= 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) \\[2mm]
-                     &\approx \Phi(-2.86) + \left(1 - \Phi(2.86)\right) \\
-                     &= 2 - 2\Phi(2.86) \approx 0{,}424\text{\%}
+                     &\overset{\widetilde{X} := \frac{X - \mu}{\sigma} }{=\joinrel=\joinrel=\joinrel=} P\left(\widetilde{X} < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+                        + P\left(\widetilde{X} > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
+                     &\approx \Phi(-2{,}86) + \left(1 - \Phi(2{,}86)\right) \\
+                     &= 2 - 2\Phi(2{,}86) \approx 0{,}424\text{\%}
             \end{align*}
             \column{\kitthreecolumns}
             \centering
@@ -359,8 +665,8 @@
                         \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))};
 
-                        \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)};
+                        \addplot +[KITblue, 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)};
                         \node at (axis cs: 4.8, 3) {$S(1-\delta)$};
                         \node at (axis cs: 5.2, 3) {$S(1+\delta)$};
 
@@ -384,17 +690,17 @@
             dass nur noch halb so viele Ladegeräte wie in a) aussortiert
             werden. Auf welchen Wert müsste er dazu $\sigma$ senken?
             \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*}
             \vspace*{-18mm}
             \begin{columns}
                 \pause\column{\kitthreecolumns}
                 \centering
                 \begin{align*}
-                    P(E_\text{b}) &\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{0{,}2}{\sigma'}\right)
-                        + P\left(Z > \frac{0{,}2}{\sigma'}\right) \\[2mm]
+                    P(E_\text{b}) &\overset{\text{a)}}{=} P\left(\widetilde{X} < \frac{S(1 - \delta) - \mu}{\sigma'}\right)
+                        + P\left(\widetilde{X} > \frac{S(1 + \delta) - \mu}{\sigma'}\right) \\[2mm]
+                    &= P\left(\widetilde{X} < -\frac{0{,}2}{\sigma'}\right)
+                        + P\left(\widetilde{X} > \frac{0{,}2}{\sigma'}\right) \\[2mm]
                     &= \Phi\left(-\frac{0{,}2}{\sigma'}\right)
                         + \left(1 - \Phi\left(\frac{0{,}2}{\sigma'} \right)\right) \\[2mm]
                     &= 2 - 2 \Phi\left(\frac{0{,}2}{\sigma'} \right)
@@ -402,20 +708,20 @@
                 \pause\column{\kitthreecolumns}
                 \centering
                 \begin{gather*}
-                    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 \\
+                    2 - 2\Phi\left(\frac{0{,}2}{\sigma'}\right) = 2{,}12\cdot 10^{-3} \\[2mm]
+                    \Rightarrow \Phi\left(\frac{0{,}2}{\sigma'}\right) \approx 0{,}9989 \\[2mm]
                     \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{,}065
                 \end{gather*}
             \end{columns}
         \pause \vspace*{-5mm}\item Durch einen Produktionsfehler verschiebt sich der
             Mittelwert $\mu$ auf $5{,}1$ Volt ($\sigma$ ist $0{,}07$ Volt).
             Wie groß ist jetzt der Prozentsatz, der aussortiert wird?
             \pause \begin{align*}
-                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(E_\text{c}) &\overset{\text{a)}}{=} P\left(\widetilde{X} < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+                    + P\left(\widetilde{X} > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
                 &\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{enumerate}
     % tex-fmt: on

src/2026-01-30/presentation.tex

@@ -0,0 +1,272 @@
+\ifdefined\ishandout
+\documentclass[de, handout]{CELbeamer}
+\else
+\documentclass[de]{CELbeamer}
+\fi
+
+%
+%
+% CEL Template
+%
+%
+
+\newcommand{\templates}{preambles}
+\input{\templates/packages.tex}
+\input{\templates/macros.tex}
+
+\grouplogo{CEL_logo.pdf}
+
+\groupname{Communication Engineering Lab (CEL)}
+\groupnamewidth{80mm}
+
+\fundinglogos{}
+
+%
+%
+% Document setup
+%
+%
+
+\usepackage{tikz}
+\usepackage{tikz-3dplot}
+\usetikzlibrary{spy, external, intersections, positioning}
+
+% \ifdefined\ishandout\else
+% \tikzexternalize
+% \fi
+
+\usepackage{pgfplots}
+\pgfplotsset{compat=newest}
+\usepgfplotslibrary{fillbetween}
+\usepgfplotslibrary{groupplots}
+
+\usepackage{enumerate}
+\usepackage{listings}
+\usepackage{subcaption}
+\usepackage{bbm}
+\usepackage{multirow}
+\usepackage{xcolor}
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{calc}
+\usepackage{amssymb}
+
+\title{WT Tutorium 6}
+\author[Tsouchlos]{Andreas Tsouchlos}
+\date[]{30. Januar 2026}
+
+%
+%
+% Custom commands
+%
+%
+
+\input{lib/latex-common/common.tex}
+\pgfplotsset{colorscheme/rocket}
+
+\newcommand{\res}{src/2026-01-16/res}
+
+\newlength{\depthofsumsign}
+\setlength{\depthofsumsign}{\depthof{$\sum$}}
+\newlength{\totalheightofsumsign}
+\newcommand{\nsum}[1][1.4]{
+    \mathop{
+        \raisebox
+        {-#1\depthofsumsign+1\depthofsumsign}
+        {\scalebox
+            {#1}
+            {$\displaystyle\sum$}%
+        }
+    }
+}
+
+% \tikzstyle{every node}=[font=\small]
+% \captionsetup[sub]{font=small}
+
+\newlength{\hght}
+\newlength{\wdth}
+
+\newcommand{\canceltotikz}[3][.5ex]{
+    \setlength{\hght}{\heightof{$#3$}}
+    \setlength{\wdth}{\widthof{$#3$}}
+    \makebox[0pt][l]{
+        \tikz[baseline]{\draw[-latex](0,-#1)--(\wdth,\hght+#1)
+            node[shift={(1mm,.5mm)}]{#2};
+        }
+    }#3
+}
+
+%
+%
+% Document body
+%
+%
+
+\begin{document}
+
+\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
+    \titlepage
+\end{frame}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section{Aufgabe 1}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Theorie Wiederholung}
+
+% TODO: Write
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Aufgabe}
+
+\begin{frame}
+    \frametitle{Aufgabe 1: Korrelationskoeffizienten}
+
+    Es ist die Zufallsvariable $X \sim \mathcal{N}(0,1)$ gegeben. Berechnen Sie
+    jeweils den Korrelationskoeffizienten $\rho_{XY}$ für
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \item $Y = aX + b \hspace{8mm}\text{mit } a, b \in R
+            \text{ und } a \neq 0$.
+        \item $Y = X^2$.
+    \end{enumerate}
+    % tex-fmt: on
+\end{frame}
+
+\begin{frame}
+    \frametitle{Aufgabe 1: Korrelationskoeffizienten}
+
+    Es ist die Zufallsvariable $X \sim \mathcal{N}(0,1)$ gegeben. Berechnen Sie
+    jeweils den Korrelationskoeffizienten $\rho_{XY}$ für
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \item $Y = aX + b \hspace{8mm}\text{mit } a, b \in R
+            \text{ und } a \neq 0$.
+            \pause \begin{gather*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+            \end{gather*}
+            \pause\begin{align*}
+                \text{cov}(X,Y) &= E(XY) - \canceltotikz[1ex]{0}{E(X)} E(Y)
+                    = E(XY) \\
+                    &= E(aX^2 + bX) = a\underbrace{E(X^2)}_{= V(X) = 1}
+                    + b\canceltotikz[1ex]{0}{E(X)} = a
+            \end{align*}
+            \pause\begin{gather*}
+                V(Y) = E\big( (Y - E(Y))^2 \big) = E\big( (aX)^2 \big)
+                    = a^2 \underbrace{E(X^2)}_{= V(X) = 1} = a^2
+            \end{gather*}
+            \pause\begin{align*}
+                \rho_{XY} = \frac{a}{\sqrt{a^2}} = \frac{a}{\lvert a \rvert}
+                = \left\{ \begin{array}{c}
+                    +1, \hspace{5mm} a > 0 \\
+                    -1, \hspace{5mm} a < 0
+                \end{array}
+                \right.
+            \end{align*}
+    \end{enumerate}
+    % tex-fmt: on
+\end{frame}
+
+\begin{frame}
+    \frametitle{Aufgabe 1: Korrelationskoeffizienten}
+
+    Es ist die Zufallsvariable $X \sim \mathcal{N}(0,1)$ gegeben. Berechnen Sie
+    jeweils den Korrelationskoeffizienten $\rho_{XY}$ für
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \setcounter{enumi}{1}
+        \item $Y = X^2$.
+            \pause \begin{gather*}
+                \rho_{XY} = \frac{\text{cov}(X,Y)}{\sqrt{V(X)V(Y)}}
+            \end{gather*}
+            \pause\begin{columns}
+                \column{\kitfourcolumns}
+                \centering
+                \begin{gather*}
+                    \text{cov}(X,Y) = E(XY) - \canceltotikz[1ex]{0}{E(X)} E(Y)
+                        = E(XY) = E(X^3)
+                \end{gather*}
+                \vspace*{-12mm}
+                \pause\begin{gather*}
+                    \hspace*{-18mm} = \int_{-\infty}^{\infty}
+                        \underbrace{x^3}_\text{ungerade}
+                        \cdot\underbrace{f_X(x)}_\text{gerade} dx = 0 \\[7mm]
+                    \rho_{XY} = 0
+                \end{gather*}
+                \column{\kittwocolumns}
+                \centering
+                \begin{figure}[H]
+                    \centering
+                    \begin{tikzpicture}
+                        \begin{axis}[
+                            domain=-3:3,
+                            width=10cm,
+                            height=6.5cm,
+                            samples=100,
+                            xtick={0},
+                            ytick={0},
+                            legend pos = south east,
+                            legend cell align = left,
+                        ]
+                            \addplot+[scol1, mark=none, line width=1pt]
+                                {1 / sqrt(2*pi) * exp(-x^2)};
+                            \addlegendentry{$f_X(x)$}
+                            \addplot+[scol2, mark=none, line width=1pt]
+                                {0.01 * x^3};
+                            \addlegendentry{$x^3$}
+                        \end{axis}
+
+                        \node at (8.7, 4.7) {\footnotemark};
+                    \end{tikzpicture}
+                \end{figure}
+            \end{columns}
+    \end{enumerate}
+    % tex-fmt: on
+
+    \footnotetext{Die zwei Kurven sind bezüglich der $y$-Achse
+    unterschiedlich skaliert.}
+\end{frame}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\section{Aufgabe 2}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Theorie Wiederholung}
+
+% TODO: Write
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{Aufgabe}
+
+\begin{frame}
+    \frametitle{Aufgabe 2: Abschätzungen von Verteilungen (ZGWS)}
+
+    Im Werk einer Zahnradfabrik werden verschiedene
+    Präzisionsmetallteile gefertigt. Während einer
+    Schicht werden 5000 Stück eines Typs A hergestellt. Bei der
+    Qualitätskontrolle werden 3% dieser
+    Teile als defekt klassifiziert und aussortiert.
+
+    % tex-fmt: off
+    \begin{enumerate}[a{)}]
+        \item Berechnen Sie näherungsweise die Wahrscheinlichkeit dafür, dass
+            während einer Schicht zwischen $125$ und $180$ Teile aussortiert
+            werden.
+        \item Die aussortierten Teile werden nach Schichtende zur
+            Wiederverwertung in einem Kessel auf einmal eingeschmolzen. Wie
+            viele Teile muss der Kessel fassen, damit er mit einer
+            Wahrscheinlichkeit von min. $0{,}98$ nicht überfüllt ist?
+        \item Der Kessel fasse maximal $200$ Teile. Es sollen nun mehr als
+            $5000$ Teile pro Schicht hergestellt werden. Wie viele Teile
+            können maximal gefertigt werden, damit der Kessel mit einer
+            Wahrscheinlichkeit von $0,98$ nicht überfüllt ist?
+    \end{enumerate}
+    % tex-fmt: on
+\end{frame}
+% TODO: Write
+
+\end{document}
+