Fix spacing

src/2025-12-19/presentation.tex

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