Begründe, dass gilt: $\mu =20\backslash mu\; =\; 20$

Aus der Abbildung kann abgelesen werden:

$P(\le 20)=\mathrm{0,5}P(\backslash leq20)=0\{,\}5$

Andererseits gilt: $P(X\le x)\approx \mathrm{\Phi }\left({\displaystyle \frac{x-\mu }{\sigma }}\right)P(X\backslash leq\; x)\backslash approx\backslash Phi\backslash left(\backslash dfrac\{x-\backslash mu\}\{\backslash sigma\}\backslash right)$

Also ist hier: $\mathrm{\Phi }\left({\displaystyle \frac{20-\mu }{\sigma }}\right)=\mathrm{0,5}\backslash Phi\backslash left(\backslash dfrac\{20-\backslash mu\}\{\backslash sigma\}\backslash right)=0\{,\}5$

Es gilt weiterhin:

$\mathrm{\Phi }(0)=\mathrm{0,5}\Rightarrow {\displaystyle \frac{20-\mu }{\sigma }}=0\Rightarrow \mu =20\backslash Phi(0)=0\{,\}5\backslash ;\backslash Rightarrow\backslash ;\backslash dfrac\{20-\backslash mu\}\{\backslash sigma\}=0\backslash ;\backslash Rightarrow\backslash ;\backslash mu=20$

Bestimme einen Näherungswert für $P(\mu -\mathrm{1,5}\sigma \le X\le \mu +\mathrm{1,5}\sigma )P(\backslash mu\; -\; 1\{,\}5\backslash sigma\; \backslash leq\; X\; \backslash leq\; \backslash mu\; +\; 1\{,\}5\backslash sigma)$

Benötigt wird der Wert von $\sigma \backslash sigma$.

Gegeben ist der Punkt $S(16\mathrm{\mid }\mathrm{0,16})S(16|0\{,\}16)$ und aus Aufgabe a) $\mu =20\backslash mu=20$.

Demnach gilt: $P(X\le 16)=\mathrm{0,16}P(X\backslash leq\; 16)=0\{,\}16$.

Also ist $\mathrm{\Phi }\left({\displaystyle \frac{16-20}{\sigma }}\right)=\mathrm{0,16}\backslash Phi\backslash left(\backslash dfrac\{16-20\}\{\backslash sigma\}\backslash right)=0\{,\}16$.

Das Argument von $\mathrm{\Phi }\backslash Phi$ ist negativ:

$\mathrm{\Phi }(-z)=1-\mathrm{\Phi }(z)=\mathrm{0,16}\Rightarrow \mathrm{\Phi }(z)=\mathrm{0,84}\Rightarrow z=1\backslash Phi(-z)=1-\backslash Phi(z)=0\{,\}16\backslash ;\backslash Rightarrow\backslash ;\backslash Phi(z)=0\{,\}84\backslash ;\backslash Rightarrow\backslash ;z=1$

Mit $z={\displaystyle \frac{4}{\sigma }}z=\backslash dfrac\{4\}\{\backslash sigma\}$ und $z=1\Rightarrow \sigma =4z=1\; \backslash ;\backslash Rightarrow\backslash ;\backslash sigma\; =4$

Setze nun $\mu =20\backslash mu=20$ und $\sigma =4\backslash sigma=4$ in $P(\mu -\mathrm{1,5}\sigma \le X\le \mu +\mathrm{1,5}\sigma )P(\backslash mu\; -\; 1\{,\}5\backslash sigma\; \backslash leq\; X\backslash leq\; \backslash mu\; +\; 1\{,\}5\backslash sigma)$ ein:

$P(20-\mathrm{1,5}\cdot 4\le X\le 20+\mathrm{1,5}\cdot 4)=P(14\le X\le 26)P(20\; -\; 1\{,\}5\backslash cdot\; 4\; \backslash leq\; X\; \backslash leq\; 20\; +\; 1\{,\}5\backslash cdot\; 4)=P(14\backslash leq\; X\backslash leq\; 26)$

Aus der Abbildung liest man ab:

$P(14\le X\le 26)=P(X\le 26)-P(X\le 14)\approx \mathrm{0,93}-\mathrm{0,07}=\mathrm{0,86}P(14\backslash leq\; X\backslash leq\; 26)=P(X\backslash leq\; 26)-P(X\backslash leq\; 14)\backslash approx0\{,\}93-0\{,\}07=0\{,\}86$