Wahrscheinlichkeitsrechnung 2 - Zettel 1

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Angabe: media:WR2_Zettel1.pdf

Contents


Beispiel 1

Probability theory was used in a famous court case: People vs. Collins (cf. M.W. Gray, “Statistics and theLaw,”, Mathematics Magazine, vol. 56 (1983), pp. 67–81). In this case a purse was snatched from an elderlyperson in a Los Angeles suburb. A couple seen running from the scene were described as a black man with abeard and a mustache and a blond girl with hair in a ponytail. Witnesses said they drove off in a partly yellowcar. Malcolm and Janet Collins were arrested. He was black and though clean shaven when arrested had evidenceof recently having had a beard and a mustache. She was blond and usually wore her hair in a ponytail. Theydrove a partly yellow Lincoln. The prosecution called a professor of mathematics as a witness who suggestedthat a conservative set of probabilities for the characteristics noted by the witnesses would be as shown below:

man with mustache 1/4
girl with blond hair 1/3
girl with ponytail 1/10
black man with beard 1/10
interracial couple in a car 1/1000
partly yellow car 1/10

The prosecution then argued that the probability that all of these characteristics are met by a randomly chosencouple is the product of the probabilities or 1/12000000, which is very small. He claimed this was proof beyonda reasonable doubt that the defendants were guilty. The jury agreed and handed down a verdict of guilty ofsecond-degree robbery.

Hinweis: People vs. Collins . . . Das Volk gegen Collins; purse . . . Handtasche; to snatch . . . entreissen, stehlen;mustache . . . Oberlippenbart; interracial . . . von unterschiedlicher Hautfarbe; defendant . . . Angeklagter; verdict. . . Schuldspruch.

There are two bad mistakes in the argument of the prosecution:

Beispiel 1 (a)

Is the given probability of 1/12000000 likely to be correct?

      Nein, da nicht unabhängig

Beispiel 1 (b)

A more devastating, but more subtle argument can be given as follows: Suppose, for example, there are 5000000 couples in the Los Angeles area and the probability that a randomly chosen couple fits the witnesses' description is 1/12000000. Then the probability that there are two such couples given that there is at least one is not at all small. Find this probability. (The California Suppreme Court overturned the initial guilty verdict.)

      Wenn P(X) gleich der Wahrscheinlichkeit x-Paare in Los Angeles 
      zu finden ist, dann ist X Binomialverteilt mit 
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \sim Binom_{n=5000000,p=\frac{1}{12000000}}
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X) = {n \choose k} p^k (1-p)^{n-k}


      Gesucht ist daher
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X=2|X \ge 1)


      Mit Hilfe des de:Satz von Bayes
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(A|B) = P(A \cap B) P(B)


      kann umgeformt werden.
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X=2|X \ge 1) = \frac{P(X=2 \cap X \ge 1)}{P(X \ge 1)}


      Da Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X = 2)\,
mehr als Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X \ge 1)
einschränkt, kann
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \frac{P(X=2 \cap X \ge 1)}{P(X \ge 1)} = \frac{P(X=2)}{P(X \ge 1)}


      dargestellt werden.
      
      Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X \ge 1) = 1 - P(X=0) = 1 -  {5000000 \choose 0} \frac{1}{12000000}^0 \left( 1-\frac{1}{12000000} \right )^{(5000000-1)}


      in R kann mit dbinom(2, 5000000, 1/12000000)/(1-dbinom(0, 5000000, 1/12000000)) 
      die Wahrscheinlichkeit ausgerechnet (ca. 17%).


Kontinuierliche Verteilungen

Exponentialverteilung

Beispiel 2.1.2 - Exponential Verteilung

Zeigen Sie, daß für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda > 0\,

die Funktion

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f(x) = \begin{cases} \lambda e^{-\lambda x}, & \mbox{falls } x \ge 0 \\ 0, & \mbox{falls } x < 0 \end{cases}


eine Dichte ist.

  Allgemein muss Folgendes gezeigt werden
  
  1. Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f(x) \ge 0
  2. stückweise stetig (d.h. keine Sprungstellen)
  3. Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \int_{-\infty}^{\infty} f(x) \, dx = 1


  Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \begin{matrix}\displaystyle \int_{-\infty}^{\infty} \lambda e^{-\lambda x} \, dx = & \underbrace{\int_{-\infty}^{0} 0 \, dx} & + \displaystyle \int_{0}^{\infty} \lambda  e^{-\lambda x} dx \\ & 0 \end{matrix}


  Der erste Teil fällt laut Angabe weg.
  
  Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):        \begin{matrix}         \displaystyle \int_{0}^{\infty} \lambda e^{-\lambda x} =        \lambda \frac{e^{-\lambda x}}{- \lambda} \Big |_{0}^{\infty} =         -e^{-\lambda x} \Big |_{0}^{\infty} =         & \underbrace{\lim_{x \to \infty} -e^{-\lambda x}} & - & \underbrace{(-e^{-\lambda \cdot 0})} \\        & x \to 0 & & 1       \end{matrix}    


Beispiel 2.1.3

Überprüfen Sie die Eigenschaft der "Gedächnislosigkeit" der Exponentatialverteilung; Ist Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \sim Exp(\lambda) \mbox{ fur } \lambda > 0\, , so ist Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X > z + x|X > z) = P(X > x)\, .

  Nach dem de:Satz von Bayes wird transformiert
  
  Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \frac{P(X > z+x) \cap P(X > z)}{P(X>z)}


  Betrachtet man die Dichtefunktion der de:Exponentialverteilung, so sieht man das Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X > z+x) < P(X > x)\,
ist. Daher
  
  Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \frac{P(X > z+x)}{P(X>z)} = \frac{1-P(X \le z+x)}{1-P(X < z)}


  Weil die Exponentialverteilung eine Dichte besitzt, gibt es keine Punktwahrscheinlichkeiten und somit 
  kann Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): <\,
mit Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \le\,
vertauscht werden. Jetzt sind wir in der Lage die obigen Ausdrücke durch die 
  Verteilungsfunktion Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): F(x) = 1 - e^{-\lambda x}\,
zu ersetzen.
  
  Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \frac{1-(1-e^{-\lambda z+x})}{1-(1-e^{-\lambda z})} = \frac{e^{-\lambda z+x}}{e^{-\lambda z}} = e^{-\lambda(z+x) + \lambda z} = e ^ {-\lambda x}


Beispiel 2.1.4

Suppose that the time (in hours) required to repair a car is an exponentially distributed random

Gamma-Verteilung

Für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha \ge 0 \,

ist die so genannte Gamma-Funktion gegeben durch 

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x} \, dx .

siehe auch de:Gamma-Verteilung

Beispiel 2.2.5

Leiten Sie die folgenden Eigenschaften der Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma -Funktion her:

  • Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(1) = 1 \,

, und Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(\alpha + 1) = \alpha \Gamma(\alpha) \,

für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha \ge 0 \,

.

  • In welcher Beziehung stehen Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(n) \,
und Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): n!\,
für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): n \in \mathbb(N) \,

?

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(1) = \int_0^\infty x^{1-1} e^{-x} \, dx = \int_0^\infty 1 e^{-x} = \lim{x \to \infty} e^{-x} - e^0 = 1 


Mit Hilfe partieller Integration Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \int f \cdot g = F \cdot g - \int F \cdot g'
kann auch die zweite Eigenschaft gezeigt werden.
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  f = e^{-x} \,
und Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  g = x^\alpha \,


Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \Gamma(\alpha + 1) = \int_0^\infty x^\alpha e^{-x} \, dx = 
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  = -e^{-x} x^\alpha \Big |_0^\infty - \int_0^\infty \alpha x^{\alpha-1} - e^{-x} \, dx
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \begin{matrix} = & \underbrace{-e^{-x} x^\alpha \Big |_0^\infty} & + \alpha & \underbrace{\int_0^\infty x^{\alpha-1} e^{-x} \, dx} \\                          & 0 & & \Gamma(\alpha) \end{matrix} 


Das der erste Ausdruck gegen 0 geht, muss man noch mit Hilfe der de:Regel_von_L'Hospital überprüfen.
Kurze Zusammenfassung von L'Hospital. Wenn Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lim_{x \to b} \frac{f'(x)}{g'(x)} \to c
dann gilt auch Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lim_{x \to b} \frac{f(x)}{g(x)} \to c


Leitet man nun Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  -e^{-x} x^\alpha \,
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha \,

-mal ab ergibt sich:

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lim_{x \to \infty} \frac{\alpha^\alpha}{(-1)^{\alpha} e^x}


Hier ist nun leicht ersichtlich das der Ausdruck gegen 0 geht.
Beziehung von Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(n) \,
und Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): n!\,

.

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(1) = 1\,
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(2) = 1 \cdot \Gamma(1) \,
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Gamma(3) = 2 \cdot \Gamma(2) = 2 \cdot 1 \cdot \Gamma(1) = \Gamma(2+1)\,
...

D.h. die Gamma-Funktion is die kontinuierliche Verallgemeinerung der Fakultät.

Beispiel 2.2.6

Zeigen Sie, daß für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda>0, \alpha>0 \,

die Funktion

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f(x) = \begin{cases}0 & \mathrm{falls} \, x <0, \\ \displaystyle \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x} & \mathrm{falls} \, x \ge 0 \end{cases}


eine Dichte ist. Die durch diese Dichte beschriebene Verteilung heißt Gamma-Verteilung mit Parametern Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda\,

und Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha \,

, kurz Gamma(Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda, \alpha ).

1. Überprüfe ob Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f(x) \ge 0 \,

. Das stimmt einmal.

2. Überprüfe ob Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \int_{-\infty}^\infty f(x) = 1 \,

.

Für Dichten heisst das:

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \int_0^\infty \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x} = 1 


Mit Hilfe von Substitution kann das Integral transformiert werden:

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  x = \frac{v}{\lambda} \,


Allgemein: Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \int f(y) \, dy = \int f(g(x)) \cdot g'(x) \, dx

. Man sollte dabei nicht auf die innere Ableitung vergessen.

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \int_0^\infty \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x} =          \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty \left( \frac{v}{\lambda} \right)^{\alpha-1} e^{-\lambda \frac{v}{\lambda}} \frac{1}{\lambda} \, dv = 


Wenn man zusammenfasst ergibt sich

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): = \frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty \frac{v^{\alpha-1}}{\lambda^\alpha} e^{-v} \, dv = 
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \begin{matrix} = \displaystyle \frac{\lambda^\alpha}{\Gamma(\alpha) \lambda^\alpha} & \underbrace{\int_0^\infty v^{\alpha-1} e^{-v} \, dv}  & = \displaystyle\frac{\Gamma(\alpha)}{\Gamma(\alpha)} = 1 \\      & \Gamma(\alpha) &        \end{matrix}  


Beispiel 2.2.7

Prüfen Sie mit Hilfe der entsprechenden Dichtefunktionen, daß gilt:

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \mathrm{Gamma}(\lambda, 1) \equiv \mathrm{Exp}(\lambda) \,


D.h. also, daß die Gamma-Verteilung mit Parametern Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda

und 1 gleich der Exponential-Verteilung mit 

Parameter Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \lambda

ist.
In die Dichte der Gamma-Verteilung eingesetzt und der Dichte der Exponential-Verteilung gleichgesetzt.
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):   \begin{matrix}    \displaystyle\frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x} & = & \lambda e^{-\lambda x}  \end{matrix}  


Setzen wir nun Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha = 1\,
er gibt sich
 
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):   \begin{matrix}    \displaystyle\frac{\lambda^1}{\Gamma(\alpha)} x^{1-1} e^{-\lambda x} & = & \lambda e^{-\lambda x} \\    \lambda e^{-\lambda x} & = & \lambda e^{-\lambda x}  \end{matrix}  


Normalverteilung

Beispiel 2.3.8

Sei Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \sim N(\mu, \sigma^2) \,

mit Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \sigma^2 > 0 \,

. Welcher Verteilung gehorcht Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha X + \beta (\alpha, \beta \in \mathbb{R} \, )?

(Hinweis: Bestimmen Sie mit Hilfe des Tranformationssatzes die Dichte von Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha X + \beta \, .)

Transformationssatz: Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f_Y(y) = f_X(g^-1(y)) \cdot \left| \frac{d}{dy} g^{-1}(y) \right| 
In Worten: Wenn die Zufallsvariable X mit einer Funktion Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): g(x)\,
transformiert werden soll (d.h. Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): Y = g(X)

)

muss die Inverse der Funktion in die Dichte Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f_X(x)\,
eingesetzt werden und mit der Ableitung der Inversen multipliziert werden.

Die Dichte einer normalverteilten Zufallsvariable X ist gegeben durch

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): f_X(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{\displaystyle -\frac{(x - \mu)^2}{2 \sigma^2}}


Die Transformation is gegeben durch

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  Y = g(X) = \alpha X + \beta \,


Um den Transformationssatz anzuwenden muss die Inverse von Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): g(x)\,


Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): g^{-1}(x) = \frac{x - \beta}{\alpha}


gebildet werden.

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):    \begin{matrix}     f_Y(x) & = & \displaystyle \frac{1}{\sigma \sqrt{2 \pi}} e^{\displaystyle -\frac{( \displaystyle\frac{x - \beta}{\alpha} - \mu)^2}{2 \sigma^2}} \left| \displaystyle \frac{1}{\alpha} \right| \\            & = & \displaystyle \frac{1}{|\alpha| \sigma \sqrt{2 \pi}}                   e^{\displaystyle -\frac{1}{2} \left( \displaystyle \frac{\frac{x-\beta-\mu \alpha}{\alpha}}{\sigma} \right)^2} \\             & = & \displaystyle \frac{1}{|\alpha| \sigma \sqrt{2 \pi}}                   e^{\displaystyle -\frac{1}{2} \left( \displaystyle \frac{x-\beta-\mu \alpha}{\alpha \sigma} \right)^2} \\            & = & \displaystyle \frac{1}{|\alpha| \sigma \sqrt{2 \pi}}                   e^{\displaystyle -\frac{1}{2} \left( \displaystyle \frac{x-(\mu \alpha + \beta)}{\alpha \sigma} \right)^2}   \end{matrix}  


Nach dieser Umformung sollte schön ersichtlich sein das Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  \alpha X + \beta \sim N(\mu \alpha + \beta , \sigma^2 \alpha^2) \,
für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \alpha \ne 0
ist.

Beispiel 2.3.9

Es sei Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \sim N(0,1) \, , und es sei Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Phi \,

die dazugehörige Verteilungsfunktion.

Zeigen Sie, daß Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(|X| > c) = 2 \Phi(-c) \,

für Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): c \ge 0 \,
ist.
Hauptbegründung: Symmetrie.

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):    \begin{matrix}     P(|X| > c) & = & P(X > c) + P(X < -c) \\                & = & 2 \Phi(-c)   \end{matrix}  


Beispiel 2.3.10

Es sei Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \sim N(75,100) \, . Bestimmen Sie Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X \le 60) \, , Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(70 < X \le 100) \, , sowie c, sodaß Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(-c \le \frac{X-75}{10} \le c) = 0.95 .

Wie in Beispiel 2.3.8 kann eine Normalverteilung sehr leicht transformiert werden. Da die Standard Normal-Verteilung tabelliert ist
möchten wir Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X \,
transformieren, sodaß Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):  g(X) = Y \sim N(0,1) \,

.

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \frac{X - 75}{10} \sim N(\frac{75}{10}-7.5, 100 \cdot \frac{1}{10^2}) = N(0,1)


Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(X \le \frac{60-75}{10}) = 0.0668


D.h. man schaut -1.5 in der Tabelle der Standard Normal-Veteilung nach.
Genauso wie eben

Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(70 < X \le 100) = P(X \le 100) - P(X \le 70) = 0.685 


Da Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): X\,
in Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P(-c \le \frac{X-75}{10} \le c) = 0.95 
bereits standardisiert ist kann aus der Tabelle
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \Phi^{-1}(0.95) = 1.96 \,
raus gesucht werden.