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\newcommand{\Ch}{Lois Continues}	
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\ddwicky{album.eps}{Tree Outside The Academy}{Thurston Moore}{Télécharger c'est tuer l'industrie, tuons les tous}{Thurston Moore (Sonic Youth)}{Thurston Moore est un musicien américain né le 25 juillet 1958 dans l'État de Floride. Membre du groupe de rock Sonic Youth au sein duquel il chante et joue de la guitare.\\
Les artistes avec qui Thurston Moore travailla tout au long de sa carrière sont nombreux et assez éclectiques pour constituer un réseau aussi puissant que rare dans l'histoire de la musique actuelle. \\
Avec Lee Ranaldo, il est connu pour l'usage de nombreuses sortes d'accordages de guitare. C'est également un adepte de la guitare préparée. En effet, il joue de la guitare avec des baguettes de batterie, avec un marteau, avec une lime de fer, avec un klaxon de vélo ou encore avec une petite barre de fer. Thurston Moore étonne et surprends par la qualité des enregistrements des disques de Sonic Youth ou de ce disque solo, artiste sur lequel le temps semble n'avoir aucune emprise. La candeur ardente et entousiasmante de Moore souffle un vent de jeunesse d'une fraicheur absolue.  Moore y va à l’économie question saturations, mais elles reviennent parfois au pas de charge, rappelant avec stridence mais sans insister les racines brutistes de tout le "Rock" engendré par Moore et ses collègues de Sonic Youth depuis vingt ans.
Jouissif.
}{0.3}{0.4}{\item Utilisation de la loi uniforme.
\item Utilisation de la loi exponentiel de paramètre $\lambda$ pour calculer une probabilité, à savoir :
$$ p\left( X\in [a;b]\right)=\int_a^b e^{-\lambda t} dt$$}

\setcounter{NumLecon}{12}
\TITRE{chap.eps}{0.45}


\begin{abstract}
Vous connaissez déjà des nombres réels particuliers : les
entiers, les rationnels, les irrationnels. D'autres catégories
(pouvant recouper les ensembles habituels) sont utilisées. Ainsi,
les nombres racines de polynômes à coefficients entiers sont
appelés \textsl{nombres algébriques} : c'est le cas de 1 (solution
de $x-1=0$), de 1/3 (solution de $3x-1=0$), de $\sqrt{2}$
(solution de $x^2-2=0$), etc. On connaît des nombres qui ne sont
pas algébriques, on les appelle les \textsl{nombres
transcendants}. C'est le cas par exemple de $\pi$ et de $e$. Au
premier abord, il semble qu'il y ait beaucoup plus de nombres
algébriques que de nombres transcendants (vous n'en connaissez que
deux !). Un jour de pur délire, on pourrait même envisager qu'il y
ait autant de nombres de chaque catégorie. Or, si on prend un
nombre au hasard dans l'intervalle $[0,1]$, vous calculerez
peut-être un jour que la probabilité d'obtenir un nombre
transcendant vaut ....1!!!! En effet, la \og~mesure~\fg de
l'ensemble des nombres algébriques est nulle. Pourtant, demandez à
un ordinateur de vous donner un milliard de nombres au hasard
entre 0 et 1, l'écran ne vous affichera que des nombres
algébriques (décimaux même !).  L'ensemble $\R$ recèle bien d'autres résultats
étonnants. 
\\
Daniel Bernoulli fut un pionnier en la matière.
\\
Vous constaterez dans ce chapitre que ce chapitre mobilisera vos connaissances acquises dans le chapire sur l'intégration, 
ce qui sera un bon moyen pour vous de réviser si vous n'êtes pas encore au point.\\\\
Les probabilités continues permettent de modéliser des phénomènes extrêmement différents.
Elles apparaissent au même moment que les probabilités modernes, autour de mathématiciens 
comme Abraham De Moivre (1667-1754). Mais c'et Gauss (1777-1855) qui donne son nom à la courbe liée à la loi normale
et qui la relie en particulier au problème des observations astronomiques.\\
Les lois continues se sont multipliées depuis (loi exponentielle, loi de Weibull, loi de Pareto ...),
ouvrant un champs immense d'applications et de recherches ...
\end{abstract}

\pagebreak
\section{Loi à densité sur un intervalle}
\subsection{Du discret au continu}

\begin{exoi}
Soit $f$ la fonction définie sur $\R$ par $f(x)=\dfrac12 e^{\frac{x}2}$ de courbe représentative $\mathscr C_f$.\\
 L'expérience consiste à choisir   un point au hasard dans le domaine $\mathscr D$ grisé ci-dessous, délimité par la courbe $\mathscr C_f$,
 l'axe des abscisses et les droites d'équations $x=0$ et $x=3$.\\
 On cherche à la probabilité que le point choisi appartienne à une certaine zone.\\
 %\textit{On admet que les réels sont uniformément répartis, donc que les issues sont équiprobables. }\\
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\begin{enumerate}
 \item \textbf{Premier constat} :\\
  Est-il plus probable que le point choisi ait une abscisse entre $0$ et $1$ ou entre $2$ et $3$ ?
\item \textbf{Calculs de certaines probabilités}
\begin{enumerate}
 \item Quelle est l'aire du domaine $\mathscr D$ ?
 \item Quelle est la probabilité que le point appartienne à la zone $\mathscr Z$ délimitée par la courbe
 $\mathscr C_f$, l'axe des abscisses et les droites d'équation $x=0$ et $x=1$ ?
 \item $a$ et $b$ sont deux nombres réels tels que $0\leq a \leq b\leq 3$, même question avec les droites d'équation $x=a$ et $x=b$.
 %\item Valider alors votre constat graphique de la question 1.
 \item Qu'obtient-on si $a=b$ ? Qu'en pensez-vous ?
 \item Déduire la probabilité de choisir le point $A$.
\end{enumerate}
\item \textbf{Tentative de Modélisation} 
\begin{enumerate}
 \item Quel est l'univers de cette expérience ?
  \item En quoi se distingue-t-il des univers déjà rencontrés ?
 %\item Rappeler ce que signifie \og définir une loi de probabilité \fg lorsque l'univers est fini.
 %\item Est-ce généralisable au cas où l'univers est infini ?
\end{enumerate}
\end{enumerate}
\end{exoi}


\begin{defi}
Lorsqu'une variable aléatoire peut prendre toutes ses valeurs dans un 
ensemble finis ou dénombrable (comme $\N$), on dit que sa loi est \textbf{discrète}.\\
Lorsqu'une variable aléatoire peut prendre toutes les valeurs d’un intervalle
de $\R$ (borné ou non), on dit que sa loi est \textbf{continue}.
\end{defi}

\Expl{
Le temps d’attente d’un bus, la durée de vie d’un appareil, la distance d’un point
d’impact au centre d’une cible admettent des lois continues.
}

\subsection{Densité et loi de probabilité sur un intervalle I}

\begin{defi}
Soit $I$ un intervalle. \\
On appelle densité de probabilité sur $I$ toute fonction $f$ continue et positive sur $I$ telle que :
$$ \int_I f(x)dx=1$$
\end{defi}

\Rq{
\begin{itemize}
\item Le nombre $\int_I f(x) dx$ est défini par :  $\int_a^b f(x)dx$ si $I=[a;b]$.
\item Si $I$ est un intervalle non borné, par exemple  $[a, +\infty[$, alors la quantité notée $\int_I f(t) dt$ désigne, lorsqu'elle
existe, la limite suivante :
$$ \int_I f(t) dt= \lim\limits_{x\to +\infty} \int_a^x f(t) dt$$
De la même manière si $I=]-\infty;a]$ alors la quantité notée $\int_I f(t) dt$ désigne, lorsqu'elle
existe, la limite suivante :
$$ \int_I f(t) dt= \lim\limits_{x\to -\infty} \int_x^a f(t) dt$$
\item Enfin si $I=\R$ alors la quantité notée $\int_I f(t) dt$ désigne, lorsqu'elle
existe, la somme des limites suivante :
$$ \int_I f(t) dt= \lim\limits_{x\to -\infty} \int_x^0 f(t) dt+\lim\limits_{x\to +\infty} \int_0^x f(t) dt$$
\end{itemize}}

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\Expl{La fonction $f$ définie sur $[1;e]$ par $f(x)=\dfrac{1}{x}$ est une densité sur $[1;e]$, en effet, $f$ en plus d'être continue et positive vérifie :
$$ \int_1^e \dfrac{1}{x}dx=[\ln x ]_1^e=\ln e-\ln 1=1$$}
\pagebreak
\begin{exoi}
\begin{enumerate}
\item Déterminer le nombre réel $\alpha$ tel que la fonction $f$ définit par $f(x)=x+\alpha$ soit une densité de probabilité sur $[0;1]$.
\item Soit $f$ une fonction constante sur un intervalle $[a, b]$ (avec $a < b$). Quelle doit être la valeur de la constante $\gamma$
pour qu'elle soit une densité  sur $[a;b]$ ?
\item Soit $\lambda$ un réel strictement positif. Démontrer que la fonction $f$ définie sur $\R^+$ par $ f(t)=\lambda e^{-\lambda t}$ est une  densité
de probabilité sur $\R^+$.
\end{enumerate}
\end{exoi}

\Sol{\begin{enumerate}
\item On doit avoir $\int_0^1 x+\alpha dx = 1\Longleftrightarrow \left[\dfrac{x^2}{2}+\alpha x\right]_0^1=1\Longleftrightarrow \dfrac{1}{2}+\alpha=1\Longleftrightarrow \alpha=\dfrac{1}{2}$
\item On doit avoir $\int_a^b \gamma dx = 1\Longleftrightarrow [\gamma x]_a^b=1\Longleftrightarrow \gamma (b-a)=1\Longleftrightarrow \gamma =\dfrac{1}{b-a}$
\item Calculons dans un premier temps : $\int_0^x f(t)dt$.
$$ \int_0^x \lambda e^{-\lambda t} dt=\left[-e^{-\lambda t}\right]_0^x=-e^{-\lambda x}+e^0=1-e^{-\lambda x}$$
Finalement :
$$ \lim\limits_{x\to +\infty} 1-e^{-\lambda x}=1$$
Ainsi on vient de montrer que :
$$ \int_I f(t) dt=1$$
ce qui implique que $f$ est une densité de probabilité.
\end{enumerate}}

\subsection{Loi de probabilité}

\begin{propo}
Soit $I$ un intervalle et $f$ une densité de probabilité sur $I$.
L'application $P$ qui, à tout sous-intervalle $[a, b]$ de $I$ associe la quantité :
$$ P([a;b])=\int_a^b f(t) dt$$ est appelée loi de probabilité sur $I$.
 \end{propo}


\underline{Illustration} : 
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\end{center} 


\Rqs{
\item $P([a;b])$ n'est rien d'autre que l'aire du domaine représenté ci-dessus. Notons que, puisque $f$ est une densité de probabilité sur $I$, on a :
$ P(I)=1$ et $0\leq P([a;b])\leq 1$. Quand on modélisera une expérience aléatoire $I$ représentera l'univers de notre expérience.
\item Un résultat étonnant : pour tout réel $x_0$ de $I$ on a :
$$ P(\{x_0\})=\int_{x_0}^{x_0}f(t)dt=0$$
i.e que l'éventualité $x_0$ est un résultat \textbf{possible} de l'expérience de probabilité \textbf{nulle}, c'est ici une différence fondamentale avec le cas discret. Prenons un exemple, si on choisit un nombre au hasard entre $0$ et $1$, la probabilité que l'on choisisse par exemple $\dfrac{1}{3}$ est nulle même si l'événement peut avoir lieu. On dit alors que $\{x_0\}$ est un événement \og quasi-impossible \fg.\\
Toujours dans cette même expérience la probabilité de choisir un nombre inférieur à $\dfrac{1}{2}$ semble naturellement égale à $\dfrac{1}{2}$ i.e 
$$ P([0;0.5])=\int_0^{0.5} f(t) dt = \dfrac{1}{2}\qquad \text{ où $f$ reste à déterminer.} $$
Nous allons dans le paragraphe suivant déterminer qu'elle est la densité de probabilité la plus à même de modéliser une telle expérience.
\item Ainsi
$$ p\big(]a;b[\big)=p\big([a;b[\big)=p\big(]a;b]\big)=p\big([a;b]\big)$$
\item Grâce aux propriétés de l'intégrale, on retrouve les trois règles de définitions d'une probabilité :
\begin{enumerate}
 \item $P\in [0;1]$
 \item $P(\Omega)=1$
 \item $P(A \cup B)=P(A)+P(B)$ avec $A\cap B=\emptyset$
\end{enumerate}
}

\Expl{La fonction $\begin{array}{ccccl}
                      f &: &[-1;7] &\longrightarrow &\R \\
                        &  & x &\longmapsto & \dfrac18
                     \end{array}$ 
est une densité de probabilité sur $[-1;7]$\\
Auquel cas il suit que  $P(]2,7])=\int_{2}^{7}\dfrac18 dt=\dfrac58$}

\subsection{Variable aléatoire continues}
Sont dites continues les variables aléatoires qui prennent leurs valeurs dans un intervalle $I$.

\begin{defi}Soit une loi de probabilité $P$ sur $I$ un intervalle de densité $f$.\\
On dit qu'une variable aléatoire $X$ à valeurs dans $I$ suit la loi de probabilité $P$ si :
$$ P(a\leq X\leq b)=\int_a^b f(t) dt \qquad \forall a, b \in I$$\end{defi}

\begin{exoi}
\begin{enumerate}
\item Montrer que $f$ est la densité d'une loi de probabilité $P$ avec : 
$$ I=[0;1]\qquad \text{et} f(x)=2x$$
\item Soit $X$ une variable aléatoire à valeurs dans $I$ qui suit la loi de probabilité $P$, déterminer $P\left(\dfrac{1}{4}\leq X\leq \dfrac{3}{4}\right)$
\item Le résultat précédent permet-il d'envisager que $P$ modélise le choix d'un nombre au hasard entre $[0;1]$ ?
\item Déterminer $ P\left(\dfrac{1}{2}\leq X\leq \dfrac{3}{4}\right)$.
\item Que penser du résultat précédent ?
\end{enumerate}\end{exoi}

\Sol{
\begin{enumerate}
\item On a :
$$ \int_0^1 2x dx= \left[x^2\right]_0^1=1-0=1$$
ce qui montre que $f$ est la densité d'une loi de probabilité $P$.
\item On  a :
$$ P\left(\dfrac{1}{4}\leq X\leq \dfrac{3}{4}\right)=\int_{\frac{1}{4}}^{\frac{3}{4}} 2xdx=\left[x^2\right]_{\frac{1}{4}}^{\frac{3}{4}}=\dfrac{9}{16}-\dfrac{1}{16}=\dfrac{1}{2}$$
\item oui.
\item $$ P\left(\dfrac{1}{2}\leq X\leq \dfrac{3}{4}\right)=\int_{\frac{1}{2}}^{\frac{3}{4}} 2xdx=\left[x^2\right]_{\frac{1}{2}}^{\frac{3}{4}}=\dfrac{9}{16}-\dfrac{1}{4}=\dfrac{5}{16}$$
\item Cette loi de probabilité ne modélise pas l'expérience consistant à choisir un nombre au hasard entre $0$ et $1$, en effet le résultat attendu à la question précédente est $\dfrac{1}{4}$.
\end{enumerate}}

\begin{defi}
 L'espérance et la variance d'une variable aléatoire $X$ suivant une loi de probabilité 
 de densité $f$ sur $I$ sont définies par :
 $$E(X)=\int_a^b tf(t)dt \qquad \qquad \text{et} \qquad \qquad V(X)=E\Big(\big(X-E(x)\big)^2\Big)=E\left(X^2\right)-E^2(X)$$
\end{defi}


\begin{exoi}
Un entrepôt accueille tous les matins des camions de livraison sur un créneau de deux heures d'ouverture,
de 7h30 à 9h30. On s'intéresse à l'heure d'arrivée $T$ d'un camion qui se présente tous les matins à l'entrepôt aux
heures d'ouverture. \\
On admet que $T$ suit la loi de probabilité de densité $f : x \longmapsto |x-8.5|$ sur l'intervalle $I=[7.5;9.5]$.
\begin{itemize}
 \item Tracer la représentation graphique de la fonction $f$ sur l'intervalle $I$.
 \item Vérifier que la fonction $f$ est une densité de probabilité sur $[7.5;9.5]$.
  \item Calculer la probabilité d'arrivée du camion entre 9h00 et 9h30.
  \item En déduire la probabilité que le camion arrive avant 9h00.
 \item Calculer la probabilité d'arrivée du camion entre 8h00 et 9h00.
 \item Calculer l'espérance de la variable $T$.
 \end{itemize}
\end{exoi}

\newpage
\section{Deux exemples de lois continues}
\subsection{La loi uniforme sur $[a;b]$}

\begin{exoi}
  Soit $f$ une fonction constante sur un intervalle $I=[a;b]$ . 
 \begin{enumerate}
  \item Quelle doit être sa valeur pour que $f$ soit une densité sur cet intervalle ? 
 \item Soit $X$ une variable aléatoire qui suit la loi de densité $f$ sur $[a,b]$. 
Calculer $P (0 \leq X \leq 0.25 )$ puis $P (c \leq X \leq d )$ pour tout $c\leq d$ appartenant à $I$.
\item Calculer l'espérance de $X$.
\end{enumerate}
\end{exoi}


\begin{defi}[proposition]
On appelle \textbf{loi uniforme} sur $I=[a;b]$, la loi de probabilité 
dont la densité est la fonction constante $f$ définie sur $I$ par 
$$f(t)=\dfrac{1}{b-a}$$
\end{defi}

\Dem{
  $\int_a^b \dfrac{1}{b-a}dt = \left[\dfrac{t}{b-a}\right]_a^b=1$.\\
 $f$ est bien une densité sur $[a;b]$. 
}

\Rqs{\item 
Soit $J=[\alpha;\beta]$ un sous intervalle de $I$ on a alors : 
$$ P(J)=\int_\alpha^\beta \dfrac{1}{b-a}dt=\dfrac{\beta-\alpha}{b-a}$$
\item Dans le cas où $I=[0;1]$, la probabilité du sous-intervalle de $I$ $[\alpha; \beta]$ est donc égale à :
$$ P(J)=\dfrac{\beta-\alpha}{1-0}=\beta-\alpha$$
Il s'agit de la loi de probabilité que nous choisissons pour modéliser le choix aléatoire d'un nombre entre $0$ et $1$. \\
On obtient dans ce cas, 
$$ P([0;0.5])=0,5-0=\dfrac{1}{2}\qquad \text{c'est bien le résultat attendu}$$
ou encore :
$$ P\left(\left[\dfrac{1}{2};\dfrac{3}{4}\right]\right)=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4} \qquad \text{c'est naturel !!}$$
\item La loi uniforme généralise au continu ce que la loi équirépartie \footnote{celle pour laquelle chaque éventualité à la même probabilité} est au discret. Elle modélise par conséquent de nombreuses situations, en particulier elle intervient \og dans le choix d'un point au hasard sur un segment (ou d'un nombre au hasard dans un intervalle $I$) \fg. }

\Exop{Dans la journée, un métro passe toutes les 6 minutes à la station n°14.\\
Soit $X$ le temps d'attente d'une personne à cette station.\\
On suppose que $X$ suit la loi uniforme sur $[0 ; 6]$.\\
Quelle est la probabilité que cette personne attende entre $3$ et $5$ minutes ?}

\Sol{On a  : 
$$ P(3\leq X\leq 5)=\int_3^5 \dfrac{1}{6} dt= \dfrac{5-3}{6}=\dfrac{1}{3}$$}

\Applic{Un point $M$ est pris au hasard sur le demi-cercle $\Gamma$ de diamètre $[AB]$, de centre $O$ et de rayon $1$.\\
Quelle est la probabilité $p$ qe le triangle $AOM$ soit d'aire inférieure ou égale à $0,25$ ?}

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\psdots[dotsize=1pt 0,dotstyle=*](0.41,0.31)
\rput[bl](0.37,0.32){$M$}
\end{pspicture*}
\end{center}

\underline{Note} : On admet que \og choisir un point $M$ au hasard sur le demi-cercle $\Gamma$ \fg revient à dire que \og l'angle $\theta=\widehat{AOM}$ suit la loi uniforme sur $[0;\pi]$ \fg.\\

Calculons dans un premier temps l'aire $\mathscr A$ du triangle $AOM$, en notant $H$ le pied de la hauteur issue de $M$ dans le triangle $AOM$ on a :
$$ \mathscr A=\dfrac{OA\times HM}{2}=\dfrac{HM}{2}$$
Déterminons $HM$, en raisonnant dans le triangle rectangle $HOM$. Exploitons le sinus de l'angle $\widehat{HOM}$ qui vaut suivant la position de $M$ soit $\theta$ soit $\pi-\theta$. On a $\sin (\pi-\theta)=\sin \theta$, par conséquent :
$$ \sin \theta=\dfrac{HM}{OM}=HM$$
Au final $$\mathscr A=\dfrac{\sin \theta}{2}$$
Ainsi $$ 0\leq \mathscr A\leq \dfrac{1}{4}\Longleftrightarrow 0\leq \dfrac{\sin \theta}{2}\leq \dfrac{1}{4}\Longleftrightarrow 0\leq \sin\theta\leq \dfrac{1}{2}\Longleftrightarrow 0\leq \theta\leq \dfrac{\pi}{6}\quad \text{ou}\quad \dfrac{5\pi}{6}\leq \theta \leq \pi$$
De plus $$P\left(0\leq \theta\leq \dfrac{\pi}{6}\right)=\int_0^{\frac{\pi}{6}} \dfrac{1}{\pi} dt= \dfrac{\frac{\pi}{6}}{\pi}=\dfrac{1}{6}$$
et $$P\left(\dfrac{5\pi}{6}\leq \theta \leq \pi\right)=\int_{\frac{5\pi}{6}}^{\pi} \dfrac{1}{\pi} dt= \dfrac{\pi-\frac{5\pi}{6}}{\pi}=\dfrac{1}{6}$$
Au final, comme les événements $\left(0\leq \theta\leq \dfrac{\pi}{6}\right)$ et $\left(\dfrac{5\pi}{6}\leq \theta \leq \pi\right)$ sont disjoints on obtient :
$$ P(\mathscr A\leq 0,25)=\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{1}{3}$$


\begin{propo}
Soit $X$ une variable aléatoire qui suit une loi uniforme sur l'intervalle $\Omega=[a;b]$. L'espérance de $X$ est :
 $$E(X)=\int_a^b t\times \dfrac{1}{b-a}dt=\dfrac{a+b}2$$
\end{propo}


\Rqs{
\item La loi uniforme  modélise le choix aléatoire 
d'un nombre réel $X$ de l'intervalle $\Omega$, tous les nombres ayant le même \og poids \fg , autrement dit
les nombres étant \textbf{uniformément} répartis.
\item La probabilité d'un intervalle $I \subset \Omega$ est proprotionnelle à sa taille.
}


\Dem{ $E(X)=\int_a^b t\times \dfrac{1}{b-a}dt=\left[\dfrac{1}{2(b-a)}\times t^2 \right]_a^b = \dfrac{b^2-a^2}{2(b-a)}
  =\dfrac{(b-a)(b+a)}{2(b-a)}=\dfrac{a+b}2$
}

\Expl{Dans la journée, un métro passe toutes les 6 minutes à la station n$^{\circ}$14.\\
Soit $X$ le temps d'attente d'une personne à cette station.
On suppose que $X$ suit la loi uniforme sur $[0 ; 6]$.\\
\\Quel est le temps moyen d'attente ?
}

\subsection{La loi exponentielle de paramètre $\lambda$}




\begin{defi}[Proposition]
 Soit $\lambda>0$ un réel.\\
On appelle \textbf{loi exponentielle de paramètre} $\lambda$ la loi de
probabilité dont la densité est la fonction $f$ définie sur
$[0,+\infty[$ par $$f(t)=\lambda e^{-\lambda t}$$
\end{defi}

\Rqs{
\item On a déjà montré que $f$ est bien une densité de probabilité sur $\R^+$.
\item Une variable aléatoire $T$ à valeurs dans $\R^{+}$ suit une loi exponentielle de paramètre $\lambda$ lorsque la probabilité de l'événement $(0\leq T\leq t)$ noté simplement $(T\leq t)$ est $ \int_0^t \lambda e^{-\lambda s}ds$.\\
On a vu alors que $$ P(T\leq t)=1-e^{-\lambda t}$$
et donc par conséquent $$ P(T>t)= e^{-\lambda t}$$
}


\begin{exoi}
  Soit $f$ une fonction exponentielle définie sur un intervalle $I=[O;+\infty[$ 
  du type $f(t)=\lambda e^{-\lambda t}$ avec $\lambda>0$ un réel fixé. 
 \begin{enumerate}
  \item Vérifier que pour tout $\lambda\in\R^+$, $f$ est une densité sur $I$. \\
  \item On note désormais $T$ la variable aléatoire qui suit la loi exponentielle de paramètre $\lambda$  sur $I$. 
 \begin{enumerate}
  \item Calculer $P (5 \leq T \leq 10 )$.
  \item Calculer $P (T \leq 15 )$. En déduire $P(T\geq 15)$.
  \item Pour tout réel $c\in I$ , calculer $P(T\leq c)$ et en déduire $P(T \geq c)$.
  \item Pourquoi ne pouvez-vous pas calculer l'espérance de $T$ ?
  \item \begin{enumerate}
   \item  Déterminer les réels $c$ et $d$ tels que la fonction 
 $$G : t \longmapsto (ct+d)e^{-\lambda t}$$
 soit une primitive sur $[0;+\infty[$ de la fonction $g : x \longmapsto \lambda \times te^{-\lambda t}$.
      \item En déduire pour tout $x\in\R^+$, la valeur de l'intégrale $\int_0^x tf(t)dt$.
   \item Déterminer alors $E(T)$.  
  \end{enumerate}
 \end{enumerate}
\end{enumerate}
\end{exoi}

\begin{prop}
 Soit $T$ est une variable aléatoire suivant la loi exponentielle de paramètre $\lambda>0$. 
 Pour tout $a\geq 0$ on a :
 \begin{itemize}
  \item $P(T\leq a)=P(T<a)=1-e^{- \lambda a}$
  \item $P(T\geq a)=P(T>a)=e^{-\lambda a}$
  \item $E(T)=\Lim{+\infty}\int_0^x tf(t)dt=\dfrac1{\lambda}$
   \end{itemize}
\end{prop}

\Dem{
  \begin{itemize}
  \item Pour tout $a\in\R^+$ on a 
  $P(T\leq a)=P(T<a)=\int_0^a\lambda e^{-\lambda t}dt=\left[-e^{-\lambda t}\right]_0^a = -e^{-\lambda a}+e^{-\lambda 0}
  1-e^{- \lambda a}$
  \item $P(T\geq a)=P(T>a)=1-P(T\leq a) = e^{-\lambda a}$
  \item \textbf{Type ROC :}
  Soit $G$ la fonction définie sur $\R+$ par 
   $$G(t)= -\left(t+\dfrac{1}{\lambda}\right)e^{-\lambda t}$$
 alors 
 $$G'(t)=-\left(e^{-\lambda t}+\left(t+\dfrac{1}{\lambda}\right)\times \left(-\lambda e^{-\lambda t}\right)\right)
 =t\times \lambda e^{-\lambda t} =tf(t)$$
 Ainsi 
 $$G'(t)= tf(t)  \iff \int_0^xtf(t)dt=\left[G(t)\right]_0^x
 =-\left(\left(x+\dfrac1{\lambda}\right)e^{-\lambda x}-\dfrac1{\lambda}\right) \quad \forall x\in\R^+$$
 Donc $E(T)=\Lim{+\infty} \int_0^xtf(t)dt=\dfrac1{\lambda}$ 
  \end{itemize}
}


\Rq{
La loi exponentielle modélise des durées de vie particulière.
}

\Expl{
On considère un matériel électronique dont le temps de fonctionnement exprimé en semestres, 
est modélisé par une variable aléatoire $T$ prenant ses valeurs dans $[0;+\infty[$.\\
On a regardé  $5000$ temps de fonctionnement de ce matériel et 
on a regroupée les données en classes d'amplitude $5$ unité de temps. \\
Il n'y a aucun appareil qui a duré plus de 125 semestres. On a donc 25 classes.\\
Sur Géogébra, on a alors réalisé l'histogramme associé à cet échantillon, ainsi que la courbe 
$\mathscr C_f$ représentative de la 
fonction $f$ définie sur $\R^+$ par $f(t)=0.07e^{-0.07t}$.
}
  \begin{center}
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\end{pspicture*}
  \end{center}
  On constate que la forme globale de l'histogramme ressemble à celle de la courbe $\mathscr C_f$.\\
  Si l'on diminue l'amplitude des classes, elle s'en rapproche encore plus.\\
  
\textit{On rappelle que sur un histogramme, les probabilités sont données par les aires des rectangles}.\\
Ainsi avec l'histogramme on trouve
$P(T\leq 5)\simeq 0.059 \times 5= 0.295$ et $P(5\leq T \leq 15)\simeq 0.042\times 5+0.03 \times 5 = 0.36$\\
Avec la fonction densité on trouve :
$$P(T\leq 5)=\int_0^{5}f(t)dt=\int_0^{5}0.07 e^{-0.07t}dt=\left[-e^{-0.07t}\right]_0^{5}=1-e^{-0.35}\simeq 0.295 $$
$$P(5\leq T\leq 15)=\int_5^{15}f(t)dt=\int_5^{15}0.07 e^{-0.07t}dt=\left[-e^{-0.07t}\right]_5^{15}
=e^{-0.35}-e^{-1.05}\simeq 0.355 $$


\begin{exoi}
On suppose que la durée de vie $X$ d'une voiture suit une loi exponentielle de paramètre $0.1$.
\begin{enumerate}
 \item Calculer la probabilité qu'une voiture dure entre 5 et 8 ans.
 \item Calculer la probabilité qu'une voiture dépasse 10 ans de durée de vie.
 \item On sait qu'une voiture a duré déjà 10 ans.\\
Quelle est la probabilité qu'elle dépasse 12 ans de durée de vie ?
 \item Comparer le résultat précédent avec la probabilité que la durée de vie de la voiture dépasse 
 deux ans.\\
\textit{On constate que la probabilité que la voiture dure deux ans de plus ne dépend pas de son âge.
On dit que X est une loi de durée de vie \textbf{sans vieillissement}.}
\end{enumerate}
\end{exoi}

\begin{prop}
  Soit $T$ est une variable aléatoire suivant la loi exponentielle de paramètre $\lambda>0$. \\
  Pour tous réels $t$ et $h$ strictement positifs, on a :
 $$P_{(T\geq t)}(T\geq t+h)=P(T\geq h)$$
\end{prop}

\Dem{
  Pour tous réels positifs $t$ et $h$ on a $P(T\geq t+h)=e^{-\lambda(t+h)}$ et $P(T\geq t)=e^{-\lambda t}$.\\
  On a $P_{(T\geq t)}(T\geq t+h)=\dfrac{P\big((T\geq t)\cap (T\geq t+h)\big)}{P(T\geq t)}=
  \dfrac{P\big(T\geq t+h\big)}{P(T\geq t)}=\dfrac{e^{-\lambda(t+h)}}{e^{-\lambda t}}=e^{-\lambda h}=P(T\geq h)$
}

\Rq{Cette propriété s'appelle \og durée de vie sans vieillissement \fg, car elle montre 
que la durée de vie $T$ sur un laps de temps $h$,
ne dépend pas de l'âge $t$ à partir duquel on considère cet événement.\\
On parle aussi de \textbf{loi sans mémoire}.\\
Les lois exponentielles modélisent des phénomènes dont la durée de vie n'est pas affectée par l'âge,
comme par exemple celle d'un atome radioactif, ou celle de certaines tribus (dont les maladies et autres
n'entraînent que des morts non dues à l'âge des personnes), etc
}

\subsection{Loi de durée de vie sans vieillissement [Aux limites du programme]}

\begin{defi}Soit $X$ la variable aléatoire correspondant à la durée de vie d'un individu ou d'un objet.
On dit que $X$ suit la loi de durée de vie sans vieillissement lorsque la probabilité que l'individu (ou l'objet) soit
vivant (ou fonctionne) à l'instant$ t + h$ sachant qu'il est vivant (ou qu'il fonctionne) à l'instant $t$ ne dépend pas
de son âge $t$ i.e 
$$ P_{X>t}(X>t+h)=P(X>h)$$\end{defi}

\Rq{ la loi de durée de vie sans vieillissement s'applique-t-elle aux humains ? Non, ce n'est pas un
modèle pertinent à long terme. En effet, un bébé à la naissance peut raisonnablement espérer vivre plusieurs
dizaines d'années alors qu'on ne peut en dire autant d'un vieillard. Le modèle semble plus proche de la réalité
lorsque $h$ est petit (dans ce cas). Par exemple la probabilité de vivre encore une minute semble comparable indépendamment
de l'âge. Mais cette loi s'applique plutôt à des composants électroniques par exemple.}

\subsubsection{Une variable aléatoire qui suit une loi exponentielle suit une loi de durée sans vieillissement}

Si $X$ suit une loi exponentielle de paramètre $\lambda>0$ sur $\R^+$ alors on a vu que :
$$P(X>t)= e^{-\lambda t}$$
Soit $h$ un réel strictement positif on a :
$$ P_{X>t}(X>t+h)=\dfrac{P\left((X>t)\cap (X>t+h)\right)}{P(X>t)}=\dfrac{P(X>t+h)}{P(X>t)}=\dfrac{e^{-\lambda (t+h)}}{e^{-\lambda t}}=e^{-\lambda h}=P(X>h)$$
Ainsi $X$ suit bien  une loi de durée sans vieillissement.\\
Examinons désormais la réciproque.

\subsubsection{Une variable aléatoire qui suit une loi de durée sans vieillissement suit une loi exponentielle}

Réciproquement, soit $X$ une variable aléatoire suivant une loi de durée de vie sans vieillissement.
Alors pour tout réel $t$ de $\R^+$ et tout réel $h$ de $\R^+$ :
$$ P_{X>t}(X>t+h)=P(X>h)\Longleftrightarrow \dfrac{P(X>t+h)}{P(X>t)}=P(X>h)\Longleftrightarrow P(X>t+h)=P(X>t)P(X>h)$$

Notons $\phi$ la fonction définie sur $\R^+$ par :
$$ \phi(t)=1-P(X\leq t)=P(X>t)$$

Montrons que :
\begin{itemize}
\item $\phi(0)=1$ 
\item $\phi$ est dérivable sur $\R^+$
\end{itemize}

Pour cela notons $F$ la fonction définie sur $\R^+$ par $F(t)=P(X\leq t)=\int_0^t f(x)dx$. \\
$F$ est la primitive de la densité $f$ qui s'annule en $0$ i.e $F'=f$ et $F(0)=0$, or :
$$ \phi(t)=1-F(t)$$
Par conséquent $\phi$ est dérivable sur $\R^+$ et $\phi(0)=1-0=1$.
\\
On sait que $\phi(t+h)=\phi(t)\phi(h)$, que $\phi$ est dérivable et vérifie $\phi(0)=1$. On en déduit  (leçon sur les équations différentielles du type  $y' = ky$ et la fonction
exponentielle) qu'il existe un réel $a$ tel que pour tout réel $t$ de $\R^+$ :
$$ \phi(t)=e^{at}$$
Mais comme $\phi$ est en fait une probabilité, on a pour tout $t$ de $\R^+$ :
$$ \phi(t)\leq 1\Longleftrightarrow e^{at}\leq 1\Longleftrightarrow at\leq \ln 1=0\Longleftrightarrow a\leq 0$$
Ainsi posons $\lambda =-a>0$ on obtient :
$$ \phi(t)=e^{-\lambda t}\Longleftrightarrow 1-F(t)=e^{-\lambda t}$$
En dérivant membre à membre :
$$ -f(t)=-\lambda e^{-\lambda t}\Longleftrightarrow f(t)=\lambda e^{-\lambda t}$$

La variable aléatoire $X$ suit donc une loi exponentielle de paramètre $\lambda$.

Nous avons démontré le résultat suivant :

\begin{theo}Une variable aléatoire  $X$ suit la loi de durée de vie sans vieillissement si seulement si elle suit une loi
exponentielle.\end{theo}
\newpage

\section{Approximation de la loi binomiale centrée réduite par une loi continue}
\subsection{Standardiser des variables aléatoires}
\begin{exoi}
En Syldavie, la proportion de personnes pratiquant la danse sous-marine en scaphandre est $p=0.35$. On prélève au hasard un échantillon de taille $n$ dans la population Syldave (celle-ci étant assez grande 
pour considérer qu'il s'agit de tirages avec remise).\\
On désigne par $X$ la variable aléatoire associant à un échantillon de taille $n$ le nombre de syldaves
de l'échantillon pratiquant la danse sous-marine en scaphandre (DSMS).
\begin{enumerate}
 \item  \begin{enumerate}
        \item Justifier que $X$ suit une loi binomiale dont on précisera les paramètres.
        \item Déterminer en fonction de $n$, l'espérance $\mu$ et l'écart-type $\sigma$ de $X$.
        \item Norbert a prélevé un échantillon de $50$ Syldaves et a obtenu 21 syldaves pratiquant
        la DSMS.\\
        Simone, sur un échantillon de taille $200$, a obtenu 79 syldaves pratiquant la DSMS.\\
        Fabrice quant à lui a obtenu sur un échantillon de $500$ syldaves, $151$ pratiquant la DSMS.\\
        A votre avis, lequel des trois a l'échantillon le plus représentatif de la population Syldave ?
\end{enumerate}
 \item On a représenté les histogrammes de la loi $B(n,0.35)$ pour les trois valeurs de $n$ précédentes.
\begin{enumerate}
 \item Que se passe-t-il lorsque $n$ varie ?
 \item Quelle est la somme des aires des rectangles pour un $n$ donné ?
\end{enumerate}
\end{enumerate}
 \end{exoi}
 
        \begin{center}
\textbf{Histogrammes de la loi B($n$, 0.35)}\\
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\\
\end{center}
\textit{On rappelle que dans un histogramme, les probabilités de chaque valeur possible pour $X$ sont données 
 par \textbf{une aire de rectangle} : l'aire de chaque rectangle 
centré autour d'une valeur $k$ vaut $P(X=k)$.
}


\textbf{Quand $n$ varie, on obtient des histogrammes qui diffèrent par leurs positions et par leurs 
dispersions. }\\

\Cadre[Conclusion]{
Les distributions de probabilité de deux variables aléatoires peuvent être très différentes, notamment en raison 
de la disparité de leur espérance ou de leur écart-type.\\
Il est alors difficile de les comparer.\\
En centrant et en réduisant les variables, on les rend indépendantes de l'unité choisie pour leurs valeurs, avec une 
espérance et un écart-type de standardisés de $0$ et $1$.\\
La comparaison de plusieurs variables aléatoires s'en trouve ainsi facilitée.
}

\subsection{Centrer l'histogramme d'une loi binomiale}
\begin{exoi}
On reprend le contexte de l'activité précédente et on considère la variable aléatoire $Y=X-\mu$. 
 \begin{enumerate}
       \item Déterminer $y$ tel que $P(X=k)=P(Y=y)$.
	\item Calculer $E(Y)$ et $\sigma(Y)$.\\
	  \textbf{Pour cette raison on dit que la variable $Y$ est \underline{centrée}.}
	\item On a représenté les histogrammes de la loi de $Y$ pour les trois valeurs de $n$ précédente.
    \begin{enumerate}
  \item Expliquer.\\
        \textbf{Valeurs prises par $Y$ ? écart entre deux valeurs consécutives de $y$ ?
        hauteur des rectangles : $P(X=k)= P(Y=y)$ pr un $y=k-\mu$}
  \item Que se passe-t-il lorsque $n$ varie ?
 \item Quelle est la somme des aires des rectangles pour un $n$ donné ?
 \end{enumerate}
\end{enumerate}
\end{exoi}
        

\begin{center}
 \textbf{Histogrammes de la loi de probabilité $Y=X-\mu$ pour $p=0.35$}\\
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%$n=50 \qquad \qquad\qquad \qquad \qquad \qquad \qquad   n=200 \qquad\qquad \qquad\qquad \qquad \qquad\qquad \qquad n=500 \qquad \qquad$
\end{center}

\textbf{On constate que dans $n$ varie, la position des histogramme ne varie plus (ils restent
centrés en $0$), 
par contre ils diffèrent toujours dans leur dispersion}\\

\Rqs{
\item Une variable aléatoire est dite centrée lorsque son espérance vaut $0$.
\item Si $X$ désigne une variable aléatoire sur $\Omega$
la variable $Y=X-E(X)$ est centrée et $\forall k\in\Omega$ on a $$P(Y=k-E(X))=P(X=k)$$
}

\subsection{Réduire}

\begin{exoi}
On reprend encore le contexte de l'activité précédente. Reste donc de la dispersion à stabiliser. \\
Considérons pour cela la variable aléatoire $Z=\dfrac{X-\mu}{\sigma}$
 \begin{enumerate}
\item Déterminer $z$ tel que $P(X=k)=P(Z=z)$.
 \item Calculer $E(Z)$ et $\sigma(Z)$.\\
 \textbf{Pour cette raison on dit que la variable $Z$ est \underline{centrée réduite}.}
 \item On a représenté les histogrammes de la loi de $Z$ pour les trois valeurs de $n$ précédentes.
 \begin{enumerate}
  \item Expliquer.\\
        \textbf{Valeurs prises par $Z$ ? écart entre deux valeurs consécutives de $z$ : $1/\sigma$, 
        hauteur des rectangles : $P(X=k)= P(Z=z)$ pr un $z=\dfrac{x-\mu}{\sigma}$}
  \item Que se passe-t-il lorsque $n$ varie ?
  \item Quelle est la somme des aires des rectangles pour un $n$ donné ?
 \end{enumerate}
 \item Dire alors lequel de Norbert, Simone ou Fabrice a l'échantillon le plus représentative de la population ?
\end{enumerate}
 \end{exoi}
        
        
\newpage
\begin{center}
 \textbf{Histogrammes des lois de probabilité de $Z=\dfrac{X-\mu}{\sigma}$ pour $p=0.35$}\\
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 \end{center}

\textbf{On constate que désormais quand $n$ varie, l'allure des histogrammes change peu. 
On peut même dire que lorsque 
$n$ devient grand, on aurait envie de l'approcher par la courbe d'une fonction continue... }\\

\Rqs{
\item Une variable aléatoire est dite centrée réduite lorsque son espérance vaut $0$ et son écart-type $1$.
 \item Si $X$ désigne une variable aléatoire sur $\Omega$
la variable $Z=\dfrac{X-E(X)}{\sigma(X)}$ est centrée réduite et $\forall k\in\Omega$ on a 
$$P\left(Z=\dfrac{k-E(X)}{\sigma(X)}\right)=P(X=k)$$
}

 
\begin{center}
\textbf{Résumé : Histogrammes des lois de probabilité de $X\hookrightarrow B(100,0.35)$, $Y=X-\mu$ et $Z=\dfrac{X-\mu}{\sigma}$}\\
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\end{center}


\pagebreak
\subsection{Théorème de Moivre-Laplace}



\begin{exoi} 
Les lois de probabilité discrètes donnant lieu à des calculs fastidieux 
dans certaines situations, 
on cherche à approcher les résultats par ceux de calculs effectués avec des variables aléatoires 
continues à densité. \\
Dans le cadre des programmes de Terminale, ce problème est traité pour les lois binomiales.\\% approchées par des lois normales.\\
Reprenons alors l'exemple précédent. 
On introduit la fonction $f$ définie sur $\R$ par $f(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.\\
On a représenté ci-dessous la courbe de cette fonction sur les trois histogrammes de loi de $Z$.
\begin{enumerate}
 \item Que constate-t-on ?
 \item Conjecturer la valeur de $I=\Lim{-\infty} \int_x^0f(t)dt + \Lim{+\infty}\int_0^x f(t)dt$, 
 puis les valeurs de $I_1=\Lim{-\infty} \int_x^0f(t)dt$ et $I_2=\Lim{+\infty}=\int_0^x f(t)dt$.
  \item Que donne Géogébra ou votre calculatrice comme valeur approchée de $J=\int_{-1}^2f(t)dt$ ?
  \item Interpréter graphiquement cette intégrale, puis en termes de probabilités sur $X$.
  \end{enumerate}
\end{exoi}

\begin{center}
 \textbf{Courbes de Gauss et Histogrammes des lois de probabilité de $Z=\dfrac{X-\mu}{\sigma}$ pour $p=0.35$}\\
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\Cadre[Rappel]{
  Une variable aléatoire $X$ suit une loi binomiale de paramètres $n$ et $p$, 
  lorsque $X$ compte les succès lors de $n$  répétitions indépendantes d'une épreuve de Bernoulli 
  de paramètre $p$.\\
  Dans ce cas, $E(X)=np$ et $\sigma(X)=\sqrt{np(1-p)}$.
}

\begin{theo}[Moivre-Laplace, Admis]
 Soit $X_n$ une variable aléatoire qui suit la loi binomiale $B(n;p)$. On pose $Z_n=\dfrac{X_n-np}{\sqrt{np(1-p)}}$.\\
 Pour tous réels $a$ et $b$, avec $a<b$ on a :
 $$\Lim[n]{+\infty} P\left(a\leq Z_n \leq b\right)=\int_a^b\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$$
\end{theo}

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\end{center}


\section{La loi normale centrée réduite}
\subsection{Définition}

\begin{propo}[Admise]
 Soit $f$ la fonction définie sur $\R$ par $f(x)=\dfrac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$. On a  :
 $$\Lim[t]{-\infty} \int_t^0f(x)dx=\dfrac12 \qquad \text{ et } \qquad \Lim[t]{+\infty}\int_0^t f(x)dx=\dfrac12$$
 Ainsi, la fonction $f$ étant de plus clairement continue est positive sur $\R$, 
 est une densité de probabilité sur $\R$.
\end{propo}

\begin{defi}
 La loi normale centré réduite $\mathscr N(0,1)$ est la loi de probabilité ayant pour densité la fonction $f$ définie sur $\R$ par 
 $$f(x)=\dfrac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$$
\end{defi}


\Rqs{\item La loi normale modélise des phénomènes naturels très fréquents (d'où son nom), 
qui résultent de l'addition de plusieurs causes indépendantes, comme la taille d'individus, 
le taux de cholestérol, les erreurs de mesure ...
  \item  Ainsi, si une variable aléatoire $X$ suit une loi $\mathscr N(0,1)$, pour tous réels $a$ et $b$ tels que $a<b$ on a :
 $$P(a\leq X \leq b)=\int_a^b \dfrac1{\sqrt{2\pi}}e^{-\dfrac{t^2}2}dt$$
 \item On sait donc que $P(X\leq 0)=P(X\geq 0 ) = \dfrac12$
 \item La fonction $f$ est paire, sa courbe est donc symétrique par rapport à l'axe des ordonnées. D'où :
 \begin{itemize}
  \item Pour tout réel $a$ on a $P(X\leq -a)=P(X\geq a) = 1-\P(X< a)=1-P(X\leq a)$
  \item Si $a\geq 0$ on a aussi $P(-a \leq X \leq a)=2P(0\leq X \leq a) = 1-2P(X > a )=2P(X\leq a)-1$
 \end{itemize}
 \begin{center}
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\qquad 
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 \end{center}
}

\begin{prop}
 Soit $X$ une variable aléatoire qui suit une loi $\mathscr N(0,1)$. On a :
 $$E(X)=\Lim{-\infty} \int_x^0tf(t)dt+ \Lim{+\infty}\int_0^x tf(t)dt=0 \qquad \text{ et }
 \qquad V(X)=E\left(\left(X-E(X)\right)^2\right)=1$$
 Ainsi une variable aléatoire qui suit une loi normale de paramètre $0$ et $1$ est centrée réduite.
\end{prop}

\Dem{
  $\int_x^0tf(t)dt=  \dfrac{1}{\sqrt{2\pi}}\int_x^0 te^{-\frac{t^2}2}dt = 
  \dfrac{1}{\sqrt{2\pi}} \left[-e^{-\frac{t^2}2}\right]_x^0 = 
  \dfrac{1}{\sqrt{2\pi}} \left(-1+e^{-\frac{x^2}2}\right)$\qquad 
    D'où $\Lim[t]{-\infty} \int_x^0tf(t)dt = -\dfrac{1}{\sqrt{2\pi}}$\\
  De même  $\int_0^xtf(t)dt= \dfrac{1}{\sqrt{2\pi}} \left(e^{-\frac{x^2}2}-1\right)$\qquad 
      D'où $\Lim{+\infty} \int_0^xtf(t)dt = \dfrac{1}{\sqrt{2\pi}}$\\
      Donc $E(X)=0$
}



\subsection{Approximation de la loi binomiale}
Le théorème de Moivre-Laplace implique donc, si une variable $X_n$ suit une loi $\mathscr B(n,p)$
 et que la loi de probabilité de la variable centrée réduite associée $Z_n=\dfrac{X_n-np}{\sqrt{np(1-p)}}$, 
 \og tend vers \fg la loi normale centrée réduite, lorsque $n$ devient grand .\\


 Dans la pratique, on estime que quand $X$ suit une loi $\mathscr B(n,p)$ avec $n\geq 30$, $np\geq 5$ et $n(1-p)\geq 5$
 on peut approximer la variable $Z=\dfrac{X-np}{np(1-p)}$ par une variable $Z$ de loi $\mathscr N(0,1)$ puis 
 se ramèner à la loi binomiale de $X$.
 
 
 \Expl{Si $X$ suit la loi binomiale $\mathscr B(50;0.2)$ on a 
 $$P(15\leq X \leq 30)
 =P\left(\dfrac{15-10}{10\times 40} \leq \dfrac{X-10}{10\times 40} \leq \dfrac{30-10}{10\times 40}\right)
 =P\left(0.0125\leq Z \leq 0.05\right)\simeq \int_{0.0125}^{0.05}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}2}dx
 \simeq 0.0150$$
 Ce qui est bien plus rapide que d'ajouter chacun des $P(X=k)$ pour $k$ variant de $15$ à $30$ !\\
 Evidemment, actuellement, vous ne connaissez encore que la méthode fastidieuse, car vous ne connaissez pas 
 de primitive à la densité de la loi normale ...
 Pour calculer cette intégrale, on a besoin de la 
 calculatrice ou encore de tables de valeurs fournies par les énoncés. \\
 La plupart des calculs liés à la loi normale seront donc des estimations.
}


\begin{center}
\textbf{Quelques approximations}\\
  \begin{tabular}{|>{\centering}m{5.5cm}*{5}{|>{\centering}m{1.8cm}}|}\hline  
   \textbf{$n$} & 15 & 50 & 100 & 100 & 100   \tabularnewline\hline
  \textbf{ $p$}& 0.4 & 0.4 & 0.4 & 0.02 & 0.98   \tabularnewline\hline
  %\textbf{$np$} &  6 & 20 & 40 & 2 & 98     \tabularnewline\hline
   %$n(1-p)$ & 9 & 30 & 60 & 98 & 2  \tabularnewline\hline
   $P(3\leq X_n \leq 12)$ & 0.97 & 0.013 & $6\times 10^{-10}$ & 0.32 & $2\times 10^{-135}$  \tabularnewline\hline
   $P\left(\dfrac{3-np}{\sqrt{np(1-p)}}\leq Z \leq \dfrac{12-np}{\sqrt{np(1-p)}}\right)$ & 0.94 & 0.010 & $55\times 10^{-10}$ & 0.23 & $4\times10^{-10}$ \tabularnewline\hline
  \end{tabular}
\end{center}
 
On constate l'écart qui peut être important lorsque les critères de validité ne sont pas remplis ...
 

\Expl{\textbf{Pondichéry 2013}\\
Dans une entreprise, on s'intéresse à la probabilité qu'un salarié soit absent durant 
une période d'épidémie de grippe.\\
Cette entreprise emploie 220 salariés. 
On admet que la probabilité pour qu'un salarié soit malade une semaine donnée durant 
cette période d'épidémie est égale à $p = 0,05$.\\ 
On suppose que l'état de santé d'un salarié ne dépend pas de l'état de santé de ses collègues.\\ 
On désigne par $X$ la variable aléatoire qui donne le nombre de salariés malades une semaine donnée. 
	\begin{enumerate}
		\item Justifier que la variable aléatoire $X$ suit une loi binomiale dont on donnera 
		les paramètres.\\		 
Calculer l'espérance mathématique $\mu$ et l'écart type $\sigma$ de la variable aléatoire $X$. 
		\item On admet que l'on peut approcher la loi de la variable aléatoire 
		$\dfrac{X - \mu}{\sigma}$ 		
		par la loi normale  centrée réduite c'est-à-dire de paramètres $0$ et $1$.\\		 
On note $Z$ une variable aléatoire suivant la loi normale centrée réduite. \\
Le tableau suivant donne les probabilités de l'évènement $Z < x$ pour quelques valeurs du nombre réel $x$.

\medskip
\begin{tabularx}{\linewidth}{|c|*{11}{>{\scriptsize \centering \arraybackslash}X|}}\hline
$x$& $-1,55$ &$-1,24$ &$-0,93$ &$- 0,62$ &$- 0,31$ &0,00 &0,31 &0,62 &0,93 &1,24 &1,55\\ \hline 
$P(Z < x)$& 0,061 &0,108 &0,177 &0,268 &0,379 &0,500 &0,621 &0,732 &0,823 &0,892 &0,939\\ \hline
\end{tabularx}

\medskip
 
Calculer, au moyen de l'approximation proposée en question b., une valeur approchée à $10^{-2}$ 
près de la probabilité de l'évènement : \og le nombre de salariés absents dans l'entreprise au cours 
d'une semaine donnée est supérieur ou égal  à 7 et inférieur ou égal à 15 \fg. 
\end{enumerate}
}

\subsection{A la calculatrice}
\subsubsection{TI82-83-84}

\noindent \begin{tabular}{|>{\centering\columncolor{vert!22}\bfseries}m{4cm}|>{\centering}m{4cm}|>{\centering}m{9.5cm}|}\hline
\rowcolor{vert!22} \textbf{Objectif} & \textbf{Affichage voulu}  & \textbf{Méthode à suivre} \tabularnewline\hline
  Calculer $P(a\leq X \leq b)$\\
  où $X \hookrightarrow \mathscr  N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=1]{82N01.eps}}  \\~\\
\fbox{\includegraphics[scale=1]{82Nms.eps}}  
&   Appuyer sur \ToucheTInd + \ToucheTInoire{var}  
  pour obtenir \ToucheTIsh{orange}{distrib}\\
  Puis choisir    \texteTI{2:normalFRép(} \\
Compléter ensuite dans l'ordre les paramètres $a$ et $b$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.\\~\\
  \textit{La calculatrice demande $\sigma$ et non $\sigma^2$ en argument !}
  \tabularnewline\hline
  
  Déterminer $\alpha$ tel que  $P(X\leq \alpha) =p$ \\
  où $X \hookrightarrow \mathscr N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=1]{82fracN01.eps}}\\~\\
\fbox{\includegraphics[scale=1]{82fracNms.eps}}
& Aller dans  \ToucheTIsh{orange}{distrib}\\
  Puis choisir   \texteTI{3:FracNormale(} \\
 Compléter ensuite le paramètre $p\in [0;1]$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.
   \tabularnewline\hline
\end{tabular}

\subsubsection{TI89}


\noindent \begin{tabular}{|>{\centering\columncolor{vert!22}\bfseries}m{3.5cm}|>{\centering}m{4.8cm}|>{\centering}m{9.4cm}|}\hline
\rowcolor{vert!22} \textbf{Objectif} & \textbf{Affichage voulu}  & \textbf{Méthode à suivre} \tabularnewline\hline
  Calculer 
   $P(a\leq X \leq b)$
  où $X \hookrightarrow \mathscr N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=0.8]{89N01.eps}}  \\~\\
\fbox{\includegraphics[scale=0.8]{89Nms.eps}}  
&   Dans \toucheTInoire{CATALOG} ouvrir l'onglet \ongletTI{F3}{AppFlash}\\
   Appuyer sur \toucheTIg{6}  %correspondant à la lettre \texttt{B} 
pour aller à la lettre N.\\
   Choisir \quad \texttt{\textbf{normFdR(....TIStat}} \\ ~\\
 Compléter ensuite dans l'ordre les paramètres $a$ et $b$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.\\~\\
   \textit{La calculatrice demande $\sigma$ et non $\sigma^2$ en argument !}
  \tabularnewline\hline
  
  Déterminer $\alpha$ tel que  $P(X\leq \alpha) =p$, 
  où $X \hookrightarrow \mathscr N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=0.8]{89fracN01.eps}}  \\~\\
\fbox{\includegraphics[scale=0.8]{89fracNms.eps}}
& Dans \toucheTInoire{CATALOG} ouvrir l'onglet \ongletTI{F3}{AppFlash}\\
   Appuyer sur \toucheTIg{9}  %correspondant à la lettre \texttt{B} 
pour aller à la lettre I.\\
   Choisir \quad \texttt{\textbf{invNorm(....TIStat}} \\ ~\\
 Compléter ensuite le paramètre $p\in [0;1]$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.
   \tabularnewline\hline
\end{tabular}

\subsubsection{Nspire CX Cas}


\noindent \begin{tabular}{|>{\centering\columncolor{vert!22}\bfseries}m{3.5cm}|>{\centering}m{5.3cm}|>{\centering}m{9cm}|}\hline
\rowcolor{vert!22} \textbf{Objectif} & \textbf{Affichage voulu}  & \textbf{Méthode à suivre} \tabularnewline\hline
  Calculer 
   $P(a\leq X \leq b)$
  où $X \hookrightarrow \mathscr N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=0.5]{N01.eps}} 
& Dans l'onglet \ongletTInsp{2}{$\int \sum$} du catalogue  \\
Ouvrir la catégorie \texteTI{Probabilités}.\\
Puis la sous-catégorie \texteTI{Distributions}\\
Choisir \texteTI{Normale FdR}\\ ~\\
 Compléter ensuite dans l'ordre les paramètres $a$ et $b$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.\\~\\
   \textit{La calculatrice demande $\sigma$ et non $\sigma^2$ en argument !}
  \tabularnewline\hline
  
  Déterminer $\alpha$ tel que  $P(X\leq \alpha) =p$, 
  où $X \hookrightarrow \mathscr N(\mu,\sigma^2)$
& \fbox{\includegraphics[scale=0.5]{fracN01.eps}} 
& Dans l'onglet \ongletTInsp{2}{$\int \sum$} du catalogue  \\
Ouvrir la catégorie \texteTI{Probabilités}.\\
Puis la sous-catégorie \texteTI{Distributions}\\
Choisir \texteTI{Inverse Normale}\\ ~\\
 Compléter ensuite le paramètre $p\in [0;1]$.\\
 Préciser éventuellement les paramètres $\mu$ et $\sigma$ si la loi normale
 n'est pas centrée réduite.
   \tabularnewline\hline
\end{tabular}



 \Rqs{
 \item Pour calculer $P(X\leq 1)$, il faudra penser à écrire $P(X\leq 1)=P(X\leq 0) + P(0 \leq X \leq 1)=\dfrac12 + P(0\leq X \leq 1)$, 
 ce qui s'obtient désormais à la calculatrice. On trouve environ $0.84$.\\
 \item De même, pour avoir $P(X \leq -1)$ on écrira $P(X\leq -1)=P(X\leq 0)-P(-1 \leq X \leq 0)=\dfrac12 - P(-1\leq X \leq 0)$.
 On trouve environ $0.16$.\\
 \item Visualiser les aires sous la courbe aide à faire ses transformations.\\
 De manière générale, on peut voir que pour calculer $P(X\leq a)$ :
 \begin{center}
  Si $a\geq 0$ on utilise $P(X\leq a) = 0.5 + P(0 \leq X \leq a)$ \qquad
  Si $a\leq 0$ on utilise $P(X\leq a) = 0.5 - P(a \leq X \leq 0)$\\
\includegraphics[scale=0.6]{Probas.eps}  
 \end{center}
 }

\subsection{Répartition de l'aire sous la courbe}

\begin{theo}
 Soit $X$ une variable aléatoire suivant la loi normale centrée réduite $\mathscr N(0,1)$.\\
 Pour tout réel $\alpha \in ]0;1[$, il existe un unique réel positif $u_{\alpha}$ tel que 
 $$P\left(-u_{\alpha} \leq X \leq u_{\alpha}\right) = 1- \alpha$$
\end{theo}


\Dem{
On considère la fonction $g$ définie sur $\R^+$ par $g(t)=P(-t\leq X\leq t)= \int_{-t}^t f(x)dx$
où $f$ est la fonction densité de la loi normale centrée réduite.\\
On cherche à applique le TVI à $g$.
Comme $f$ est paire, on a pour tout $t$ positif, $g(t)=2\int_0^t f(x)dx$\\
De plus, $f$ est continue et positive, donc $2f$ aussi, et $g$ est la primitive de $2f$ qui s'annule en $0$ (d'après le 
théorème fondamental).\\
Ainsi $g$ est dérivable sur $\R^+$ donc continue sur $\R^+$.\\
Ensuite, il est clair que $2f$ strictement positive implique que $g$ est strictement 
croissante sur $\R^+$.\\
Enfin, comme $\alpha\in]0;1[$ on a $1-\alpha \in]0,1[$ et 
on a $g(0)=0$ et $\Lim[t]{+\infty}g(t)=\Lim[t]{+\infty}2\int_0^tf(x)dx=2\times \dfrac 12 = 1$.\\
D'après le corollaire du TVI, il existe un uniquement réel $u_{\alpha} \in]0;1[$ tel que $g\left(u_{\alpha}\right)=1-\alpha$.
}

\begin{coro}
Avec les notations précédentes, on retiendra deux valeurs : $u_{0.05}\simeq 1.96$ et $u_{0.01}\simeq 2.58$.\\
  Ainsi, \quad \fbox{$P(-1.96 \leq X \leq 1.96) \simeq 0.95$} \quad et \quad \fbox{$P(-2.58 \leq X \leq 2.58) \simeq 0.99$}
\end{coro}

\Rq{
  Cela donne une idée de la répartition : 
  \begin{itemize}
   \item Environ $95\%$ des réalisations de $X$ sont entre $-1.96$ et $1.96$.
   \item Environ $99\%$ des réalisations de $X$ sont entre $-2.58$ et $2.58$.
  \end{itemize}  
}

\begin{center}
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\rput[br](-0.1,-0.04){$ 0 $}
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\rput[bl](0.2,0.15){$95\%$}
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\psline{->}(2.8,0.1)(2.3,0.01)
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\rput[bl](2.6,0.12){$2.5\%$}
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\quad
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\end{center}

\section{La loi normale $\mathscr N(\mu ; \sigma^2)$}
\subsection{Définition}
\begin{defi}
 Soient un réel $\mu$ et un réel strictement positif $\sigma$.\\
 On dit qu'une variable aléatoire $X$ suit la loi normale $\mathscr N(\mu , \sigma^2)$ si la variable aléatoire
 $\dfrac{X-\mu}{\sigma}$ suit la loi normale centrée réduite $\mathscr N(0,1)$.
\end{defi}

\Rq{
  Dans la pratique, grâce au théorème de Moivre-Laplace, on approxime une loi binomiale $\mathscr B(n,p)$
  par une loi $\mathscr N (\mu , \sigma^2)$ avec $\mu=np$ et $\sigma^2=np(1-p)$.\\
  On considère que l'approximation est satisfaisante dès que $n\geq 30$, $np\geq 5$ et $n(1-p)\geq 5$.
}

\Expl{
  \begin{center}
\textbf{Loi normale $\mathscr N(1,4)$ : $\mu=1$ et $\sigma=2$}\\
\medskip
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\psline[linewidth=2pt,linecolor=blue](1,-0.03)(1,0.22)
\rput[bl](1.1,0.2){\blue{$x=1$}}
\end{pspicture*}
\end{center}
}

\begin{prop}
 Soit une variable aléatoire $X$ qui suit la loi normale $\mathscr N(\mu , \sigma^2)$. On a :
 $$E(X)=\mu \qquad , \qquad V(X)=\sigma^2 \qquad \text{ et } \qquad \sigma(X)=\sigma$$
\end{prop}


\Dem{
Par définition, $\dfrac{X-\mu}{\sigma}$ suit la loi $\mathscr N(0,1)$. On a donc 
$$E\left(\dfrac{X-\mu}{\sigma}\right)=0 \iff \dfrac1{\sigma}\left(E(X)-\mu\right)=0 \iff E(X)-\mu=0 \iff E(X)=\mu$$
De même, 
$$V\left(\dfrac{X-\mu}{\sigma}\right)=1 \iff \dfrac1{\sigma^2}V(X)=1 \iff V(X)=\sigma^2 \iff \sigma(X)=|\sigma|=\sigma $$
}

\Rqs{
  \item $\mu$ est un paramètre de position et $\sigma$ un paramètre de dispersion.
\item La densité de la loi normale $\mathscr N(\mu , \sigma)$ est représentée par une courbe en cloche, dont
l'axe de symétrie vertical est $x=\mu$.\\
Plus $\sigma$ est petit, et moins la dispersion est grande, donc plus la cloche est resserrée autour de son axe de symétrie.
}

Reconnaître les lois $\mathscr N(0,1)$, $\mathscr N(0,1/4)$, $\mathscr N(0,2)$ et $\mathscr N(5,1)$ parmi 
les quatre courbes représentées ci-dessous.
\begin{center}
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\end{pspicture*}
\end{center}




\subsection{Méthodes de calcul}
$X \hookrightarrow \mathscr N(\mu , \sigma^2)$. \qquad Par exemple $X \hookrightarrow \mathscr N(1 , 4)$
\begin{enumerate}
 \item Calculer $P(a\leq X\leq b)$. \qquad Par exemple $P(-1\leq X\leq 3)$. 
 \begin{itemize}
  \item On se ramène à une loi normale  centrée réduite en écrivant :\\
$P(-1\leq X\leq 3)=P(-2\leq X-1\leq 2)= P\left(-1\leq \dfrac{X-1}2 \leq 1\right)=P(-1 \leq Z \leq 1)$ où $Z \hookrightarrow \mathscr N(0,1)$.\\
A la calculatrice, on trouve $P(-1\leq X\leq 3)\simeq 0.683$.
\item Ou alors on calcule directement cette valeur sur la calculatrice en rajoutant les paramètres $\mu$ et $\sigma$
aux commandes déjà connues.
\end{itemize}
\item Déterminer $\alpha$ tel que $P(X\leq \alpha)=p$, avec $p$ donné appartenant à $[0;1]$.\\
Par exemple, on cherche $\alpha$ tel que $P(X\leq \alpha) = 0.9$.
\begin{itemize}
 \item On se ramène à une loi normale centrée réduite en écrivant :\\
$P( X\leq \alpha)=P(\leq X-1\leq \alpha-1)= P\left(\dfrac{X-1}2 \leq \dfrac{\alpha-1}2\right)=P(Z \leq \beta)$ 
où $Z \hookrightarrow \mathscr N(0,1)$ et $\beta=\dfrac{\alpha-1}2$.\\
A la calculatrice, on trouve $\beta\simeq 1.28$ donc $\alpha \simeq 3.56$.
\item Ou alors on trouve directement $\alpha$ sur la calculatrice en rajoutant les paramètres $\mu$ et $\sigma$
aux commandes déjà connues.
\end{itemize}
\end{enumerate}


 \Att{
  Les calculatrices demandent $\mu$ et $\sigma$ en arguments, et non $\sigma^2$.\\
  Si on ne les précise pas, elle considère que $\mu=0$ et $\sigma=1$ (donc elle fait les calculs pour une loi
  normale centrée réduite).
 }

\Expl{
  La masse en kg des nouveaux nés à la naissance est une variable aléatoire qui peut-être modélisée par une loi 
  normale\footnote{Le poids d'un nouveau né ne prend pas de valeurs négatives ni trop grandes, mais on peut vérifier que $P(X<0)$ 
  et $P(X>5)$ sont négligeable.} de moyenne $\mu=3.3$ et d'écart-type $\sigma=0.5$.
  \begin{enumerate}
   \item Déterminer $P(X<2.5)$ en vous ramenant à une loi normale centrée réduite.\\
  Retrouver $P(X<2.5)$ en utlisant la symétrie de la courbe représentative
de la densité de la loi $\mathscr N(3.3,0.5)$.
  \end{enumerate}
}

\subsection{Répartition de l'aire sous la courbe}
\begin{prop}
 Soit une variable aléatoire $X$ qui suit la loi normale $\mathscr N(\mu , \sigma^2)$. On a :
 \begin{itemize}
   \item $P(\mu-\sigma \leq X \leq \mu + \sigma)\simeq 0.683$
  \item $P(\mu-2\sigma \leq X \leq \mu + 2\sigma)\simeq 0.954$
  \item $P(\mu-3\sigma \leq X \leq \mu + 3\sigma)\simeq 0.997$
 \end{itemize}
\end{prop}

\begin{center}
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\quad
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\begin{center}
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\Dem{
$$P(\mu-\sigma \leq X \leq \mu + \sigma)=P(-\sigma \leq X -\mu\leq  \sigma) = 
P\left(-1 \leq \dfrac{X-\mu}{\sigma} \leq 1\right) \simeq 0.683$$
car $\dfrac{X-\mu}{\sigma}$ suit la loi  $\mathscr N(0,1)$. \\
De même pour les autres approximations.

}

 
\Expl{ 
  Une compagnie d'empaquetage de jus possède un équipement versant une quantité $X$ de jus 
  dans chacune des bouteilles de format $1000$ ml. Cette quantité peut varier un peu d'une bouteille à l'autre.\\
  On considère alors que $X$ suit une loi normale de paramètre $\mu$ et $\sigma^2=25$ ml$^2$ et 
  on appelle $Z$ la variable aléatoire $Z=\dfrac{X-\mu}{\sigma}$. 
  \begin{enumerate}
   \item Dans un premier temps, on considère $\mu=1000$.
   \begin{enumerate}
    \item Déterminer la probabilité qu'une bouteille contiennent moins de $1000$ ml.
    \item Déterminer l'intervalle dans lequel se situe $99\%$ de la production.
   \end{enumerate}
    \item Reprendre les deux questions précédentes avec $\mu=1005$ ml.
    \item La législation impose qu'il y ait moins de $1\%$ des bouteilles contiennent moins de $1000$ ml.
        \begin{enumerate}
     \item Déterminer $\mu$ pour que l'embouteilleur respecte la législation et interpréter.
     \item La contenance des bouteilles étant de $1020$ ml, quelle est la probabilité qu'une bouteille déborde lors du remplissage ?
     \item Un inspecteur choisit un échantillon de $20$ bouteilles, au hasard et avec remise, parmi la production 
   de la journée.\\
   L'embouteilleur recevra une amende si l'inspecteur trouve au moins une bouteille contenant moins de $1000$ ml.\\
   On appelle $Y$ la variable aléatoire qui compte le nombre de bouteilles dans l'échantillon qui contiennent
   moins de $1000$ ml. Quelle est la probabilité que l'embouteilleur reçoive une amende ?
      \end{enumerate}
   \end{enumerate}
}

\Sol{\begin{enumerate}
      \item \begin{enumerate}
             \item Par symétrie de la courbe, on sait que  $P(X\leq 1000)=\dfrac12$
             \item On sait que 
  \item $P(\mu-3\sigma \leq X \leq \mu + 3\sigma)\simeq 0.99$ donc $X\in[985 ; 1015]$
            \end{enumerate}
  \item \begin{enumerate}
         \item On a $P(X\leq 1000)=0.5-P(1000 \leq X\leq 1005)=0.5-\dfrac12 P(1000\leq X\leq 1010)
         \simeq 0.5-\dfrac12 0.683 = 0.1585$\\
         Ou encore, à la calculatrice, en calculant $P(1000 \leq X\leq 1005)$, on trouve environ $0.1587$, donc presque pareil.
         \item On sait que $X\in[1005-15 ; 1005+15]$ ie $X\in[990 ; 1020]$ dans $99\%$ des cas.
        \end{enumerate}
  \item \begin{enumerate}
   \item On sait que $Z$ suit une loi normale
   $\mathscr N(0;1)$.
   On cherche alors $\mu$ tel que 
   $$P(X \leq 1000)=0.01 \iff P\left(\dfrac{X-\mu}{\sigma}\leq \dfrac{1000-\mu}{\sigma}\right)=0.01 \iff 
   P\left(Z\leq \dfrac{1000-\mu}{5}\right)=0.01$$
   On cherche donc $\alpha$ tel que $P(Z\leq \alpha)=0.01$.\\
   Grâce à la calculatrice, on trouve $\alpha \simeq -2.326$. Ainsi 
   $$\dfrac{1000-\mu}{5}\simeq -2.326 \iff \mu \simeq 1011.632$$
   L'embouteilleur doit verser en moyenne $1012$ ml par bouteille pour que environ $1\%$ des bouteilles 
   contiennent moins de $1000$ ml.\\
\textit{On peut vérifier en cherchant à la calculatrice $\alpha$ tel que $P(X\leq \alpha)=0.01$ pour la loi de $X$.\\
On doit évidemment trouver une valeur proche de $1000$.}
  \item A la calculatrice, avec $\mu \simeq 1011.6$, on trouve $P(X>1020)=0.5-P(1011.6<X<1020)\simeq 0.046$.
   \item On répète $20$ fois de manière identique et indépendante, l'épreuve de Bernoulli 
   qui consiste à choisir une bouteille, le succès étant 
   \og la bouteille contient moins de $1000$ ml \fg, de probabilité $0.01$ d'après ce qui précède. \\
   $Y$ compte le nombre de succès, donc $Y$ suit une loi binomiale de paramètres $20$ et $0.01$.\\
   On cherche $P(Y \leq 1)=1-P(Y=0) = 1-0.99^20 \simeq 0.18$\\
   La probabilité que l'embouteilleur reçoive une amende est de $0.18$.   
  \end{enumerate}
     \end{enumerate}
}

\end{document}