Compare commits
4 Commits
tut5-v1.0
...
d898c48619
| Author | SHA1 | Date | |
|---|---|---|---|
| d898c48619 | |||
| 91cf66a544 | |||
| 34769737b0 | |||
| 51718e4749 |
4
.gitignore
vendored
4
.gitignore
vendored
@@ -1,5 +1 @@
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build/
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||||
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src/*/.latexmkrc
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src/*/lib
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src/*/src
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@@ -6,7 +6,6 @@ RUN apt update -y && apt upgrade -y
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RUN apt install make texlive latexmk texlive-pictures -y
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RUN apt install texlive-publishers texlive-science texlive-fonts-extra texlive-latex-extra -y
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RUN apt install biber texlive-bibtex-extra -y
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RUN apt install texlive-lang-german -y
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RUN apt install python3 python3-pygments -y
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24
Makefile
24
Makefile
@@ -1,21 +1,11 @@
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PRESENTATIONS := $(patsubst src/%/presentation.tex,build/presentation_%.pdf,$(wildcard src/*/presentation.tex))
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HANDOUTS := $(patsubst build/presentation_%.pdf,build/presentation_%_handout.pdf,$(PRESENTATIONS))
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all:
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mkdir -p build/build
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.PHONY: all
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all: $(PRESENTATIONS) $(HANDOUTS)
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TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk src/template/presentation.tex
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mv build/presentation.pdf build/presentation_template.pdf
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build/presentation_%.pdf: src/%/presentation.tex build/prepared
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TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk $<
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mv build/presentation.pdf $@
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build/presentation_%_handout.pdf: src/%/presentation.tex build/prepared
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TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk -pdflatex='pdflatex %O "\def\ishandout{1}\input{%S}"' $<
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mv build/presentation.pdf $@
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build/prepared:
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mkdir -p build
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touch build/prepared
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.PHONY: clean
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TEXINPUTS=./lib/cel-slides-template-2025:$$TEXINPUTS latexmk src/2025-11-07/presentation.tex
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mv build/presentation.pdf build/presentation_2025-11-07.pdf
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clean:
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rm -rf build
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18
README.md
18
README.md
@@ -1,18 +0,0 @@
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# WT Tutorial Presentations
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This repository contains the latex source files for the WT Tutorial slides.
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## Build
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### Local Environment
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```bash
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$ make
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```
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### With Docker
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```bash
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$ docker build . -t wt-tut
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$ docker run --rm -u `id -u`:`id -g` -w $PWD -v $PWD:$PWD wt-tut make
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```
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0
src/2025-11-07/presentation.bib
Normal file
0
src/2025-11-07/presentation.bib
Normal file
@@ -1,8 +1,4 @@
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\ifdefined\ishandout
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\documentclass[de, handout]{CELbeamer}
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\else
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\documentclass[de]{CELbeamer}
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\fi
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%
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%
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@@ -30,7 +26,8 @@
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\input{lib/latex-common/common.tex}
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\pgfplotsset{colorscheme/rocket}
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\newcommand{\res}{src/2025-11-07/res}
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%TODO: Fix path
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\newcommand{\res}{src/template/res}
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% \tikzstyle{every node}=[font=\small]
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% \captionsetup[sub]{font=small}
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@@ -60,7 +57,7 @@
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\title{WT Tutorium 1}
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\author[Tsouchlos]{Andreas Tsouchlos}
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\date[]{7. November 2025}
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\date[]{\today}
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%
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%
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@@ -74,177 +71,74 @@
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\titlepage
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\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Struktur des Tutoriums}
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\begin{frame}
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\frametitle{Struktur des Tutoriums}
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\begin{itemize}
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\item Ziele
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\begin{itemize}
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\item Üben/Verstehen der Herangehensweisen Aufgaben zu lösen
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\item Wiederholung der für die Aufgaben wichtigsten Teile
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der Theorie
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\end{itemize}
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\item Struktur der Tutorien
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\begin{table}
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\begin{tabular}{l||c}
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Abschnitt & Dauer \\\hline\hline
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Aufgabe 1: Theorie Wiederholung & $\SI{10}{\minute}$ \\
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Aufgabe 1: Selbstrechenphase & $\SI{20}{\minute}$ \\
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Aufgabe 1: Besprechung der Lösung &
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$\SI{10}{\minute}$ \\\hline
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Aufgabe 2: Theorie Wiederholung & $\SI{10}{\minute}$ \\
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Aufgabe 2: Selbstrechenphase & $\SI{20}{\minute}$ \\
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Aufgabe 2: Besprechung der Lösung &
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$\SI{10}{\minute}$ \\\hline
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Zusammenfassung & $\SI{10}{\minute}$ \\
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\end{tabular}
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\end{table}
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\end{itemize}
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\end{frame}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Aufgabe 1}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Theorie Wiederholung}
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\subsection{Theorie}
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\begin{frame}{Ereignisse \& Laplace}
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\vspace*{-15mm}
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\begin{itemize}
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\item Ereignisse
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\begin{columns}
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\column{\kitthreecolumns}
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\begin{align*}
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\text{Ergebnisraum: } & \hspace{5mm} \Omega =
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\mleft\{ \omega_1, \ldots, \omega_N \mright\}\\
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\text{Ergebnis: } & \hspace{5mm} \omega_i\\
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\text{Ereignis: } & \hspace{5mm} A \subseteq \Omega
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\end{align*}
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\column{\kitthreecolumns}
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\begin{lightgrayhighlightbox}
|
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Beispiel: Würfeln mit einem Würfel
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\begin{align*}
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\Omega &= \mleft\{ 1, \ldots, 6 \mright\}\\
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A &= \mleft\{ 1, 6 \mright\}
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\end{align*}\\[1em]
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\vspace*{-12mm}
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\end{lightgrayhighlightbox}
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\begin{lightgrayhighlightbox}
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Beispiel: Würfeln mit zwei Würfeln
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\begin{align*}
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\Omega &= \mleft\{(i,j): i,j \in \mleft\{
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1,\ldots, 6 \mright\}\mright\} \\
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A &= \mleft\{ (1,1),(1,2), \ldots, (6,6) \mright\}
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\end{align*}
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\vspace*{-12mm}
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\end{lightgrayhighlightbox}
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\vspace*{0mm}
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\end{columns}\pause
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\item Laplace'sches Zufallsexperiment
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% tex-fmt: off
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\begin{gather*}
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\text{Voraussetzungen: }\hspace{5mm} \left\{
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\begin{array}{l}
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\lvert\Omega\rvert \text{ endlich}\\
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P(\omega_i) = \frac{1}{\lvert\Omega\rvert}
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\end{array}
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\right.\\[1em]
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P(A) = \frac{\lvert A \rvert}{\lvert \Omega \rvert} =
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\frac{\text{Anzahl ``günstiger''
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Möglichkeiten}}{\text{Anzahl Möglichkeiten}}
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\end{gather*}
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% tex-fmt: on
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\end{itemize}
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\end{frame}
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% TODO: Replace slide content with relevant stuff
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\begin{frame}
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\frametitle{Relevante Theorie I}
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\begin{frame}{Kombinationen und Hypergeometrische\\ Verteilung}
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\begin{itemize}
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\item Kombinationen: Ziehen ohne zurücklegen, ohne
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Betrachtung der Reihenfolge
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\vspace*{5mm}
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\begin{columns}
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\column{\kitthreecolumns}
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||||
\begin{gather*}
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||||
\lvert C_N^{(K)} \rvert = \binom{N}{K} =
|
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\frac{N!}{(N-K)!K!}
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\end{gather*}
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\column{\kitthreecolumns}
|
||||
\begin{lightgrayhighlightbox}
|
||||
Beispiel: Wie viele mögliche Ergebnisse gibt
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es beim Lotto ``6 aus 49''?
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\vspace*{0mm}
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\begin{align*}
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\begin{array}{c}
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N = 49 \\
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K = 6
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\end{array} \hspace{5mm} \rightarrow
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\hspace{5mm} \binom{49}{6} = 13983816
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\end{align*}
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\vspace*{-8mm}
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\end{lightgrayhighlightbox}
|
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\end{columns}
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\pause
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\item Hypergeometrische Verteilung
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\begin{columns}
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\column{\kitthreecolumns}
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\begin{gather*}
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P_r = \frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}
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\end{gather*}
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\column{\kitthreecolumns}
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||||
\begin{lightgrayhighlightbox}
|
||||
Beispiel: In einer Urne sind N Kugeln, davon
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R rot. Wie groß ist die Wahrscheinlichkeit
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||||
beim ziehen von n Kugeln (ohne Zurücklegen)
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||||
genau r rote zu erwischen?
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||||
\end{lightgrayhighlightbox}
|
||||
\end{columns}
|
||||
\end{itemize}
|
||||
\end{frame}
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\begin{frame}{Zusammenfassung}
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\begin{columns}
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\column{\kitthreecolumns}
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\begin{greenblock}{Laplace'sches Zufallsexperiment}%
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\begin{greenblock}{Zufallsvariablen (ZV)}%
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\vspace*{-6mm}
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\begin{gather*}
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P(A) = \frac{\lvert A \rvert}{\lvert \Omega \rvert} =
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||||
\frac{\text{Anzahl ``günstiger''
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Möglichkeiten}}{\text{Anzahl Möglichkeiten}}
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||||
f_X(x) := \frac{d}{dx} F_X(x) \\
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||||
P(X \le x) = F_X(x) = \int_{-\infty}^{x} f_X(t) dt \\
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||||
E(X) = \int_{-\infty}^{\infty} x\cdot f_X(x) dx
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||||
\end{gather*}
|
||||
\end{greenblock}
|
||||
|
||||
\column{\kitthreecolumns}
|
||||
\begin{greenblock}{Kombinationen}%
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||||
\begin{greenblock}{Important Equations}%
|
||||
\vspace*{-6mm}
|
||||
\begin{gather*}
|
||||
\lvert C_N^{(K)}\rvert = \binom{N}{K} =
|
||||
\frac{N!}{(N-K)!K!}
|
||||
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{\kitonecolumn}
|
||||
\column{\kitthreecolumns}
|
||||
\begin{greenblock}{Hypergeometrische Verteilung}%
|
||||
\vspace*{-6mm}
|
||||
\begin{gather*}
|
||||
P_R = \frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}
|
||||
\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{greenblock}
|
||||
\column{\kitonecolumn}
|
||||
|
||||
\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{Aufgabe 1: Ergebnisraum \&
|
||||
Hypergeometrische\\ Verteilung}
|
||||
\frametitle{Aufgabe 1: Ergebnisraum \& Hypergeometrische\\ Verteilung}
|
||||
|
||||
Bei einem Kartenspiel erhält ein Spieler 5 Karten aus einem Deck
|
||||
von 52 Karten (bestehend aus
|
||||
@@ -261,180 +155,61 @@
|
||||
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}
|
||||
\frametitle{Aufgabe 1: Ergebnisraum \&
|
||||
Hypergeometrische\\ Verteilung}
|
||||
|
||||
Bei einem Kartenspiel erhält ein Spieler 5 Karten aus einem Deck
|
||||
von 52 Karten (bestehend aus
|
||||
13 Arten mit je 4 Farben). Wie groß ist die Wahrscheinlichkeit,
|
||||
dass der Spieler
|
||||
|
||||
% tex-fmt: off
|
||||
\begin{enumerate}[a{)}]
|
||||
\item mindestens ein Ass hat?\pause
|
||||
\begin{gather*}
|
||||
P(\text{mindestens ein Ass}) = 1 - P(\text{kein Ass})
|
||||
= 1 - \frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}} \approx 0{,}341
|
||||
\end{gather*}\pause\vspace*{-5mm}
|
||||
\item genau ein Ass hat?\pause
|
||||
\begin{gather*}
|
||||
P(\text{genau ein Ass}) = \frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}} \approx 0{,}299
|
||||
\end{gather*}\pause
|
||||
\item mindestens zwei Karten der gleichen Art (“Paar”) hat?\pause
|
||||
\begin{align*}
|
||||
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{\binom{13}{5}\cdot 4^5}{\binom{52}{5}} \approx 0{,}493
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
% tex-fmt: on
|
||||
|
||||
\end{frame}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\section{Aufgabe 2}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\subsection{Theorie Wiederholung}
|
||||
\subsection{Theorie}
|
||||
|
||||
% TODO: Replace slide content with relevant stuff
|
||||
\begin{frame}
|
||||
\frametitle{Kombinatorik}
|
||||
\frametitle{Relevante Theorie II}
|
||||
|
||||
\vspace*{-18mm}
|
||||
|
||||
\begin{itemize}
|
||||
\item Potenzmenge
|
||||
\vspace*{-2mm}
|
||||
\begin{columns}
|
||||
\column{\kitfourcolumns}
|
||||
\begin{align*}
|
||||
\mathcal{P}\mleft( \Omega \mright) = \mleft\{ A:
|
||||
A \subseteq \Omega \mright\} \hspace{10mm}
|
||||
\left(\text{``Menge aller
|
||||
Teilmengen von $\Omega$''}\right)
|
||||
\end{align*}
|
||||
\column{\kittwocolumns}
|
||||
\begin{lightgrayhighlightbox}
|
||||
Beispiel
|
||||
\begin{gather*}
|
||||
\Omega = \{ A, B, C \}
|
||||
\end{gather*}%
|
||||
\vspace*{-15mm}%
|
||||
\begin{align*}
|
||||
\mathcal{P}(\Omega) = \{ &\emptyset,
|
||||
\mleft\{ A \mright\}, \mleft\{ B \mright\},
|
||||
\mleft\{ C \mright\}, \mleft\{ A, B \mright\},\\
|
||||
&\mleft\{ A, C \mright\},
|
||||
\mleft\{ B, C \mright\}, \mleft\{ A, B, C
|
||||
\mright\} \}
|
||||
\end{align*}%
|
||||
\vspace*{-14mm}%
|
||||
\end{lightgrayhighlightbox}
|
||||
\end{columns}
|
||||
\vspace*{-3mm}
|
||||
\item \pause Variationen und Kombinationen
|
||||
\setlength\extrarowheight{2mm}
|
||||
\begin{table}
|
||||
\begin{tabular}{r||l|l}
|
||||
& Mit Zurücklegen & Ohne Zurücklegen
|
||||
\\\hline\hline Mit Reihenfolge
|
||||
(\textit{Variationen}) & $\lvert
|
||||
\widetilde{V}_N^{(K)} \rvert = N^K$ & $\lvert
|
||||
V_N^{(K)}\rvert = \frac{N!}{(N-K)!} $ \\\hline
|
||||
Ohne Reihenfolge (\textit{Kombinationen}) &
|
||||
$\lvert \widetilde{C}_N^{(K)} \rvert =
|
||||
\binom{N+K-1}{K} $ & $\lvert C_N^{(K)} \rvert
|
||||
= \binom{N}{K} $
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
\item \pause Permutationen
|
||||
\begin{columns}
|
||||
\column{\kitfourcolumns}
|
||||
\begin{gather*}
|
||||
\Pi_N = \mleft\{ \mleft( a_1, \ldots, a_N
|
||||
\mright) \in \Omega : a_i \neq a_j, i \neq j
|
||||
\mright\}\\
|
||||
\begin{array}{r}
|
||||
\text{Alle Elemente von $\Omega$ unterscheidbar:} \\
|
||||
\text{Jeweils $L_1, L_2, \ldots, L_M$ der Elemente
|
||||
sind gleich:}
|
||||
\end{array}
|
||||
\hspace{5mm}
|
||||
\begin{array}{rl}
|
||||
\lvert \Pi_N \rvert &= N! \\
|
||||
\lvert \Pi_N^{(L_1,
|
||||
L_2, \ldots, L_M)} \rvert &=
|
||||
\frac{N!}{L_1!L_2!\cdots L_M!}
|
||||
\end{array}
|
||||
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*}
|
||||
\column{\kittwocolumns}
|
||||
\begin{lightgrayhighlightbox}
|
||||
Beispiel:
|
||||
\begin{gather*}
|
||||
\Omega = \{A, B, C\}\\
|
||||
\Pi_N = \{ (A,B,C), (A,C,B), (B,A,C),\\
|
||||
(B,C,A), (C,A,B), (C,B,A)\}
|
||||
\end{gather*}
|
||||
\vspace*{-14mm}%
|
||||
\end{lightgrayhighlightbox}
|
||||
\end{columns}
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}
|
||||
\frametitle{Zusammenfassung}
|
||||
\begin{figure}
|
||||
\centering
|
||||
|
||||
\begin{columns}
|
||||
\column{\kitthreecolumns}
|
||||
\begin{greenblock}{Potenzmenge}
|
||||
\vspace*{-6mm}
|
||||
\begin{subfigure}[c]{0.5\textwidth}
|
||||
\centering
|
||||
\begin{gather*}
|
||||
\mathcal{P}\mleft( \Omega \mright) = \mleft\{ A:
|
||||
A \subseteq \Omega \mright\}
|
||||
\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{greenblock}
|
||||
\column{\kitthreecolumns}
|
||||
\begin{greenblock}{Permutationen}
|
||||
\vspace*{-6mm}
|
||||
\begin{align*}
|
||||
\lvert \Pi_N \rvert &= N! \\
|
||||
\lvert \Pi_N^{(L_1, L_2, \ldots, L_M)} \rvert &=
|
||||
\frac{N!}{L_1!L_2!\cdots L_M!}
|
||||
\end{align*}
|
||||
\end{greenblock}
|
||||
\end{columns}
|
||||
\begin{columns}
|
||||
\column{\kitonecolumn}
|
||||
\column{\kitfourcolumns}
|
||||
\begin{greenblock}{Variationen \& Kombinationen }
|
||||
\begin{table}
|
||||
\begin{tabular}{r||l|l}
|
||||
& Mit Zurücklegen & Ohne Zurücklegen
|
||||
\\\hline\hline Mit Reihenfolge
|
||||
(\textit{Variationen}) & $\lvert
|
||||
\widetilde{V}_N^{(K)} \rvert = N^K$ & $\lvert
|
||||
V_N^{(K)}\rvert = \frac{N!}{(N-K)!} $ \\\hline
|
||||
Ohne Reihenfolge (\textit{Kombinationen}) &
|
||||
$\lvert \widetilde{C}_N^{(K)} \rvert =
|
||||
\binom{N+K-1}{K} $ & $\lvert C_N^{(K)} \rvert
|
||||
= \binom{N}{K} $
|
||||
\end{tabular}
|
||||
\end{table}
|
||||
\end{greenblock}
|
||||
\column{\kitonecolumn}
|
||||
\end{columns}
|
||||
\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{Aufgabe 2: Variationen \& Permutationen}
|
||||
|
||||
Aufgabe 2: Variationen \& Permutationen
|
||||
Ein Burgerrestaurant bietet verschiedene Burger mit den
|
||||
Zutaten Salat
|
||||
Ein Burgerrestaurant bietet verschiedene Burger mit den Zutaten Salat
|
||||
(S), Käse (K), Tomate (T)
|
||||
und Patty (P) an. Diese werden zufällig für die Zubereitung eines
|
||||
Burgers ausgewählt.
|
||||
@@ -460,66 +235,48 @@
|
||||
|
||||
\end{frame}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\section{Zusammenfassung}
|
||||
|
||||
% TODO: Replace slide content with relevant stuff
|
||||
\begin{frame}
|
||||
\frametitle{Aufgabe 2: Variationen \& Permutationen}
|
||||
\frametitle{Zusammenfassung}
|
||||
|
||||
Aufgabe 2: Variationen \& Permutationen
|
||||
Ein Burgerrestaurant bietet verschiedene Burger mit den
|
||||
Zutaten Salat
|
||||
(S), Käse (K), Tomate (T)
|
||||
und Patty (P) an. Diese werden zufällig für die Zubereitung eines
|
||||
Burgers ausgewählt.
|
||||
|
||||
% tex-fmt: off
|
||||
\begin{enumerate}[a{)}]
|
||||
\item Die Ergebnismenge sei $\Omega = \{S, K, T, P\}$. Wie lautet die
|
||||
Potenzmenge $P(\Omega)$?\pause
|
||||
\begin{align*}
|
||||
\mathcal{P}(\Omega) = \{ &\emptyset, \mleft\{ S \mright\}, \mleft\{ K \mright\}, \mleft\{ T \mright\}, \mleft\{ P \mright\},\\
|
||||
&\mleft\{ S, K \mright\}, \mleft\{ S, T \mright\}, \mleft\{ S, P \mright\}, \mleft\{ K, T \mright\}, \mleft\{ K,P \mright\}, \mleft\{ T, P \mright\}, \\
|
||||
&\mleft\{ S, K, T \mright\}, \mleft\{ S, K, P \mright\}, \mleft\{ S, T, P \mright\}, \mleft\{ K, T, P \mright\}, \mleft\{ S, K, T, P \mright\}\}
|
||||
\end{align*}%
|
||||
\item \pause Für einen normalen Burger werden 3 der 4 möglichen Zutaten
|
||||
ausgewählt und in einer
|
||||
bestimmten Reihenfolge auf das Burgerbrötchen gelegt. Wie viele
|
||||
verschiedene normale
|
||||
Burger gibt es?\pause
|
||||
\begin{gather*}
|
||||
\lvert V_N^{(K)} \rvert = \frac{4!}{1!} = 24
|
||||
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{enumerate}
|
||||
% tex-fmt: on
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}
|
||||
\frametitle{Aufgabe 2: Variationen \& Permutationen}
|
||||
\begin{figure}
|
||||
\centering
|
||||
|
||||
Aufgabe 2: Variationen \& Permutationen
|
||||
Ein Burgerrestaurant bietet verschiedene Burger mit den
|
||||
Zutaten Salat
|
||||
(S), Käse (K), Tomate (T)
|
||||
und Patty (P) an. Diese werden zufällig für die Zubereitung eines
|
||||
Burgers ausgewählt.
|
||||
|
||||
% tex-fmt: off
|
||||
\begin{enumerate}[a{)}]
|
||||
\setcounter{enumi}{2}
|
||||
\item Ein Burger ``Spezial'' besteht ebenfalls aus 3 Zutaten. Jedoch
|
||||
können Tomate und Salat
|
||||
doppelt vorkommen. Wie viele verschiedene Burger „Spezial“ gibt es?\pause
|
||||
\begin{align*}
|
||||
n_\text{Burger} &= n_\text{Burger,alle Unterschiedlich} + n_{\text{Burger,2}\times\text{Salat}} + n_{\text{Burger,2}\times\text{Tomate}} \\
|
||||
&= 24 + 3\cdot 3 + 3\cdot 3 = 42
|
||||
\end{align*}
|
||||
\item \pause Der Burger „Jumbo“ enthält die folgende Menge an Zutaten: $\{S, S,
|
||||
T, T, K, K, K, P, P, P\}$
|
||||
die alle verwendet werden. Wie viele mögliche Belegungen des Burgers
|
||||
``Jumbo'' gibt es?\pause
|
||||
\begin{subfigure}[c]{0.5\textwidth}
|
||||
\centering
|
||||
\begin{gather*}
|
||||
\lvert \Pi_N^{L_1,L_2,L_3,L_4} \rvert = \frac{10!}{2!2!3!3!} = 25200
|
||||
\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{enumerate}
|
||||
% tex-fmt: on
|
||||
\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}
|
||||
|
||||
|
||||
@@ -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}
|
||||
@@ -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}
|
||||
|
||||
@@ -1,731 +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/2025-12-19/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}
|
||||
|
||||
File diff suppressed because it is too large
Load Diff
3
src/template/.latexmkrc
Normal file
3
src/template/.latexmkrc
Normal file
@@ -0,0 +1,3 @@
|
||||
$pdflatex="pdflatex -shell-escape -interaction=nonstopmode -synctex=1 %O %S -cd ./../..";
|
||||
$out_dir = "build";
|
||||
$pdf_mode = 1;
|
||||
1
src/template/lib
Symbolic link
1
src/template/lib
Symbolic link
@@ -0,0 +1 @@
|
||||
/home/andreas/Documents/kit/wt-tut/presentations/lib
|
||||
0
src/template/presentation.bib
Normal file
0
src/template/presentation.bib
Normal file
304
src/template/presentation.tex
Normal file
304
src/template/presentation.tex
Normal file
@@ -0,0 +1,304 @@
|
||||
\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}
|
||||
|
||||
1
src/template/src
Symbolic link
1
src/template/src
Symbolic link
@@ -0,0 +1 @@
|
||||
/home/andreas/Documents/kit/wt-tut/presentations/src
|
||||
Reference in New Issue
Block a user