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Replace Z with X for the standart normal distribution

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This commit is contained in:
Andreas Tsouchlos 2025-12-17 15:16:56 +01:00
parent 7bea062e6a
commit b815a88361
1 changed files with 32 additions and 27 deletions
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src/2025-12-19/presentation.tex
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@ -168,7 +168,7 @@
\begin{frame}
\frametitle{Zusammenfassung}
\begin{columns}[c]
\begin{columns}[t]
\column{\kitthreecolumns}
\centering
\begin{greenblock}{Verteilungsfunktion (kontinuierlich)}
@ -463,7 +463,6 @@
\end{columns}
\end{frame}
% TODO: Are Z/z notation used in the lecture?
\begin{frame}
\frametitle{Rechnen mithilfe der Standardnormalverteilung}
@ -476,16 +475,16 @@
\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)
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
$z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ \\
$x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ \\
\hline
\hline
0{,}00 & 0{,}500000 & 0{,}10 & 0{,}539828 & 0{,}20 & 0{,}579260 \\
@ -512,7 +511,7 @@
\end{gather*}
\end{itemize}
\vspace*{5mm}
\vspace*{3mm}
\pause
\begin{lightgrayhighlightbox}
@ -521,13 +520,13 @@
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
= 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}
% TODO: Are Z/z notation used in the lecture?
\begin{frame}
\frametitle{Zusammenfassung}
@ -539,9 +538,9 @@
\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)
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}
@ -562,14 +561,20 @@
% \cdots
\begin{tabular}{|c|c||c|c||c|c||c|c|}
\hline
$z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ & $z$ & $\Phi(z)$ \\
$x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$ & $\Phi(x)$ & $x$
& $\Phi(x)$ \\
\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 \\
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
@ -631,8 +636,8 @@
P(E_\text{a}) &= P \Big( \big( X < S(1-\delta) \big)
\cup \big( X > S(1 + \delta) \big) \Big) \\
&= P(X < S(1 - \delta)) + P(X > S(1 + \delta)) \\[2mm]
&= P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+ P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
&\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*}
@ -687,10 +692,10 @@
\pause\column{\kitthreecolumns}
\centering
\begin{align*}
P(E_\text{b}) &\overset{\text{a)}}{=} P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma'}\right)
+ P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma'}\right) \\[2mm]
&= P\left(Z < -\frac{0{,}2}{\sigma'}\right)
+ P\left(Z > \frac{0{,}2}{\sigma'}\right) \\[2mm]
P(E_\text{b}) &\overset{\text{a)}}{=} P\left(\widetilde{X} < \frac{S(1 - \delta) - \mu}{\sigma'}\right)
+ P\left(\widetilde{X} > \frac{S(1 + \delta) - \mu}{\sigma'}\right) \\[2mm]
&= P\left(\widetilde{X} < -\frac{0{,}2}{\sigma'}\right)
+ P\left(\widetilde{X} > \frac{0{,}2}{\sigma'}\right) \\[2mm]
&= \Phi\left(-\frac{0{,}2}{\sigma'}\right)
+ \left(1 - \Phi\left(\frac{0{,}2}{\sigma'} \right)\right) \\[2mm]
&= 2 - 2 \Phi\left(\frac{0{,}2}{\sigma'} \right)
@ -708,8 +713,8 @@
Mittelwert $\mu$ auf $5{,}1$ Volt ($\sigma$ ist $0{,}07$ Volt).
Wie groß ist jetzt der Prozentsatz, der aussortiert wird?
\pause \begin{align*}
P(E_\text{c}) &\overset{\text{a)}}{=} P\left(Z < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+ P\left(Z > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
P(E_\text{c}) &\overset{\text{a)}}{=} P\left(\widetilde{X} < \frac{S(1 - \delta) - \mu}{\sigma}\right)
+ P\left(\widetilde{X} > \frac{S(1 + \delta) - \mu}{\sigma}\right) \\[2mm]
&\approx \Phi(-4{,}29) + (1 - \Phi(1{,}43)) \\
& = 2 - \Phi(4{,}29) - \Phi(1{,}43) \approx 7{,}78 \text{\%}
\end{align*}