7 Commits

34 changed files with 342 additions and 7532 deletions

4
.gitignore vendored
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@@ -1,5 +1 @@
build/ build/
src/*/.latexmkrc
src/*/lib
src/*/src

5
.gitmodules vendored
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@@ -1,3 +1,6 @@
[submodule "lib/latex-common"]
path = lib/latex-common
url = ssh://git@git.mercurial-manifold.eu:2224/an.tsouchlos/latex-common.git
[submodule "lib/cel-slides-template-2025"] [submodule "lib/cel-slides-template-2025"]
path = lib/cel-slides-template-2025 path = lib/cel-slides-template-2025
url = git@gitlab.kit.edu:kit/cel/misc/cel-slides-template-2025.git url = ssh://git@git.mercurial-manifold.eu:2224/an.tsouchlos/cel-slides-template-2025.git

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

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@@ -6,7 +6,6 @@ RUN apt update -y && apt upgrade -y
RUN apt install make texlive latexmk texlive-pictures -y RUN apt install make texlive latexmk texlive-pictures -y
RUN apt install texlive-publishers texlive-science texlive-fonts-extra texlive-latex-extra -y RUN apt install texlive-publishers texlive-science texlive-fonts-extra texlive-latex-extra -y
RUN apt install biber texlive-bibtex-extra -y RUN apt install biber texlive-bibtex-extra -y
RUN apt install texlive-lang-german -y
RUN apt install python3 python3-pygments -y RUN apt install python3 python3-pygments -y

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@@ -1,27 +1,11 @@
PRESENTATIONS := $(patsubst src/%/presentation.tex,build/presentation_%.pdf,$(wildcard src/*/presentation.tex)) all:
HANDOUTS := $(patsubst build/presentation_%.pdf,build/presentation_%_handout.pdf,$(PRESENTATIONS)) mkdir -p build/build
RC_PDFLATEX := $(shell grep '$$pdflatex' .latexmkrc \ TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk src/template/presentation.tex
| sed -e 's/.*"\(.*\)".*/\1/' -e 's/%S//' -e 's/%O//') mv build/presentation.pdf build/presentation_template.pdf
.PHONY: all TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk src/2025-11-07/presentation.tex
all: $(PRESENTATIONS) $(HANDOUTS) mv build/presentation.pdf build/presentation_2025-11-07.pdf
build/presentation_%.pdf: src/%/presentation.tex build/prepared
TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
latexmk -outdir=build/$* $<
cp build/$*/presentation.pdf $@
build/presentation_%_handout.pdf: src/%/presentation.tex build/prepared
TEXINPUTS=./lib/cel-slides-template-2025:$(dir $<):$$TEXINPUTS \
latexmk -outdir=build/$*_handout \
-pdflatex='$(RC_PDFLATEX) %O "\def\ishandout{1}\input{%S}"' $<
cp build/$*_handout/presentation.pdf $@
build/prepared:
mkdir build
touch build/prepared
.PHONY: clean
clean: clean:
rm -rf build rm -rf build

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@@ -1,18 +0,0 @@
# WT Tutorial Presentations
This repository contains the latex source files for the WT Tutorial slides.
## Build
### Local Environment
```bash
$ make
```
### With Docker
```bash
$ docker build . -t wt-tut
$ docker run --rm -u `id -u`:`id -g` -w $PWD -v $PWD:$PWD wt-tut make
```

1
lib/latex-common Submodule

Submodule lib/latex-common added at bded242752

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@@ -1 +0,0 @@
build/

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@@ -1,11 +0,0 @@
SRCS=$(shell find examples -type f -name "*.tex")
PDFS=$(shell echo $(SRCS:.tex=.pdf) | sed -e "s/examples\\///g")
all: $(PDFS)
%.pdf: examples/%.tex
latexmk $<
clean:
rm -rf build

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@@ -1,33 +0,0 @@
# latex-common
Repository containing common latex code that can be shared across multiple projects.
## Usage
Put
```latex
\input{/path/to/common.tex}
```
in your preamble. See the `examples` folder for usage examples.
## Build examples
### Build manually
```bash
$ make
```
### Build using docker
1. Build docker image
```bash
$ docker build -f dockerfiles/Dockerfile.alpine . -t latex-common
```
2. Build examples
```bash
$ docker run --rm -v $PWD:$PWD -w $PWD -u `id -u`:`id -g` latex-common make
```

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@@ -1,317 +0,0 @@
% Author: Andreas Tsouchlos
%
% Collection of useful commands and definitions
%
% ||====================================================================||
% || WARNING ||
% ||====================================================================||
% || The following packages have to be included before using this file: ||
% || amsmath ||
% || pgfplots ||
% ||====================================================================||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Math Symbols %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DeclareMathOperator*{\argmin}{\arg\!\min}
\DeclareMathOperator*{\argmax}{\arg\!\max}
\DeclareMathOperator\sign{sign}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Data Manipulation %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Filters for Pgfplots
% Source: https://tex.stackexchange.com/a/58563 (modified)
%
\pgfplotsset{
discard if/.style 2 args={
x filter/.append code={
\edef\tempa{\thisrow{#1}}
\edef\tempb{#2}
\ifx\tempa\tempb
\def\pgfmathresult{inf}
\fi
}
},
discard if not/.style 2 args={
x filter/.append code={
\edef\tempa{\thisrow{#1}}
\edef\tempb{#2}
\ifx\tempa\tempb
\else
\def\pgfmathresult{inf}
\fi
}
},
discard if gt/.style 2 args={
x filter/.append code={
\edef\tempa{\thisrow{#1}}
\edef\tempb{#2}
\ifdim\tempa pt > \tempb pt
\def\pgfmathresult{inf}
\fi
}
},
discard if lt/.style 2 args={
x filter/.append code={
\edef\tempa{\thisrow{#1}}
\edef\tempb{#2}
\ifdim\tempa pt < \tempb pt
\def\pgfmathresult{inf}
\fi
}
}
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Graphics & Plotting %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Colors
%
% KIT Colors
\definecolor{kit-green100}{rgb}{0,.59,.51}
\definecolor{kit-green70}{rgb}{.3,.71,.65}
\definecolor{kit-green50}{rgb}{.50,.79,.75}
\definecolor{kit-green30}{rgb}{.69,.87,.85}
\definecolor{kit-green15}{rgb}{.85,.93,.93}
\definecolor{KITgreen}{rgb}{0,.59,.51}
\definecolor{KITpalegreen}{RGB}{130,190,60}
\colorlet{kit-maigreen100}{KITpalegreen}
\colorlet{kit-maigreen70}{KITpalegreen!70}
\colorlet{kit-maigreen50}{KITpalegreen!50}
\colorlet{kit-maigreen30}{KITpalegreen!30}
\colorlet{kit-maigreen15}{KITpalegreen!15}
\definecolor{KITblue}{rgb}{.27,.39,.66}
\definecolor{kit-blue100}{rgb}{.27,.39,.67}
\definecolor{kit-blue70}{rgb}{.49,.57,.76}
\definecolor{kit-blue50}{rgb}{.64,.69,.83}
\definecolor{kit-blue30}{rgb}{.78,.82,.9}
\definecolor{kit-blue15}{rgb}{.89,.91,.95}
\definecolor{KITyellow}{rgb}{.98,.89,0}
\definecolor{kit-yellow100}{cmyk}{0,.05,1,0}
\definecolor{kit-yellow70}{cmyk}{0,.035,.7,0}
\definecolor{kit-yellow50}{cmyk}{0,.025,.5,0}
\definecolor{kit-yellow30}{cmyk}{0,.015,.3,0}
\definecolor{kit-yellow15}{cmyk}{0,.0075,.15,0}
\definecolor{KITorange}{rgb}{.87,.60,.10}
\definecolor{kit-orange100}{cmyk}{0,.45,1,0}
\definecolor{kit-orange70}{cmyk}{0,.315,.7,0}
\definecolor{kit-orange50}{cmyk}{0,.225,.5,0}
\definecolor{kit-orange30}{cmyk}{0,.135,.3,0}
\definecolor{kit-orange15}{cmyk}{0,.0675,.15,0}
\definecolor{KITred}{rgb}{.63,.13,.13}
\definecolor{kit-red100}{cmyk}{.25,1,1,0}
\definecolor{kit-red70}{cmyk}{.175,.7,.7,0}
\definecolor{kit-red50}{cmyk}{.125,.5,.5,0}
\definecolor{kit-red30}{cmyk}{.075,.3,.3,0}
\definecolor{kit-red15}{cmyk}{.0375,.15,.15,0}
\definecolor{KITpurple}{RGB}{160,0,120}
\colorlet{kit-purple100}{KITpurple}
\colorlet{kit-purple70}{KITpurple!70}
\colorlet{kit-purple50}{KITpurple!50}
\colorlet{kit-purple30}{KITpurple!30}
\colorlet{kit-purple15}{KITpurple!15}
\definecolor{KITcyanblue}{RGB}{80,170,230}
\colorlet{kit-cyanblue100}{KITcyanblue}
\colorlet{kit-cyanblue70}{KITcyanblue!70}
\colorlet{kit-cyanblue50}{KITcyanblue!50}
\colorlet{kit-cyanblue30}{KITcyanblue!30}
\colorlet{kit-cyanblue15}{KITcyanblue!15}
% Matplotlib Colors
\definecolor{Mpl1}{HTML}{1f77b4}
\definecolor{Mpl2}{HTML}{ff7f0e}
\definecolor{Mpl3}{HTML}{2ca02c}
\definecolor{Mpl4}{HTML}{d62728}
\definecolor{Mpl5}{HTML}{9467bd}
\definecolor{Mpl6}{HTML}{8c564b}
\definecolor{Mpl7}{HTML}{e377c2}
\definecolor{Mpl8}{HTML}{7f7f7f}
\definecolor{Mpl9}{HTML}{bcbd22}
\definecolor{Mpl10}{HTML}{17becf}
%
% Color Schemes
%
% Define colormaps
\pgfplotsset{
colormap={mako}{
rgb=(0.18195582, 0.11955283, 0.23136943)
rgb=(0.25307401, 0.23772973, 0.48316271)
rgb=(0.21607792, 0.39736958, 0.61948028)
rgb=(0.20344718, 0.56074869, 0.65649508)
rgb=(0.25187832, 0.71827158, 0.67872193)
rgb=(0.54578602, 0.8544913, 0.69848331)
},
colormap={rocket}{
rgb=(0.20973515, 0.09747934, 0.24238489)
rgb=(0.43860848, 0.12177004, 0.34119475)
rgb=(0.67824099, 0.09192342, 0.3504148)
rgb=(0.8833417, 0.19830556, 0.26014181)
rgb=(0.95381595, 0.46373781, 0.31769923)
rgb=(0.96516917, 0.70776351, 0.5606593)
},
colormap={cividis}{
rgb=(0.130669, 0.231458, 0.43284)
rgb=(0.298421, 0.332247, 0.423973)
rgb=(0.42512, 0.431334, 0.447692)
rgb=(0.555393, 0.537807, 0.471147)
rgb=(0.695985, 0.648334, 0.440072)
rgb=(0.849223, 0.771947, 0.359729)
},
colormap={cel}{
color=(KITred!90!black);
color=(kit-blue100);
color=(kit-green70);
color=(kit-yellow70!80!kit-orange70);
},
colormap={matplotlib}{
% Source: https://github.com/matplotlib/matplotlib/blob/e5a85f960b2d47eac371cff709b830d52c36d267/lib/matplotlib/_cm.py#L1114
rgb=(0.2298057, 0.298717966, 0.753683153)
rgb=(0.26623388, 0.353094838, 0.801466763)
rgb=(0.30386891, 0.406535296, 0.84495867 )
rgb=(0.342804478, 0.458757618, 0.883725899)
rgb=(0.38301334, 0.50941904, 0.917387822)
rgb=(0.424369608, 0.558148092, 0.945619588)
rgb=(0.46666708, 0.604562568, 0.968154911)
rgb=(0.509635204, 0.648280772, 0.98478814 )
rgb=(0.552953156, 0.688929332, 0.995375608)
rgb=(0.596262162, 0.726149107, 0.999836203)
rgb=(0.639176211, 0.759599947, 0.998151185)
rgb=(0.681291281, 0.788964712, 0.990363227)
rgb=(0.722193294, 0.813952739, 0.976574709)
rgb=(0.761464949, 0.834302879, 0.956945269)
rgb=(0.798691636, 0.849786142, 0.931688648)
rgb=(0.833466556, 0.860207984, 0.901068838)
rgb=(0.865395197, 0.86541021, 0.865395561)
rgb=(0.897787179, 0.848937047, 0.820880546)
rgb=(0.924127593, 0.827384882, 0.774508472)
rgb=(0.944468518, 0.800927443, 0.726736146)
rgb=(0.958852946, 0.769767752, 0.678007945)
rgb=(0.96732803, 0.734132809, 0.628751763)
rgb=(0.969954137, 0.694266682, 0.579375448)
rgb=(0.966811177, 0.650421156, 0.530263762)
rgb=(0.958003065, 0.602842431, 0.481775914)
rgb=(0.943660866, 0.551750968, 0.434243684)
rgb=(0.923944917, 0.49730856, 0.387970225)
rgb=(0.89904617, 0.439559467, 0.343229596)
rgb=(0.869186849, 0.378313092, 0.300267182)
rgb=(0.834620542, 0.312874446, 0.259301199)
rgb=(0.795631745, 0.24128379, 0.220525627)
rgb=(0.752534934, 0.157246067, 0.184115123)
rgb=(0.705673158, 0.01555616, 0.150232812)
}
}
% Define cycle lists
\pgfplotscreateplotcyclelist{mako}{%
[samples of colormap={4} of mako]%
}
\pgfplotscreateplotcyclelist{rocket}{%
[samples of colormap={4} of rocket]%
}
\pgfplotscreateplotcyclelist{cividis}{%
[samples of colormap={4} of cividis]%
}
\pgfplotscreateplotcyclelist{viridis}{%
[samples of colormap={4} of viridis]%
}
\pgfplotscreateplotcyclelist{cel}{%
[samples of colormap={4} of cel]%
}
\pgfplotscreateplotcyclelist{matplotlib}{%
{Mpl1},{Mpl2},{Mpl3},{Mpl4}
}
% Define 'scolX' colors
\makeatletter
\def\extractcolormapcolor#1#2{%
\expandafter\pgfplotscolormapaccess\expandafter[\pgfplotspointmetatransformedrange]%
[1.0]%
{#2}%
{\pgfkeysvalueof{/pgfplots/colormap name}}%
\def\pgfplots@loc@TMPb{\pgfutil@definecolor{#1}{\csname pgfpl@cm@\pgfkeysvalueof{/pgfplots/colormap name}@colspace\endcsname}}%
\expandafter\pgfplots@loc@TMPb\expandafter{\pgfmathresult}%
}%
\def\getcolorbyvalue#1{
\csname pgfpl@cm@\pgfkeysvalueof{/pgfplots/colormap name}@colspace\endcsname
}
\makeatother
\def\setschemecolorsfrommap{
\extractcolormapcolor{scol0}{0}
\extractcolormapcolor{scol1}{333}
\extractcolormapcolor{scol2}{666}
\extractcolormapcolor{scol3}{1000}
}
\newcommand{\setschemecolorsmanually}[4]{
\colorlet{scol0}{#1}
\colorlet{scol1}{#2}
\colorlet{scol2}{#3}
\colorlet{scol3}{#4}
}
% Define color schemes
\pgfplotsset{
/pgfplots/colorscheme/cel/.style={
colormap name={cel},
cycle list name={cel},
/utils/exec={\setschemecolorsfrommap},
},
/pgfplots/colorscheme/rocket/.style={
colormap name={rocket},
cycle list name={rocket},
/utils/exec={\setschemecolorsfrommap},
},
/pgfplots/colorscheme/viridis/.style={
colormap name={viridis},
cycle list name={viridis},
/utils/exec={\setschemecolorsfrommap},
},
/pgfplots/colorscheme/mako/.style={
colormap name={mako},
cycle list name={mako},
/utils/exec={\setschemecolorsfrommap},
},
/pgfplots/colorscheme/cividis/.style={
colormap name={cividis},
cycle list name={cividis},
/utils/exec={\setschemecolorsfrommap},
},
/pgfplots/colorscheme/matplotlib/.style={
colormap name={matplotlib},
cycle list name={matplotlib},
/utils/exec={\setschemecolorsmanually{Mpl1}{Mpl2}{Mpl3}{Mpl4}},
},
}

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FROM alpine:3.19
RUN apk update && apk upgrade
RUN apk add make texlive texmf-dist-pictures

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@@ -1,8 +0,0 @@
FROM archlinux:latest
RUN pacman-key --init
RUN pacman-key --populate archlinux
RUN pacman -Sy archlinux-keyring --noconfirm && pacman -Su --noconfirm
RUN pacman -Syu --noconfirm
RUN pacman -S make perl texlive texlive-binextra texlive-pictures --noconfirm

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@@ -1,6 +0,0 @@
FROM ubuntu:22.04
ARG DEBIAN_FRONTEND=noninteractive
RUN apt update -y && apt upgrade -y
RUN apt install make texlive latexmk texlive-pictures -y

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@@ -1,103 +0,0 @@
\documentclass{article}
% Packages necessary for common.tex
\usepackage{amsmath}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
% Other packages
\usepackage{float}
\usepackage{subcaption}
\usepackage[a4paper, total={5in, 9in}]{geometry}
\usetikzlibrary{positioning}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%% Set common options %%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\input{common.tex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%% Actual Document %%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Colorschemes}
\author{}
\date{}
\begin{document}
\maketitle
\foreach \x in {cel, rocket, viridis, mako, cividis, matplotlib}{
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Set colorscheme
\pgfplotsset{colorscheme/\x}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Preview colorscheme
\noindent\begin{minipage}{\textwidth}
\colorbox{gray!30}{\texttt{colorscheme/\x}}
\begin{figure}[H]
\centering
\tikzstyle{colornode} = [draw, inner sep=0pt, minimum width=1cm, minimum height=0.5cm]
\pgfplotsset{scaled y ticks=false}
\begin{subfigure}[c]{0.45\textwidth}
\begin{tikzpicture}
\begin{axis}[
domain=0:10,
view={0}{90},
yticklabel=\empty,xticklabel=\empty,
width=\textwidth,
height=0.75\textwidth,
]
\addplot3+[surf]
{x + y};
\end{axis}
\end{tikzpicture}
\end{subfigure}%
\begin{subfigure}[c]{0.45\textwidth}
\begin{tikzpicture}
\begin{axis}[
domain=0:10,
ymin=-1e4,ymax=1,
legend pos=south west,
yticklabel=\empty,xticklabel=\empty,
width=\textwidth,
height=0.75\textwidth,
]
\foreach \i in {1,2,4,8}{
\addplot+ [mark=none, line width=1pt]
{-\i*exp(x)};
%\expandafter\addlegendentry\expandafter{$e^{-\text{\i} x}$}
\addlegendentryexpanded{$e^{-\i x}$}
}
\end{axis}
\end{tikzpicture}
\end{subfigure}%
\begin{subfigure}[c]{0.1\textwidth}
\begin{tikzpicture}
\node[colornode, fill=scol3] (color3) {};
\node[colornode, fill=scol2, above=1mm of color3] (color2) {};
\node[colornode, fill=scol1, above=1mm of color2] (color1) {};
\node[colornode, fill=scol0, above=1mm of color1] (color0) {};
\end{tikzpicture}
\end{subfigure}%
\end{figure}
\vspace{\parskip}
\end{minipage}
}
\end{document}

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@@ -1,177 +0,0 @@
\documentclass{article}
% Packages necessary for common.tex
\usepackage{amsmath}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
% Other packages
\usepackage{float}
\usepackage{subcaption}
\usepackage[a4paper, total={6.5in, 9in}]{geometry}
\usetikzlibrary{positioning}
\usepackage{ifthen}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%% Set common options %%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\input{common.tex}
\pgfplotsset{colorscheme/rocket}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%% Actual Document %%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Manipulation of CSV data}
\author{}
\date{}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 'discard if lt' & 'discard if gt'
\begin{figure}[H]
\centering
\begin{subfigure}[t]{0.49\textwidth}
%%%%%%%%%%%%%%%%%
% Crop along x axis
\begin{tikzpicture}
\begin{axis}[
width=\textwidth,
height=0.55\textwidth,
xmin=-5, xmax=132,
ymin=-3, ymax=6,
]
% Plot all data as reference
\addplot+[mark=none, line width=1pt]
table[col sep=comma, x=x, y=y]
{res/random.csv};
\addlegendentry{All data}
% Crop and plot desired data
\addplot+[scol2, mark=none, line width=1pt]
table[col sep=comma, x=x, y=y, discard if lt={x}{40},
discard if gt={x}{80}]
{res/random.csv};
\addlegendentry{Cropped data}
\end{axis}
\end{tikzpicture}
\caption{\texttt{discard if lt/gt} used to crop along $x$-axis}
\end{subfigure}%
\hfill%
\begin{subfigure}[t]{0.49\textwidth}
%%%%%%%%%%%%%%%%%
% Crop along y axis
\begin{tikzpicture}
\begin{axis}[
width=\textwidth,
height=0.55\textwidth,
xmin=-5, xmax=132,
ymin=-3, ymax=6,
]
% Plot all data as reference
\addplot+[mark=*]
table[col sep=comma, x=x, y=y]
{res/random.csv};
\addlegendentry{All data}
% Crop and plot desired data
\addplot+[scol2, only marks]
table[col sep=comma, x=x, y=y, discard if gt={y}{1},
discard if lt={y}{-1}]
{res/random.csv};
\addlegendentry{Cropped data}
\end{axis}
\end{tikzpicture}
\caption{\texttt{discard if lt/gt} used to crop along $y$-axis}
\end{subfigure}
\caption{\texttt{discard if lt} and \texttt{discard if gt}}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 'discard if' & 'discard if not'
\begin{figure}[H]
\centering
\begin{subfigure}[t]{0.49\textwidth}
%%%%%%%%%%%%%%%%%
% Select single datastream
\begin{tikzpicture}
\begin{axis}[
width=\textwidth,
height=0.5\textwidth,
ymin=-4,ymax=14,
]
% Plot all data as reference
\foreach \i in {0.0, 3.0, 6.0, 9.0} {
\addplot[scol0!20, mark=none, line width=1pt, forget plot]
table[col sep=comma, x=x, y=y, discard if not={mu}{\i}]
{res/random_multiple.csv};
}
\addlegendimage{scol0!20, mark=none, line width=1pt, forget plot}
\addlegendentry{All data}
% Select and plot desired datastream
\addplot+[scol2, mark=none, line width=1pt]
table[col sep=comma, x=x, y=y, discard if not={mu}{3.0}]
{res/random_multiple.csv};
\addlegendentry{$\mu=3.0$}
\end{axis}
\end{tikzpicture}
\caption{\texttt{discard if not} used to select single datastream}
\end{subfigure}%
\begin{subfigure}[t]{0.49\textwidth}
%%%%%%%%%%%%%%%%%
% Discard single datastream
\begin{tikzpicture}
\begin{axis}[
width=\textwidth,
height=0.5\textwidth,
ymin=-4,ymax=14,
]
% Plot all data as reference
\addplot+[scol0!20, only marks, point meta=\thisrow{mu}]
table[col sep=comma, x=x, y=y]
{res/random_multiple.csv};
\addlegendentry{All data}
% Discard datastream and plot desired data
\addplot+[scol1, only marks, point meta=\thisrow{mu}]
table[col sep=comma, x=x, y=y, discard if={mu}{6.0}]
{res/random_multiple.csv};
\addlegendentry{All except discarded}
\end{axis}
\end{tikzpicture}
\caption{\texttt{discard if} used to discard single datastream}
\end{subfigure}
\caption{\texttt{discard if} and \texttt{discard if not}}
\end{figure}
\end{document}

View File

@@ -1,129 +0,0 @@
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View File

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0,10.074507194453178,9.0
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1 x y mu
2 0 0.734523348833303 0.0
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4 2 0.12740396798342196 0.0
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@@ -1,8 +1,4 @@
\ifdefined\ishandout
\documentclass[de, handout]{CELbeamer}
\else
\documentclass[de]{CELbeamer} \documentclass[de]{CELbeamer}
\fi
% %
% %
@@ -137,26 +133,24 @@
\begin{align*} \begin{align*}
\Omega &= \mleft\{(i,j): i,j \in \mleft\{ \Omega &= \mleft\{(i,j): i,j \in \mleft\{
1,\ldots, 6 \mright\}\mright\} \\ 1,\ldots, 6 \mright\}\mright\} \\
A &= \mleft\{ (1,1),(1,2), \ldots, (6,6) \mright\} A &= \mleft\{ (1,1),(2,2), \ldots, (6,6) \mright\}
\end{align*} \end{align*}
\vspace*{-12mm} \vspace*{-12mm}
\end{lightgrayhighlightbox} \end{lightgrayhighlightbox}
\vspace*{0mm} \vspace*{0mm}
\end{columns}\pause \end{columns}\pause
\item Laplace'sches Zufallsexperiment \item Laplace'sches Zufallsexperiment
% tex-fmt: off
\begin{gather*} \begin{gather*}
\text{Voraussetzungen: }\hspace{5mm} \left\{ \text{Voraussetzungen: }\hspace{5mm} \left\{
\begin{array}{l} \begin{array}{l}
\lvert\Omega\rvert \text{ endlich}\\ \lvert\Omega\rvert \text{ endlich}\\
P(\omega_i) = \frac{1}{\lvert\Omega\rvert} P(\omega_i) = \frac{1}{\lvert\Omega\rvert}
\end{array} \end{array}
\right.\\[1em] \right.\\[1em]
P(A) = \frac{\lvert A \rvert}{\lvert \Omega \rvert} = P(A) = \frac{\lvert A \rvert}{\lvert \Omega \rvert} =
\frac{\text{Anzahl ``günstiger'' \frac{\text{Anzahl ``günstiger''
Möglichkeiten}}{\text{Anzahl Möglichkeiten}} Möglichkeiten}}{\text{Anzahl Möglichkeiten}}
\end{gather*} \end{gather*}
% tex-fmt: on
\end{itemize} \end{itemize}
\end{frame} \end{frame}
@@ -275,17 +269,17 @@
\item mindestens ein Ass hat?\pause \item mindestens ein Ass hat?\pause
\begin{gather*} \begin{gather*}
P(\text{mindestens ein Ass}) = 1 - P(\text{kein Ass}) P(\text{mindestens ein Ass}) = 1 - P(\text{kein Ass})
= 1 - \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} \approx 0{,}341 = 1 - \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} \approx 0.341
\end{gather*}\pause\vspace*{-5mm} \end{gather*}\pause\vspace*{-5mm}
\item genau ein Ass hat?\pause \item genau ein Ass hat?\pause
\begin{gather*} \begin{gather*}
P(\text{genau ein Ass}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} \approx 0{,}299 P(\text{genau ein Ass}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} \approx 0.299
\end{gather*}\pause \end{gather*}\pause
\item mindestens zwei Karten der gleichen Art (“Paar”) hat?\pause \item mindestens zwei Karten der gleichen Art (“Paar”) hat?\pause
\begin{align*} \begin{align*}
P(\text{mindestens zwei gleiche Karten}) &= 1 - P(\text{alle Karten unterschiedlich}) \\ P(\text{mindestens zwei gleiche Karten}) &= 1 - P(\text{alle Karten unterschiedlich}) \\
&= 1 - \frac{\text{Anzahl Möglichkeiten mit nur unterschiedlichen Karten}}{\text{Anzahl Möglichkeiten}}\\ &= 1 - \frac{\text{Anzahl Möglichkeiten mit nur unterschiedlichen Karten}}{\text{Anzahl Möglichkeiten}}\\
&= 1 - \frac{\binom{13}{5}\cdot 4^5}{\binom{52}{5}} \approx 0{,}493 &= 1 - \frac{\binom{13}{5}\cdot 4^5}{\binom{52}{5}} \approx 0.493
\end{align*} \end{align*}
\end{enumerate} \end{enumerate}
% tex-fmt: on % tex-fmt: on
@@ -326,8 +320,7 @@
\mleft\{ A \mright\}, \mleft\{ B \mright\}, \mleft\{ A \mright\}, \mleft\{ B \mright\},
\mleft\{ C \mright\}, \mleft\{ A, B \mright\},\\ \mleft\{ C \mright\}, \mleft\{ A, B \mright\},\\
&\mleft\{ A, C \mright\}, &\mleft\{ A, C \mright\},
\mleft\{ B, C \mright\}, \mleft\{ A, B, C \mleft\{ B, C \mright\}, \mleft\{ A, B, C \mright\} \}
\mright\} \}
\end{align*}% \end{align*}%
\vspace*{-14mm}% \vspace*{-14mm}%
\end{lightgrayhighlightbox} \end{lightgrayhighlightbox}
@@ -372,7 +365,7 @@
\begin{lightgrayhighlightbox} \begin{lightgrayhighlightbox}
Beispiel: Beispiel:
\begin{gather*} \begin{gather*}
\Omega = \{A, B, C\}\\ \Omega = {A, B, C}\\
\Pi_N = \{ (A,B,C), (A,C,B), (B,A,C),\\ \Pi_N = \{ (A,B,C), (A,C,B), (B,A,C),\\
(B,C,A), (C,A,B), (C,B,A)\} (B,C,A), (C,A,B), (C,B,A)\}
\end{gather*} \end{gather*}
@@ -522,4 +515,10 @@
% tex-fmt: on % tex-fmt: on
\end{frame} \end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Zusammenfassung}
% TODO: Do we even want this section?
\end{document} \end{document}

View File

@@ -1,497 +0,0 @@
\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{}
%
%
% Custom commands
%
%
\input{lib/latex-common/common.tex}
\pgfplotsset{colorscheme/rocket}
\newcommand{\res}{src/2025-11-21/res}
% \tikzstyle{every node}=[font=\small]
% \captionsetup[sub]{font=small}
%
%
% Document setup
%
%
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{spy, external, intersections, positioning}
%\tikzexternalize[prefix=build/]
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepgfplotslibrary{fillbetween}
\usepackage{enumerate}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{bbm}
\usepackage{multirow}
\usepackage{xcolor}
\title{WT Tutorium 2}
\author[Tsouchlos]{Andreas Tsouchlos}
\date[]{21. November 2025}
%
%
% Document body
%
%
\begin{document}
\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
\titlepage
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\frametitle{Bedingte Wahrscheinlichkeiten \& Bayes}
\vspace*{-10mm}
\begin{columns}
\column{\kitthreecolumns}
\begin{itemize}
\item Definition der bedingten Wahrscheinlichkeit
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\item Formel von Bayes
\begin{gather*}
P(A\vert B) = \frac{P(B\vert A) P(A)}{P(B)}
\end{gather*}
\end{itemize}
\column{\kitthreecolumns}
\begin{figure}
\centering
\begin{tikzpicture}
\node[rectangle, minimum width=8cm, minimum height=5cm,
draw, line width=1pt, fill=black!20] at (0,0) {};
\node [circle, minimum size = 4cm,
draw, line width=1pt, fill=KITgreen,
fill opacity = 0.5] at (1.25cm,0) {};
\draw[line width=1pt, fill=KITblue,
fill opacity = 0.5, rounded corners=5mm]
(-2.4cm, -2.25cm) -- (-2.4cm, 2.25cm) -- (1.1cm,0) -- cycle;
\node[left] at (4cm, 2cm) {\Large $\Omega$};
\node at (-1.8cm, 0) {$A$};
\node at (1.8cm, 0) {$B$};
\node at (0, 0) {$AB$};
\end{tikzpicture}
\end{figure}
\end{columns}
\vspace*{1cm}
\pause
\begin{columns}
\column{\kitthreecolumns}
\begin{itemize}
\item Satz der totalen Wahrscheinlichkeit
% tex-fmt: off
\begin{gather*}
\text{Voraussetzungen: }\hspace{5mm} \left\{
\begin{array}{l}
A_1, A_2, \ldots \text{ disjunkt}\\
\displaystyle\sum_{n} A_n = \Omega
\end{array}
\right.\\[1em]
P(B) = \sum_{n} P(B\vert A_n)P(A_n)\\
\end{gather*}
% tex-fmt: on
\end{itemize}
\column{\kitthreecolumns}
\begin{figure}
\centering
\begin{tikzpicture}
\newcommand{\hordist}{1.2cm}
\newcommand{\vertdist}{2cm}
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm] (root) at (0, 0) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and
2.4*\hordist of root] (n1) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and
2.4*\hordist of root] (n2) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and \hordist
of n1] (n11) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and \hordist
of n1] (n12) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below left=\vertdist and \hordist
of n2] (n21) {};
\node[circle, fill=KITgreen, inner sep=0pt,
minimum size=3mm, below right=\vertdist and \hordist
of n2] (n22) {};
\draw[-{Latex}, line width=1pt] (root) -- (n1);
\draw[-{Latex}, line width=1pt] (root) -- (n2);
\draw[-{Latex}, line width=1pt] (n1) -- (n11);
\draw[-{Latex}, line width=1pt] (n1) -- (n12);
\draw[-{Latex}, line width=1pt] (n2) -- (n21);
\draw[-{Latex}, line width=1pt] (n2) -- (n22);
\node[left] at ($(root)!0.4!(n1)$) {$P(A_1)$};
\node[right] at ($(root)!0.4!(n2)$) {$P(A_2)$};
\node[left] at ($(n1)!0.4!(n11)$) {$P(B\vert A_1)$};
\node[right] at ($(n1)!0.2!(n12)$) {$P(C\vert A_1)$};
\node[left] at ($(n2)!0.6!(n21)$) {$P(B\vert A_2)$};
\node[right] at ($(n2)!0.4!(n22)$) {$P(C\vert A_2)$};
\node[below] at (n11) {$P(BA_1)$};
\node[below] at (n12) {$P(CA_1)$};
\node[below] at (n21) {$P(BA_2)$};
\node[below] at (n22) {$P(CA_2)$};
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}
\column{\kitthreecolumns}
\begin{greenblock}{Bedingte Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\begin{greenblock}{Formel von Bayes}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(B\vert A) P(A)}{P(B)}
\end{gather*}
\end{greenblock}
\end{columns}
\begin{columns}
\column{\kitonecolumn}
\column{\kitthreecolumns}
\begin{greenblock}{Satz der totalen Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(B) = \sum_{n} P(B\vert A_n)P(A_n)
\end{gather*}
\end{greenblock}
\column{\kitonecolumn}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 1: Bedingte Wahrscheinlichkeiten \\\& Bayes}
In einer Population von gelben Animationsfiguren, den Minions,
werden zwei Merkmale unterschieden: Augenzahl und Körpergröße. Es gilt:
\begin{itemize}
\item $80\%$ der Minions haben zwei Augen, $20\%$ nur eines.
\item Von den zweiäugigen Minions sind $20\%$ groß, $70\%$
mittelgroß und $10\%$ klein.
\item Von den einäugigen Minions sind $5\%$ groß, $60\%$
mittelgroß und $35\%$ klein.
\end{itemize}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Bestimmen Sie die Wahrscheinlichkeit, dass ein zufällig
ausgewähltes Minion klein, mittelgroß
oder groß ist.
\item Ein zufällig ausgewähltes Minion ist nicht klein. Mit
welcher Wahrscheinlichkeit ist es
einäugig?
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 1: Bedingte Wahrscheinlichkeiten \\\& Bayes}
In einer Population von gelben Animationsfiguren, den Minions,
werden zwei Merkmale unterschieden: Augenzahl und Körpergröße. Es gilt:
\begin{itemize}
\item $80\%$ der Minions haben zwei Augen, $20\%$ nur eines.
\item Von den zweiäugigen Minions sind $20\%$ groß, $70\%$
mittelgroß und $10\%$ klein.
\item Von den einäugigen Minions sind $5\%$ groß, $60\%$
mittelgroß und $35\%$ klein.
\end{itemize}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Bestimmen Sie die Wahrscheinlichkeit, dass ein zufällig
ausgewähltes Minion klein, mittelgroß
oder groß ist.
\pause\begin{align*}
P(K) &= P(K\vert N_1)P(N_1) + P(K\vert N_2)P(N_2) = 0{,}35\cdot 0{,}2 + 0{,}1\cdot 0{,}8 = 0{,}15\\
P(M) &= P(M\vert N_1)P(N_1) + P(M\vert N_2)P(N_2) = \cdots = 0{,}68\\
P(G) &= P(G\vert N_1)P(N_1) + P(G\vert N_2)P(N_2) = \cdots = 0{,}17
\end{align*}
\item \pause Ein zufällig ausgewähltes Minion ist nicht klein. Mit
welcher Wahrscheinlichkeit ist es
einäugig?
\pause\begin{align*}
P(N_1 \vert \overline{K})
= \frac{P(\overline{K} \vert N_1)P(N_1)}{P(\overline{K})}
= \frac{\left[ 1 - P(K\vert N_1) \right] P(N_1)}{1 - P(K)}
= \frac{(1 - 0{,}35)\cdot 0{,}2}{1 - 0{,}15} \approx 0{,}153
\end{align*}
\end{enumerate}
% tex-fmt: on
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\frametitle{Zusätzliche Bedingungen und Unabhängigkeit}
\begin{itemize}
\item Erweiterte Definition der bedingten Wahrscheinlichkeit
\begin{gather*}
P(A\vert BC) = \frac{P(AB\vert C)}{P(B\vert C)}
\end{gather*}
\item Satz von Bayes mit zusätzlichen Bedingungen
\begin{gather*}
P(A\vert BC) = \frac{P(B\vert AC) P(A\vert C)}{P(B\vert C)}
\end{gather*}
\pause
\item Unabhängigkeit
\begin{gather*}
A,B \text{ Unabhängig} \hspace{5mm}
\Leftrightarrow\hspace{5mm} P(AB) = P(A) P(B)
\hspace{5mm} \Leftrightarrow \hspace{5mm} P(A\vert B) = P(A)
\end{gather*}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}
\column{\kitthreecolumns}
\begin{greenblock}{Bedingte Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(AB)}{P(B)}
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\begin{greenblock}{Formel von Bayes}
\vspace*{-6mm}
\begin{gather*}
P(A\vert B) = \frac{P(B\vert A) P(A)}{P(B)}
\end{gather*}
\end{greenblock}
\end{columns}
\begin{columns}
\column{\kitthreecolumns}
\begin{greenblock}{Satz der totalen Wahrscheinlichkeit}
\vspace*{-6mm}
\begin{gather*}
P(B) = \sum_{n} P(B\vert A_n)P(A_n)
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\begin{greenblock}{Unabhängigkeit von Ereignissen}
\vspace*{-6mm}
\begin{gather*}
P(AB) = P(A) P(B)
\end{gather*}
\end{greenblock}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 2: Bayes \& Unabhängigkeit}
\vspace*{-18mm}
Bei einer Qualitätskontrolle können Werkstücke zwei Fehler
aufweisen: Fehler $A$, Fehler $B$, oder
beide Fehler gleichzeitig. Die folgenden Wahrscheinlichkeiten
sind bekannt:
\begin{itemize}
\item mit Wahrscheinlichkeit $0{,}05$ hat ein Werkstück den Fehler $A$
\item mit Wahrscheinlichkeit $0{,}01$ hat ein Werkstück beide Fehler
\item mit Wahrscheinlichkeit $0{,}03$ hat ein Werkstück nur den
Fehler $B$ und nicht Fehler $A$.
\end{itemize}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Berechnen Sie die Wahrscheinlichkeit für das Auftreten von
Fehler $B$ und dafür, dass ein
Werkstück fehlerfrei ist.
\item Ist das Auftreten von Fehler $A$ unabhängig von Fehler $B$?
\end{enumerate}
% tex-fmt: on
Bei der Kontrolle wird unerwartet ein zusätzlicher, dritter Fehler $C$
beobachtet. Der Fehler tritt
mit der Wahrscheinlichkeit $0{,}01$ ein, wenn weder Fehler $A$ noch $B$
eingetreten sind und mit der
Wahrscheinlichkeit $0{,}02$, wenn sowohl Fehler $A$ als auch $B$ eingetreten
sind. In allen anderen
Fällen tritt der Fehler $C$ nicht auf.
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{2}
\item Berechnen Sie die Wahrscheinlichkeit für das Auftreten von
Fehler $C$.
\item Sie beobachten, dass ein Werkstück den Fehler $C$ hat. Mit
welcher Wahrscheinlichkeit hat es auch Fehler $A$?
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Bayes \& Unabhängigkeit}
\vspace*{-10mm}
Bei einer Qualitätskontrolle können Werkstücke zwei Fehler
aufweisen: Fehler $A$, Fehler $B$, oder
beide Fehler gleichzeitig. Die folgenden Wahrscheinlichkeiten
sind bekannt:
\begin{itemize}
\item mit Wahrscheinlichkeit $0{,}05$ hat ein Werkstück den Fehler $A$
\item mit Wahrscheinlichkeit $0{,}01$ hat ein Werkstück beide Fehler
\item mit Wahrscheinlichkeit $0{,}03$ hat ein Werkstück nur den
Fehler $B$ und nicht Fehler $A$.
\end{itemize}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Berechnen Sie die Wahrscheinlichkeit für das Auftreten von
Fehler $B$ und dafür, dass ein
Werkstück fehlerfrei ist.
\pause\begin{gather*}
P(B) = P(B\vert A)P(A) + P(B\vert \overline{A})P(\overline{A}) = P(AB) + P(\overline{A}B) = 0{,}01 + 0{,}03 = 0{,}04
\end{gather*}\pause
\vspace*{-15mm}\begin{gather*}
P(\overline{A}\cap \overline{B}) = 1 - P(A\cup B) = 1 - \left[P(A) + P(B) - P(A\cap B)\right] = 1 - \left(0{,}05 + 0{,}04 - 0{,}01\right) = 0{,}92
\end{gather*}
\vspace*{-12mm}\pause \item Ist das Auftreten von Fehler $A$ unabhängig von Fehler $B$?
\pause\begin{gather*}
\left. \begin{array}{l}
P(AB) = 0{,}01 \\
P(A)P(B) = 0{,}05\cdot 0{,}04 = 0{,}002
\end{array}\right\}
\hspace{5mm} \Rightarrow \hspace{5mm} P(AB) \neq P(A)P(B) \hspace{5mm}\Rightarrow\hspace{5mm}A,B \text{ nicht unabhängig}
\end{gather*}
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Bayes \& Unabhängigkeit}
\vspace*{-13mm}
Bei der Kontrolle wird unerwartet ein zusätzlicher, dritter Fehler $C$
beobachtet. Der Fehler tritt
mit der Wahrscheinlichkeit $0{,}01$ ein, wenn weder Fehler $A$ noch $B$
eingetreten sind und mit der
Wahrscheinlichkeit $0{,}02$, wenn sowohl Fehler $A$ als auch $B$ eingetreten
sind. In allen anderen
Fällen tritt der Fehler $C$ nicht auf.
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{2}
\item Berechnen Sie die Wahrscheinlichkeit für das Auftreten von
Fehler $C$.
\pause\begin{align*}
P(C) &= P(C\vert AB)P(AB) + \overbrace{P(C\vert A \overline{B})}^{0}P(A \overline{B})
+ \overbrace{P(C\vert \overline{A}B)}^{0}P(\overline{A} B)
+ P(C\vert \overline{A}\overline{B})P(\overline{A}\overline{B}) \\
&= 0{,}02\cdot 0{,}01 + 0{,}01\cdot 0{,}92 = 0{,}0094
\end{align*}
\vspace*{-12mm}\pause \item Sie beobachten, dass ein Werkstück den Fehler $C$ hat. Mit
welcher Wahrscheinlichkeit hat es auch Fehler $A$?
\pause\hspace*{-5mm}\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
P(A\vert C) &= \frac{P(AC)}{P(C)}\\[5mm]
P(AC) &= P(ACB) + P(AC \overline{B})\\
&= P(C\vert AB)P(AB) + \overbrace{P(C\vert A \overline{B})}^{0}P(A \overline{B})\\
&= 0{,}02\cdot 0{,}01 = 0{,}0002\\[5mm]
P(A\vert C) &= \frac{0{,}0002}{0{,}0094} \approx 0{,}0213
\end{align*}
\end{minipage}%
\hspace*{-10mm}
\begin{minipage}{0.06\textwidth}
\centering
\begin{tikzpicture}
\draw[line width=1pt] (0,0) -- (0,6cm);
\end{tikzpicture}
\end{minipage}%
\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
P(A\vert C) &= \frac{P(C\vert A)P(A)}{P(C)}\\[5mm]
P(C\vert A) &= P(C\vert AB)P(B\vert A)
+ \overbrace{P(C\vert \overline{A} B)}^{0}P(\overline{A}B) \\
&= P(C\vert AB)\frac{P(AB)}{P(A)} = 0{,}02 \cdot \frac{0{,}01}{0{,}05} = 0{,}004\\[5mm]
P(A\vert C) &= \frac{0{,}004\cdot 0{,}05}{0{,}0094} \approx 0{,}0213
\end{align*}
\end{minipage}
\end{enumerate}
% tex-fmt: on
\end{frame}
\end{document}

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@@ -1,928 +0,0 @@
\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{}
%
%
% Custom commands
%
%
\input{lib/latex-common/common.tex}
\pgfplotsset{colorscheme/rocket}
\newcommand{\res}{src/2025-12-05/res}
% \tikzstyle{every node}=[font=\small]
% \captionsetup[sub]{font=small}
%
%
% Document setup
%
%
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{spy, external, intersections, positioning}
%\tikzexternalize[prefix=build/]
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepgfplotslibrary{fillbetween}
\usepackage{enumerate}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{bbm}
\usepackage{multirow}
\usepackage{xcolor}
\title{WT Tutorium 3}
\author[Tsouchlos]{Andreas Tsouchlos}
\date[]{5. Dezember 2025}
%
%
% Document body
%
%
\begin{document}
\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
\titlepage
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\frametitle{Zufallsvariablen \& Verteilungen}
\vspace*{-10mm}
\begin{itemize}
\item Zufallsvariablen (ZV)
\begin{minipage}{0.33\textwidth}
\centering
\begin{gather*}
\text{Idee: ``Wegabstrahieren'' von Ergebnisraum
$\Omega$} \\[1cm]
X: \Omega \mapsto \mathbb{R} \\
\underbrace{P_X(x)}_\text{Verteilung} :=
P(\underbrace{X}_\text{ZV}=\underbrace{x}_\text{Realisierung})
\end{gather*}
\end{minipage}%
\hspace*{15mm}%
\begin{minipage}{0.6\textwidth}
\centering
\begin{lightgrayhighlightbox}
Beispiel: Würfeln mit zwei Würfeln
\begin{gather*}
X := \text{\normalfont``Summe beider Augenzahlen''}\\
X: \underbrace{\left\{(i, j) : i, j \in \left\{1, \ldots
, 6\right\}\right\}}_{\Omega} \mapsto
\underbrace{\left\{2,3,
\ldots, 12\right\}}_{\in \mathbb{R}}
\end{gather*}\\
\vspace*{5mm}
\begin{tikzpicture}
\draw[line width=1pt] (0,0) -- (18cm,0);
\end{tikzpicture}
\vspace*{2mm}
\begin{gather*}
A = \text{\normalfont``Die Summe der
Augenzahlen ist 4''}
\end{gather*}
\begin{minipage}[t]{0.5\textwidth}
\centering
Direkter Weg
\begin{align*}
P(A) &= P(\mleft\{ (1,3), (2,2),
(3,1) \mright\}) \\
&= P( (1,3)) + P( (2, 2)) + P( (3,1)) \\
&= 3\cdot \frac{1}{36} = \frac{1}{12}
\end{align*}
\end{minipage}%
\begin{minipage}[t]{0.5\textwidth}
\centering
Über ZV
\begin{gather*}
P(A) = P_X(4) = \cdots
\end{gather*}
\end{minipage}
\end{lightgrayhighlightbox}
\end{minipage}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Verteilungen \& Verteilungsfunktionen}
\vspace*{-18mm}
\begin{itemize}
\item Verteilungsfunktionen diskreter ZV
\vspace*{-6mm}
\begin{columns}
\column{\kitthreecolumns}
\begin{align*}
\overbrace{F_X(x)}^\text{Verteilungsfunktion} = P(X \le x)
&= \sum_{n:x_n \le x}
\overbrace{P_X(x)}^\text{Verteilung}\\
&= \sum_{n:x_n \le x} P(X=x)
\end{align*}
\begin{gather*}
P(a < X \le b) = F_X(b) - F_X(a)
\end{gather*}
\column{\kitthreecolumns}
\begin{lightgrayhighlightbox}
Beispiel: Würfeln mit zwei Würfeln
\begin{gather*}
X := \text{\normalfont``Summe beider Augenzahlen''}
\end{gather*}
\vspace*{-10mm}
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=2,xmax=12,
ymin=-0.2,ymax=1.2,
xlabel=$x$,
ylabel=$F_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[mark=none, line width=1pt]
coordinates
{
(2 , 0.02777)
(3 , 0.02777)
(3 , 0.08333)
(4 , 0.08333)
(4 , 0.16666)
(5 , 0.16666)
(5 , 0.27777)
(6 , 0.27777)
(6 , 0.41666)
(7 , 0.41666)
(7 , 0.58333)
(8 , 0.58333)
(8 , 0.72222)
(9 , 0.72222)
(9 , 0.83333)
(10 , 0.83333)
(10, 0.91666)
(11, 0.91666)
(11, 0.97222)
(12, 0.97222)
(12, 1.00000)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\vspace*{-10mm}
\end{lightgrayhighlightbox}
\end{columns}
\pause
\item Kenngrößen von Verteilungen
\vspace*{2mm}
\begin{columns}[t]
\column{\kittwocolumns}
\centering
\textbf{Erwartungswert}
\begin{gather*}
E(X) = \sum_{n=1}^{\infty} x_n P(X=x_n)
\end{gather*}%
\vspace*{-8mm}%
\begin{align*}
E(X + b) &= E(X) + b\\
E(X+Y) &= E(X) + E(Y)\\
E(aX) &= aE(X)
\end{align*}
\column{\kittwocolumns}
\centering
\textbf{Varianz}
\begin{gather*}
V(X) = E\left(\left(X - E(X)\right)^2\right)
\end{gather*}%
\vspace*{-8mm}
\begin{align*}
V(X) &= E(X^2) - \left(E(X)\right)^2\\
V(aX) &= a^2 V(x)\\
V(X+b) &= V(X)
\end{align*}
\column{\kittwocolumns}
\centering
\textbf{$p$-Quantil}
\begin{gather*}
x_p = \text{inf}\mleft\{ x\in \mathbb{R} : P(X
\le x) \ge p \mright\}
\end{gather*}
\vspace*{-8mm}
\begin{gather*}
p=0.5 \hspace{5mm} \rightarrow \hspace{5mm} x_p
\equiv \text{``Median''}
\end{gather*}
\end{columns}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Beispiele von Verteilungen}
\vspace*{-18mm}
\begin{columns}[t]
\column{\kittwocolumns}
\centering
\textbf{Bernoulli Verteilung}\\
\vspace*{10mm}
$X$ kann nur die Werte $0$ oder $1$\\ annehmen
\rule{0.9\textwidth}{0.4pt}
\begin{gather*}
X \sim \text{Bernoulli}(p)
\end{gather*}
\begin{gather*}
P(X=0) = 1-p, \hspace{5mm} P(X=1) = p
\end{gather*}
\begin{align*}
E(X) &= p\\
V(X) &= p(1-p)
\end{align*}
\column{\kittwocolumns}
\centering
\textbf{Binomialverteilung}\\
\vspace*{10mm}
$X\equiv$ ``Zählen der Treffer bei $N$ unabhängigen Versuchen''
\rule{0.9\textwidth}{0.4pt}
\begin{gather*}
X \sim \text{Bin}(N,p)
\end{gather*}
\begin{gather*}
P_X(k) = \binom{N}{k} p^k (1-p)^{N-k}
\end{gather*}
\begin{align*}
E(X) &= Np\\
V(X) &= Np(1-p)
\end{align*}
\column{\kittwocolumns}
\centering
\textbf{Poisson Verteilung}\\
\vspace*{10mm}
Binomialverteilung für $N\rightarrow \infty$ mit
$pN=\text{const.}=: \lambda$
\rule{0.9\textwidth}{0.4pt}
\begin{gather*}
X \sim \text{Poisson}(\lambda)
\end{gather*}
\begin{gather*}
P_X(k) = \frac{\lambda^k}{k!}e^{-\lambda}
\end{gather*}
\begin{align*}
E(X) &= \lambda\\
V(X) &= \lambda
\end{align*}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}[t]
\column{\kittwocolumns}
\begin{greenblock}{Verteilungsfunktion (diskret)}
\vspace*{-6mm}
\begin{gather*}
F_X(x) = P(X \le x) = \sum_{n:x_n < x} P_X(x_n)
\end{gather*}
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{Erwartungswert}
\vspace*{-6mm}
\begin{gather*}
E(X) = \sum_{n=1}^{\infty} x_n P(X=x_n)
\end{gather*}%
\vspace*{-8mm}%
\begin{align*}
E(X + b) &= E(X) + b\\
E(X+Y) &= E(X) + E(Y)\\
E(aX) &= aE(X)
\end{align*}
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{Binomialverteilung}
\vspace*{-6mm}
\begin{gather*}
P_X(k) = \binom{N}{k} p^k (1-p)^{N-k}
\end{gather*}
\begin{align*}
E(X) &= Np\\
V(X) &= Np(1-p)
\end{align*}
\end{greenblock}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 1: Diskrete Verteilungen}
\vspace*{-10mm}
Eine Polizistin führt $N = 6$ Radarkontrollen auf einer
Landstraße durch. Die Radarkontrollen
können als unabhängig angenommen werden und führen jeweils mit
der Wahrscheinlichkeit
$p = 0{,}2$ zu einem Strafzettel. Die diskrete Zufallsvariable $R :
\Omega \rightarrow \mathbb{R}$ beschreibt die Anzahl der
Strafzettel in $N = 6$ Kontrollen.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Geben Sie den Ergebnisraum $\Omega$ der diskreten Zufallsvariablen $R$ an
und bestimmen Sie deren Erwartungswert $E(R)$.
\item Wie groß ist die Wahrscheinlichkeit dafür, dass es bei $6$
Kontrollen genau $3$ Strafzettel gibt?
\item Skizzieren Sie die Verteilungsfunktion $F_R(r)$ der
Zufallsvariablen $R$.
\end{enumerate}
% tex-fmt: on
\vspace*{5mm}
\textit{Die folgenden Teilaufgaben können unabhängig von den
bisherigen Teilaufgaben bearbeitet werden.}
\vspace*{5mm}
Ein Autofahrer muss jeden Tag auf seinem Arbeitsweg über die
Landstraße und über die
Autobahn fahren. Die Wahrscheinlichkeit dafür, dass der
Autofahrer auf der Landstraße bzw.
auf der Autobahn zu schnell fährt und einen Strafzettel bekommt,
liegt bei $p_\text{L} = 0{,}2$ bzw. bei
$p_\text{A} = 0{,}3$.
\vspace*{5mm}
\textbf{Hinweis}: Es wird nur der einfache Weg (Hinweg) betrachtet.
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{3}
\item Wie groß ist die Wahrscheinlichkeit dafür, dass der Autofahrer
an einem Tag $0$, $1$ oder $2$ Strafzettel bekommt?
\item Der Autofahrer fährt an $200$ unabhängigen Tagen im Jahr über
seinen Arbeitsweg zur Arbeit. Wie viele Strafzettel sammelt der
Autofahrer innerhalb eines Jahres?
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 1: Diskrete Verteilungen}
\vspace*{-16mm}
Eine Polizistin führt $N = 6$ Radarkontrollen auf einer
Landstraße durch. Die Radarkontrollen
können als unabhängig angenommen werden und führen jeweils mit
der Wahrscheinlichkeit
$p = 0{,}2$ zu einem Strafzettel. Die diskrete Zufallsvariable $R :
\Omega \rightarrow \mathbb{R}$ beschreibt die Anzahl der
Strafzettel in $N = 6$ Kontrollen.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Geben Sie den Ergebnisraum $\Omega$ der diskreten Zufallsvariablen $R$ an
und bestimmen Sie deren Erwartungswert $E(R)$.
\pause\begin{gather*}
\Omega = \mleft\{ 0, 1\mright\}^6 \\
R \sim \text{Bin}(N=6, p=0{,}2)\hspace{5mm} \Rightarrow \hspace{5mm} E(R) = Np = 1{,}2
\end{gather*}
\vspace*{-10mm}\pause \item Wie groß ist die Wahrscheinlichkeit dafür, dass es bei $6$
Kontrollen genau $3$ Strafzettel gibt?
\pause \begin{gather*}
P(R=3) = \binom{N}{3}p^3 (1-p)^{N-3} = \binom{6}{3} \cdot 0{,}2^3\cdot 0{,}8^3 \approx 0{,}0819
\end{gather*}
\vspace*{-6mm}\pause \item Skizzieren Sie die Verteilungsfunktion $F_R(r)$ der
Zufallsvariablen $R$.
\end{enumerate}
% tex-fmt: on
\vspace*{2mm}
\pause
\begin{columns}
\column{\kitthreecolumns}
\begin{gather*}
F_R(r) = \sum_{\widetilde{r} \le r}
\binom{N}{\widetilde{r}}p^{\widetilde{r}} (1-p)^{N-\widetilde{r}}
\end{gather*}
\begin{table}
\begin{tabular}{c|ccccccc}
$r$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ \\ \hline
$F_R(r)$ & $0{,}262$ & $0{,}655$ & $0{,}901$ &
$0{,}983$ & $0{,}998$ & $0{,}999$ & $1$
\end{tabular}
\end{table}
\column{\kitthreecolumns}
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=6,
ymin=-0.2,ymax=1.2,
xlabel=$r$,
ylabel=$F_R(r)$,
width=12cm,
height=5cm,
]
\addplot+[mark=none, line width=1pt]
coordinates
{
(0,0.262)
(1,0.262)
(1,0.655)
(2,0.655)
(2,0.901)
(3,0.901)
(3,0.983)
(4,0.983)
(4,0.998)
(5,0.998)
(5,0.999)
(6,0.999)
(6,1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Aufgabe 1: Diskrete Verteilungen}
\vspace*{-16mm}
Ein Autofahrer muss jeden Tag auf seinem Arbeitsweg über die
Landstraße und über die Autobahn fahren. Die Wahrscheinlichkeit
dafür, dass der Autofahrer auf der Landstraße bzw. auf der
Autobahn zu schnell fährt und einen Strafzettel bekommt, liegt
bei $p_\text{L} = 0{,}2$ bzw. bei $p_\text{A} = 0{,}3$.
\vspace*{2mm}
\textbf{Hinweis}: Es wird nur der einfache Weg (Hinweg) betrachtet.
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{3}
\item Wie groß ist die Wahrscheinlichkeit dafür, dass der Autofahrer
an einem Tag $0$, $1$ oder $2$ Strafzettel bekommt?
\pause\begin{gather*}
R := A + L
\end{gather*}%
\vspace*{-14mm}%
\begin{align*}
P(R = 0) &= P(A = 0 \text{ und } L = 0) &&\hspace{-24mm}= (1-p_A)(1-p_L) &&\hspace{-24mm}= 0{,}56\\
P(R = 1) &= P(A=1 \text{ und } L=0) + P(A=0 \text{ und } L=1) &&\hspace{-24mm}= p_A \cdot (1-p_L) + (1-p_A)\cdot p_L &&\hspace{-24mm}= 0{,}38 \\
P(R = 2) &= P(A=1 \text{ und } L=1) &&\hspace{-24mm}= p_A\cdot p_L &&\hspace{-24mm}= 0{,}06
\end{align*}
\vspace*{-10mm}\pause \item Der Autofahrer fährt an $200$ unabhängigen Tagen im Jahr über
seinen Arbeitsweg zur Arbeit. Wie viele Strafzettel sammelt der
Autofahrer innerhalb eines Jahres?
\end{enumerate}
% tex-fmt: on
\vspace*{-6mm}
\pause
\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
E\left(\sum_{n=1}^{200} R_n\right) &= \sum_{n=1}^{200}
E\left(R_n\right) = \sum_{n=1}^{200} \left[1\cdot0{,}38 +
2\cdot 0{,}06\right]\\[2mm]
&= 200\cdot 0{,}5 = 100
\end{align*}
\end{minipage}%
\begin{minipage}{0.06\textwidth}
\centering
\begin{tikzpicture}
\draw[line width=1pt] (0,0) -- (0,4cm);
\end{tikzpicture}
\end{minipage}%
\begin{minipage}{0.48\textwidth}
\centering
\begin{align*}
E\left(\sum_{n=1}^{200} R_n\right) &=
E\Big(\overbrace{\sum_{n=1}^{200} A_n}^{\sim
\text{Bin}(N=200, p=0{,}3)} + \overbrace{\sum_{n=1}^{200}
L_n}^{\sim \text{Bin}(N=200, p=0{,}2)}\Big)\\[2mm]
&= E\left(\sum_{n=1}^{200} A_n\right) +
E\left(\sum_{n=1}^{200} L_n\right) \\[2mm]
&= 200\cdot 0{,}3 + 200 \cdot 0{,}2 = 100
\end{align*}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\frametitle{Weitere Kenngrößen von Verteilungen}
\vspace*{-10mm}
\vspace*{10mm}
\begin{columns}[t]
\column{\kitthreecolumns}
\centering
\textbf{$k$-tes Moment}
\begin{gather*}
E(X^k) = \sum_{n=1}^{\infty} x_n^k P(X=x_n)
\end{gather*}%
\column{\kitthreecolumns}
\centering
\textbf{$k$-tes zentrales Moment}
\begin{gather*}
E\left( \left(X - E(X)\right)^k \right) =
\sum_{n=1}^{\infty} \left(x_n - E(X)\right)^k P(X=x_n)
\end{gather*}%
\end{columns}
\vspace*{20mm}
\pause
\begin{columns}[t]
\column{\kitthreecolumns}
\centering
\textbf{Charakteristische Funktion (diskret)}
\begin{gather*}
\phi_X(s) = E(e^{jsX}) = \sum_{n=1}^{\infty}
e^{jsx_n} P(X=x_n)\\[5mm]
E(X^k) = \frac{\phi_X^{(k)}(0)}{j^k}
\end{gather*}
\column{\kitthreecolumns}
\centering
\textbf{Erzeugende Funktion}
\begin{gather*}
\text{Voraussetzung:} \hspace{5mm} x \in \mathbb{N}_0\\[5mm]
\psi(z) = E(z^x) = \sum_{n=1}^{\infty} z^n P(x=n)\\[5mm]
P(X=n) = \frac{\psi_X^{(n)}(0)}{n!}
\end{gather*}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Zusammenfassung}
\vspace*{-16mm}
\begin{columns}[t]
\column{\kittwocolumns}
\begin{greenblock}{Verteilungsfunktion (diskret)}
\vspace*{-6mm}
\begin{gather*}
F_X(x) = P(X \le x) = \sum_{n:x_n < x} P_X(x_n)
\end{gather*}
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{Varianz}
\vspace*{-6mm}
\begin{gather*}
V(X) = E\left(\left(X - E(X)\right)^2\right)
\end{gather*}%
\vspace*{-8mm}
\begin{align*}
V(X) &= E(X^2) - \left(E(X)\right)^2\\
V(aX) &= a^2 V(x)\\
V(X+b) &= V(X)
\end{align*}
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{$p$-Quantil}
\vspace*{-6mm}
\begin{gather*}
x_p = \text{inf}\mleft\{ x\in \mathbb{R} : P(X
\le x) \ge p \mright\}
\end{gather*}
\vspace*{-8mm}
\begin{gather*}
p=0.5 \hspace{5mm} \rightarrow \hspace{5mm} x_p
\equiv \text{``Median''}
\end{gather*}
\end{greenblock}
\end{columns}
\begin{columns}[t]
\column{\kittwocolumns}
\begin{greenblock}{$k$-tes Moment}
\vspace*{-6mm}
\begin{gather*}
E(X^k) = \sum_{n=1}^{\infty} x_n^k P(X=x_n)
\end{gather*}%
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{Charakt. Funktion (diskret)}
\vspace*{-6mm}
\begin{gather*}
\phi_X(s) = \sum_{n=1}^{\infty}
e^{jsx_n} P(X=x_n)\\[5mm]
E(X^k) = \frac{\phi_X^{(k)}(0)}{j^k}
\end{gather*}
\end{greenblock}
\column{\kittwocolumns}
\begin{greenblock}{Erzeugende Funktion}
\vspace*{-6mm}
\begin{gather*}
\psi(z) = \sum_{n=1}^{\infty} z^n P(X=n)\\[5mm]
P(X=n) = \frac{\psi_X^{(n)}(0)}{n!}
\end{gather*}
\end{greenblock}
\end{columns}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 2: Erzeugende \& Charakteristische\\ Funktion}
Gegeben ist folgende Verteilungsfunktion $F_X(x)$:
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=5.5,
ymin=0,ymax=1,
xtick={0,...,5},
ytick={0,0.2,...,1},
xlabel=$x$,
ylabel=$F_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[mark=none, line width=1pt]
coordinates
{
(0,0)
(1,0)
(1,0.2)
(2,0.2)
(2,0.6)
(3,0.6)
(3,0.7)
(4,0.7)
(4,0.9)
(5,0.9)
(5,1)
(5.5,1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\vspace*{-10mm}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Stellen Sie die Verteilung $P_X(x)$ graphisch dar.
\item Geben Sie die erzeugende Funktion $\psi_X(z)$ und die
charakteristische Funktion $\phi_X(s)$ an. Berechnen Sie mit mithilfe
von $\phi_X(s)$ die Varianz $V(X)$.
\item Vergleichen Sie den Median und den Erwartungswert von $X$. Sind
beide Kenngrößen gleich? Begründen Sie, welche Eigenschaft einer
diskreten Verteilung ausschlaggebend ist, damit beide Werte gleich
sind.
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Erzeugende \& Charakteristische\\ Funktion}
Gegeben ist folgende Verteilungsfunktion $F_X(x)$:
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=5.5,
ymin=0,ymax=1,
xtick={0,...,5},
ytick={0,0.2,...,1},
xlabel=$x$,
ylabel=$F_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[mark=none, line width=1pt]
coordinates
{
(0,0)
(1,0)
(1,0.2)
(2,0.2)
(2,0.6)
(3,0.6)
(3,0.7)
(4,0.7)
(4,0.9)
(5,0.9)
(5,1)
(5.5,1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\vspace*{-10mm}
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Stellen Sie die Verteilung $P_X(x)$ graphisch dar.
\end{enumerate}
% tex-fmt: on
\pause
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=5.5,
ymin=0,ymax=0.5,
xtick={0,...,5},
ytick={0,0.1,...,0.5},
xlabel=$x$,
ylabel=$P_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[ycomb,mark=*, line width=1pt]
coordinates
{
(1,0.2)
(2,0.4)
(3,0.1)
(4,0.2)
(5,0.1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Erzeugende \& Charakteristische\\ Funktion}
\vspace*{-12mm}
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=5.5,
ymin=0,ymax=0.5,
xtick={0,...,5},
ytick={0,0.1,...,0.5},
xlabel=$x$,
ylabel=$P_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[ycomb,mark=*, line width=1pt]
coordinates
{
(1,0.2)
(2,0.4)
(3,0.1)
(4,0.2)
(5,0.1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
\vspace*{-5mm}
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{1}
\item Geben Sie die erzeugende Funktion $\psi_X(z)$ und die
charakteristische Funktion $\phi_X(s)$ an. Berechnen Sie mit mithilfe
von $\phi_X(s)$ die Varianz $V(X)$.
\pause\begin{align*}
\psi_X(z) &= \sum_{n=1}^{5} z^n P(X=n) = 0{,}2z + 0{,}4z^2 + 0{,}1z^3
+ 0{,}2z^4 + 0{,}1z^5 \\
\phi_X(s) &= \sum_{n=1}^{5} e^{jsx_n}P(X=n) = 0{,}2e^{js}
+ 0{,}4e^{j2s} + 0{,}1e^{j3s} + 0{,}2e^{j4s} + 0{,}1e^{j5s}
\end{align*}
\pause\begin{gather*}
\left.\begin{array}{c}
V(X) = E(X^2) - \left(E(X)\right)^2\\[3mm]
E(X) = \displaystyle\frac{\phi_X'(0)}{j}
= \sum_{n=1}^{5} nP(X=n) = 2{,}6\\[5mm]
E(X^2) = \displaystyle\frac{\phi_X''(0)}{j^2}
= \sum_{n=1}^{5} n^2 P(X=n) = 8{,}4
\end{array}\right\} \Rightarrow V(X) = 8{,}4 - 2{,}6^2 = 1{,}64
\end{gather*}
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Erzeugende \& Charakteristische\\ Funktion}
\vspace*{-5mm}
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
xmin=0,xmax=5.5,
ymin=0,ymax=1,
xtick={0,...,5},
ytick={0,0.2,...,1},
xlabel=$x$,
ylabel=$F_X(x)$,
width=12cm,
height=5cm,
]
\addplot+[mark=none, line width=1pt]
coordinates
{
(0,0)
(1,0)
(1,0.2)
(2,0.2)
(2,0.6)
(3,0.6)
(3,0.7)
(4,0.7)
(4,0.9)
(5,0.9)
(5,1)
(5.5,1)
};
\end{axis}
\end{tikzpicture}
\end{figure}
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{2}
\item Vergleichen Sie den Median und den Erwartungswert von $X$. Sind
beide Kenngrößen gleich? Begründen Sie, welche Eigenschaft einer
diskreten Verteilung ausschlaggebend ist, damit beide Werte gleich
sind.
\pause\begin{align*}
x_{1/2} &= \text{inf}\mleft\{ x\in \mathbb{R}: F_X(x) \ge 1/2 \mright\} = 2\\
E(X) &= 2{,}6
\end{align*}
\vspace*{5mm}
\centering
\pause\begin{minipage}{0.7\textwidth}
Median und Erwartungswert sind gleich (bei einer diskreten
Verteilung mit ganzzahligen Stützstellen), wenn die Verteilung
symmetrisch um denselben Punkt $c$ ist, d.h.,
\begin{gather*}
P(c+k) = P(c-k) \hspace*{5mm} \forall k\in \mathbb{Z}.
\end{gather*}
\end{minipage}
\end{enumerate}
% tex-fmt: on
\end{frame}
\end{document}

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@@ -1,731 +0,0 @@
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\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{}
%
%
% Custom commands
%
%
\input{lib/latex-common/common.tex}
\pgfplotsset{colorscheme/rocket}
\newcommand{\res}{src/2025-12-19/res}
% \tikzstyle{every node}=[font=\small]
% \captionsetup[sub]{font=small}
%
%
% Document setup
%
%
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{spy, external, intersections, positioning}
%\tikzexternalize[prefix=build/]
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepgfplotslibrary{fillbetween}
\usepackage{enumerate}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{bbm}
\usepackage{multirow}
\usepackage{xcolor}
\title{WT Tutorium 4}
\author[Tsouchlos]{Andreas Tsouchlos}
\date[]{19. Dezember 2025}
%
%
% Document body
%
%
\begin{document}
\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
\titlepage
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\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}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 1: Stetige Verteilungen}
Die Zufallsvariable X besitze die Dichte
% tex-fmt: off
\begin{align*}
f_X (x) = \left\{
\begin{array}{ll}
C \cdot x e^{-ax^2}, & x \ge 0 \\
0, &\text{sonst}
\end{array}
\right.
\end{align*}
% tex-fmt: on
mit dem Parameter $a > 0$.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Bestimmen Sie den Koeffizienten $C$, sodass $f_X(x)$ eine
Wahrscheinlichkeitsdichte ist. Welche Eigenschaften muss eine
\textbf{Wahrscheinlichkeitsdichte} erfüllen? Skizzieren Sie
$f_X (x)$ für $a = 0{,}5$.
\item Welche Eigenschaften muss eine \textbf{Verteilungsfunktion}
erfüllen?
\item Berechnen und skizzieren Sie die Verteilungsfunktion $F_X (x)$.
\item Welche Wahrscheinlichkeit hat das Ereignis
$\{\omega : 1 < X(\omega) \le 2\}$?
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 1: Stetige Verteilungen}
\vspace*{-15mm}
Die Zufallsvariable X besitze die Dichte
% tex-fmt: off
\begin{align*}
f_X (x) = \left\{
\begin{array}{ll}
C \cdot x e^{-ax^2}, & x \ge 0 \\
0, &\text{sonst}
\end{array}
\right.
\end{align*}
% tex-fmt: on
mit dem Parameter $a > 0$.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Bestimmen Sie den Koeffizienten $C$, sodass $f_X(x)$ eine
Wahrscheinlichkeitsdichte ist. Welche Eigenschaften muss eine
\textbf{Wahrscheinlichkeitsdichte} erfüllen? Skizzieren Sie
$f_X (x)$ für $a = 0{,}5$.
\pause\begin{columns}
\column{\kitthreecolumns}
\begin{align*}
\text{Eigenschaften:} \hspace{5mm}
\left\{
\begin{array}{rl}
f_X(x) &\ge 0 \\[3mm]
\displaystyle\int_{-\infty}^{\infty} f_X(x) dx &= 1
\end{array}
\right.
\end{align*}
\pause\begin{gather*}
\int_{-\infty}^{\infty} f_X(x) 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
\column{\kitthreecolumns}
\pause \begin{align*}
f_X(x) =
\left\{
\begin{array}{ll}
2ax \cdot e^{-ax^2}, & x\ge 0\\
0, & \text{sonst}
\end{array}
\right.
\end{align*}
\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
domain=0:5,
width=12cm,
height=5cm,
samples=100,
xlabel={$x$},
ylabel={$f_X(x)$},
]
\addplot+[mark=none, line width=1pt]
{x * exp(-0.5*x*x)};
% {x *exp(-a*x*x)};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{columns}
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 1: Stetige Verteilungen}
\vspace*{-20mm}
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{1}
\item Welche Eigenschaften muss eine \textbf{Verteilungsfunktion}
erfüllen?
\pause\vspace{-10mm}\begin{columns}[t]
\column{\kitonecolumn}
\column{\kittwocolumns}
\centering
\begin{gather*}
\lim_{x\rightarrow -\infty} F_X(x) = 0\\
\lim_{x\rightarrow +\infty} F_X(x) = 1
\end{gather*}
\column{\kittwocolumns}
\centering
\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)
\end{gather*}
\column{\kitonecolumn}
\end{columns}
\pause\item Berechnen und skizzieren Sie die Verteilungsfunktion $F_X (x)$.
\begin{gather*}
f_X(x) = 2ax\cdot e^{-ax^2}, \hspace{5mm} x\ge 0
\end{gather*}
\pause \vspace*{-6mm}\begin{gather*}
F_X(x) = \int_{-\infty}^{x} f_X(u) du
= \left\{ \begin{array}{ll}
\displaystyle\int_{0}^{x} 2au\cdot e^{-au^2} du, & x\ge 0 \\
0, & x < 0
\end{array} \right.
\hspace{5mm} = \left\{ \begin{array}{ll}
\mleft[ -e^{-au^2} \mright]_0^{x}, & x\ge 0 \\
0, & x < 0
\end{array} \right.
\hspace{5mm} = \left\{ \begin{array}{ll}
1 - e^{-ax^2}, & x\ge 0\\
0, & x < 0
\end{array} \right.
\end{gather*}
\pause\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
domain=0:5,
width=14cm,
height=5cm,
xlabel={$x$},
ylabel={$F_X(x)$},
]
\addplot+[mark=none, line width=1pt]
{1 - exp(-0.5 * x*x)};
\end{axis}
\end{tikzpicture}
\end{figure}
\vspace*{-3mm}
\pause\item Welche Wahrscheinlichkeit hat das Ereignis
$\{\omega : 1 < X(\omega) \le 2\}$?
\pause \begin{gather*}
P(\mleft\{ \omega: 1 < X(\omega) \le 2 \mright\})
= P(1 < X \le 2) = F_X(2) - F_X(1) = e^{-a} - e^{-4a}
\end{gather*}
\end{enumerate}
% tex-fmt: on
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie Wiederholung}
\begin{frame}
\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}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
\begin{frame}
\frametitle{Aufgabe 2: Normalverteilung}
In einem Produktionsprozess werden Ladegeräte für Mobiltelefone
hergestellt. Bevor die Ladegeräte mit den Mobiltelefonen zusammen
verpackt werden, wird die Ladespannung von jedem Ladegerät einmal
gemessen. Die Messwerte der Ladespannungen der verschiedenen
Ladegeräte genüge näherungsweise einer normalverteilten
Zufallsvariablen mit $\mu = 5$ Volt und $\sigma = 0,07$ Volt. Alle
Ladegeräte, bei denen die Messung um mehr als $4$ \% vom Sollwert
$S = 5$ Volt abweicht, sollen aussortiert werden.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Wie viel Prozent der Ladegeräte werden aussortiert?
\item Der Hersteller möchte seinen Produktionsprozess so verbessern,
dass nur noch halb so viele Ladegeräte wie in a) aussortiert
werden. Auf welchen Wert müsste er dazu $\sigma$ senken?
\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?
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Normalverteilung}
\vspace*{-10mm}
In einem Produktionsprozess werden Ladegeräte für Mobiltelefone
hergestellt. Bevor die Ladegeräte mit den Mobiltelefonen zusammen
verpackt werden, wird die Ladespannung von jedem Ladegerät einmal
gemessen. Die Messwerte der Ladespannungen der verschiedenen
Ladegeräte genüge näherungsweise einer normalverteilten
Zufallsvariablen mit $\mu = 5$ Volt und $\sigma = 0,07$ Volt. Alle
Ladegeräte, bei denen die Messung um mehr als $4$ \% vom Sollwert
$S = 5$ Volt abweicht, sollen aussortiert werden.
% tex-fmt: off
\begin{enumerate}[a{)}]
\item Wie viel Prozent der Ladegeräte werden aussortiert?
\begin{columns}[c]
\column{\kitthreecolumns}
\centering
\pause \begin{gather*}
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]
&\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
\pause\begin{figure}[H]
\centering
\begin{tikzpicture}
\begin{axis}[
domain=4.6:5.3,
xmin=4.7, xmax=5.3,
width=14cm,
height=6cm,
xlabel={$x$},
ylabel={$F_X (x)$},
samples=100,
xtick = {4.6,4.7,4.8,...,5.4}
]
\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 +[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) {$S(1+\delta)$};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{columns}
\end{enumerate}
% tex-fmt: on
\end{frame}
\begin{frame}
\frametitle{Aufgabe 2: Normalverteilung}
\vspace*{-18mm}
% tex-fmt: off
\begin{enumerate}[a{)}]
\setcounter{enumi}{1}
\item Der Hersteller möchte seinen Produktionsprozess so verbessern,
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{\%} \\
\end{gather*}
\vspace*{-18mm}
\begin{columns}
\pause\column{\kitthreecolumns}
\centering
\begin{align*}
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)
\end{align*}
\pause\column{\kitthreecolumns}
\centering
\begin{gather*}
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{,}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(\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{\%}
\end{align*}
\end{enumerate}
% tex-fmt: on
\end{frame}
\end{document}

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import numpy as np
import matplotlib.pyplot as plt
from numpy.typing import NDArray
import argparse
def twodim_array_to_pgfplots_table_string(a: NDArray):
return (
" \\\\\n".join([" ".join([str(vali) for vali in val]) for val in a]) + "\\\\\n"
)
def main():
# Parse command line arguments
parser = argparse.ArgumentParser()
parser.add_argument("--correlation", "-c", type=np.float32, required=True)
parser.add_argument("-N", type=np.int32, required=True)
parser.add_argument("--plot", "-p", action="store_true")
args = parser.parse_args()
# Generate & plot data
means = np.array([0, 0])
cov = np.array([[1, args.correlation], [args.correlation, 1]])
x = np.random.multivariate_normal(means, cov, size=args.N)
print(twodim_array_to_pgfplots_table_string(x))
if args.plot:
plt.scatter(x[:, 0], x[:, 1])
plt.show()
if __name__ == "__main__":
main()

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@@ -1,40 +0,0 @@
import argparse
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import binom
def array_to_pgfplots_table_string(a):
return " ".join([f"({k}, {val})" for (k, val) in enumerate(a)]) + f" ({len(a)}, 0)"
def P_binom(N, p, k):
return binom(N, k) * p**k * (1 - p) ** (N - k)
def main():
# Parse command line arguments
parser = argparse.ArgumentParser()
parser.add_argument("-N", type=np.int32, required=True)
parser.add_argument("-p", type=np.float32, required=True)
parser.add_argument("--show", "-s", action="store_true")
args = parser.parse_args()
# Generate and show data
N = args.N
p = args.p
bars = np.array([P_binom(N, p, k) for k in range(N + 1)])
print(array_to_pgfplots_table_string(bars))
if args.show:
plt.stem(bars)
plt.show()
if __name__ == "__main__":
main()

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

1
src/template/lib Symbolic link
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/home/andreas/Documents/kit/wt-tut/presentations/lib

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\documentclass[de]{CELbeamer}
%
%
% 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{}
%
%
% Custom commands
%
%
\input{lib/latex-common/common.tex}
\pgfplotsset{colorscheme/rocket}
%TODO: Fix path
\newcommand{\res}{src/template/res}
% \tikzstyle{every node}=[font=\small]
% \captionsetup[sub]{font=small}
%
%
% Document setup
%
%
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{spy, external, intersections}
%\tikzexternalize[prefix=build/]
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\usepgfplotslibrary{fillbetween}
\usepackage{listings}
\usepackage{subcaption}
\usepackage{bbm}
\usepackage{multirow}
\usepackage{xcolor}
\title{WT Tutorium 1}
\author[Tsouchlos]{Andreas Tsouchlos}
\date[]{\today}
%
%
% Document body
%
%
\begin{document}
\begin{frame}[title white vertical, picture=images/IMG_7801-cut]
\titlepage
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie}
% TODO: Replace slide content with relevant stuff
\begin{frame}
\frametitle{Relevante Theorie I}
\begin{columns}
\column{\kitthreecolumns}
\begin{greenblock}{Zufallsvariablen (ZV)}%
\vspace*{-6mm}
\begin{gather*}
f_X(x) := \frac{d}{dx} F_X(x) \\
P(X \le x) = F_X(x) = \int_{-\infty}^{x} f_X(t) dt \\
E(X) = \int_{-\infty}^{\infty} x\cdot f_X(x) dx
\end{gather*}
\end{greenblock}
\column{\kitthreecolumns}
\begin{greenblock}{Important Equations}%
\vspace*{-6mm}
\begin{gather*}
f_X(x) := \frac{d}{dx} F_X(x) \\
P(X \le x) = F_X(x) = \int_{-\infty}^{x} f_X(t) dt \\
E(X) = \int_{-\infty}^{\infty} x\cdot f_X(x) dx
\end{gather*}
\end{greenblock}
\end{columns}
\begin{greenblock}{Normalverteilung}
\begin{columns}
\column{\kitthreecolumns}
\begin{gather*}
\text{Normalverteilung:} \hspace{8mm}
f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}
e^{-\frac{(x - \mu)^2}{2\sigma^2}}
\end{gather*}
\column{\kitthreecolumns}
\begin{figure}
\centering
\begin{tikzpicture}
\begin{axis}[
domain=-4:4,
samples=100,
width=11cm,
height=6cm,
ticks=none,
xlabel={$x$},
ylabel={$f_X(x)$}
]
\addplot+[mark=none, line width=1pt] {exp(-x^2)};
\end{axis}
\end{tikzpicture}
\end{figure}
\end{columns}
\end{greenblock}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
% TODO: Replace slide content with relevant stuff
\begin{frame}
\frametitle{2022H - Aufgabe 4}
Für die Planung und Konstruktion von Windkraftanlagen ist eine
statistische Modellierung der
Windgeschwindigkeit essentiell. Die absolute Windgeschwindigkeit
kann als Weibull-verteilte
Zufallsvariable V mit den Parametern $\beta > 0$ und $\theta > 0$
modelliert werden. Die zugehörige
Verteilungsfunktion ist%
%
\begin{gather*}
F_V(v) = 1 - exp\left( -\left( \frac{v}{\theta} \right)^\beta
\right), \hspace{3mm} v \ge 0
\end{gather*}
%
\begin{enumerate}
\item Berechnen Sie die Wahrscheinlichkeitsdichte $f_V(v)$
der Weibullverteilung.
\item Eine Windkraftanlage speist Strom in das Stromnetz ein,
wenn die absolute Windgeschwindigkeit größer als $4
m/s$, jedoch kleiner als $25 m/s$ ist. Berechnen Sie die
Wahrscheinlichkeit dafür, dass eine Windkraftanlage Strom
einspeist, wenn die Windgeschwindigkeit Weibull-verteilt
mit $\beta = 2,0$ und $\theta = 6,0$ ist.
\item Eine Zufallsvariable W genüge einer Weibullverteilung
mit $\beta = 1$ und $\theta = 3$. Ermitteln Sie den
Erwartungsvert $E(W)$.
\item Warum ist die Weibullverteilung für die Modellierung
der absoluten Windgeschwindigkeit besser geeignet als
eine Normalverteilung?
\end{enumerate}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Aufgabe 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Theorie}
% TODO: Replace slide content with relevant stuff
\begin{frame}
\frametitle{Relevante Theorie II}
\begin{gather*}
f_X(x) := \frac{d}{dx} F_X(x) \\
P(X \le x) = F_X(x) = \int_{-\infty}^{x} f_X(t) dt \\
E(X) = \int_{-\infty}^{\infty} x\cdot f_X(x) dx
\end{gather*}
\begin{figure}
\centering
\begin{subfigure}[c]{0.5\textwidth}
\centering
\begin{gather*}
\text{Normalverteilung:} \hspace{8mm}
f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}
e^{-\frac{(x - \mu)^2}{2\sigma^2}}
\end{gather*}
\end{subfigure}%
\begin{subfigure}[c]{0.4\textwidth}
\centering
\begin{tikzpicture}
\begin{axis}[
domain=-4:4,
samples=100,
width=\textwidth,
height=0.5\textwidth,
ticks=none,
xlabel={$x$},
ylabel={$f_X(x)$}
]
\addplot+[mark=none, line width=1pt] {exp(-x^2)};
\end{axis}
\end{tikzpicture}
\end{subfigure}
\end{figure}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Aufgabe}
% TODO: Replace slide content with relevant stuff
\begin{frame}
\frametitle{2022H - Aufgabe 4}
Für die Planung und Konstruktion von Windkraftanlagen ist eine
statistische Modellierung der
Windgeschwindigkeit essentiell. Die absolute Windgeschwindigkeit
kann als Weibull-verteilte
Zufallsvariable V mit den Parametern $\beta > 0$ und $\theta > 0$
modelliert werden. Die zugehörige
Verteilungsfunktion ist%
%
\begin{gather*}
F_V(v) = 1 - exp\left( -\left( \frac{v}{\theta} \right)^\beta
\right), \hspace{3mm} v \ge 0
\end{gather*}
%
\begin{enumerate}
\item Berechnen Sie die Wahrscheinlichkeitsdichte $f_V(v)$
der Weibullverteilung.
\item Eine Windkraftanlage speist Strom in das Stromnetz ein,
wenn die absolute Windgeschwindigkeit größer als $4
m/s$, jedoch kleiner als $25 m/s$ ist. Berechnen Sie die
Wahrscheinlichkeit dafür, dass eine Windkraftanlage Strom
einspeist, wenn die Windgeschwindigkeit Weibull-verteilt
mit $\beta = 2,0$ und $\theta = 6,0$ ist.
\item Eine Zufallsvariable W genüge einer Weibullverteilung
mit $\beta = 1$ und $\theta = 3$. Ermitteln Sie den
Erwartungsvert $E(W)$.
\item Warum ist die Weibullverteilung für die Modellierung
der absoluten Windgeschwindigkeit besser geeignet als
eine Normalverteilung?
\end{enumerate}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Zusammenfassung}
% TODO: Replace slide content with relevant stuff
\begin{frame}
\frametitle{Zusammenfassung}
\begin{gather*}
f_X(x) := \frac{d}{dx} F_X(x) \\
P(X \le x) = F_X(x) = \int_{-\infty}^{x} f_X(t) dt \\
E(X) = \int_{-\infty}^{\infty} x\cdot f_X(x) dx
\end{gather*}
\begin{figure}
\centering
\begin{subfigure}[c]{0.5\textwidth}
\centering
\begin{gather*}
\text{Normalverteilung:} \hspace{8mm}
f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}
e^{-\frac{(x - \mu)^2}{2\sigma^2}}
\end{gather*}
\end{subfigure}%
\begin{subfigure}[c]{0.4\textwidth}
\centering
\begin{tikzpicture}
\begin{axis}[
domain=-4:4,
samples=100,
width=\textwidth,
height=0.5\textwidth,
ticks=none,
xlabel={$x$},
ylabel={$f_X(x)$}
]
\addplot+[mark=none, line width=1pt] {exp(-x^2)};
\end{axis}
\end{tikzpicture}
\end{subfigure}
\end{figure}
\end{frame}
\end{document}

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