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\newcommand{\ch}{Chapitre 11}
\newcommand{\Ch}{Des stats sur des \'el\'ephants, des pi\`eces et des punaises}	
\newcommand{\num}{11}	
\newcommand{\Annee}{2012-2013}

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%\debutSsCadreCha
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%\vspace{3cm}

\begin{act}
En Syldavie, le sport national des éléphants est le saut en parachute. Malheureusement pour les crocodiles,
il arrive qu'ils meurent écrasés par un éléphant à son atterrissage, pas toujours contrôlé...\\
Les crocodiles savent que les éléphants gauchers, tout comme les éléphants daltoniens présentent plus de 
risque pour leur survie. \\
Or $46\%$ des éléphants pratiquant le saut en parachute sont des gauchers et 
$18\%$ sont daltoniens.\\
Aussi, lorsqu'un club vient à ouvrir près de chez eux, ils se demandent toujours 
si la fréquence de gauchers et celle de daltoniens dans ce club est anormale (pour éventuellement 
demander sa fermeture).

\partieBac{Fluctuation d'échantillonnage} 
A Gataca, un club de saut en parachute pour éléphants de 324 membres vient d'ouvrir. \\
 \textit{L'échantillon de 324 éléphants peut être assimilé à un tirage aléatoire avec remise, 
 au vu du grand nombre d'éléphants pratiquant le saut en parachute en Syldavie.} 
\begin{enumerate}
 \item Intuitivement, quelles fréquences de gauchers considéreriez-vous \og normale \fg ? de daltoniens ?\\\\
 \item On donne le programme suivant et un exemple d'affichage en sortie.
\begin{center}
 \begin{minipage}{8.5cm}
  \includegraphics[scale=0.7]{Algo_Simul.eps}
 \end{minipage}
 \begin{minipage}{8.5cm}
  \includegraphics[scale=0.8]{Simul_Echantillon0.7.eps}
 \end{minipage}
\end{center}
\begin{enumerate}
 \item Le programmer sur Algobox (régler la même échelle).
 \item Que fait-il ? \\ \\
 \item Qu'observe-t-on ?\\ \\
 \item Placer les informations relatives à l'échelle sur les axes du graphique donné.
 \item A votre avis, pour quelles fréquences de gauchers peut-on considérer \og normale \fg ? de daltoniens ?\\\\
 \item Votre critère vous semble-t-il assez convainquant pour demander ou non la fermeture du club ?
\end{enumerate}
\end{enumerate}

 \newpage
 

\partieBac{Intervalle de fluctuation de seconde et critère de décision}
Norbert, fervent militant de la défense des crocodiles, cherche à savoir s'il peut considérer les fréquences 
d'apparition des deux caractères dans le club comme exceptionnelles ou non.\\
Il émet donc l'hypothèse que la probabilité qu'un éléphant
du club soit gaucher est $p_1=0.46$ et celle qu'il soit daltonien est $p_2=0.18$ (comme dans la population totale
d'éléphants pratiquant le saut en parachute).\\
Puis il dénombre dans le club les gauchers et les daltoniens : il y a $165$ gauchers et $252$ non daltoniens.\\

 \Cadre[Rappel de seconde]{
 Pour un caractère donné, on note $p$ sa probabilité d'apparition et $f$ sa fréquence d'apparition dans un échantillon.\\
 La fréquence $f$ se situe dans au moins $95\%$ des cas dans l'intervalle 
 $$\Big[ \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots , \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots \Big]$$
 appelé \textbf{intervalle de fluctuation au seuil de} $95\%$.
 \begin{itemize}
  \item[\textbullet] \textbf{Avantages} : \\\\
  \item[\textbullet] \textbf{Inconvénients} :  \\\\
  \textit{Sans parler du fait que cet intervalle
 n'a jamais été justifié par votre enseignant ...}
 \end{itemize}
}
\Rqs{
\item Il s'agit d'un intervalle dans lequel se situent au moins $95\%$ ...\\\\\\
%des échantillons établis dans ces conditions.\\
Cet intervalle dépend évidemment de \begin{itemize}
                                     \item ~\\
                                     \item ~\\
                                    \end{itemize}
%la taille $n$ de l'échantillon et de la probabilité $p$ d'apparition du caractère dans la population.
  \item Dans la pratique, on connait $\ldots\ldots$ et on peut faire en sorte que notre échantillon soit établi ...\\
  Par contre, pour $\ldots\ldots$, à moins d'avoir regarder la population entière (et dans ce cas, ce chapitre n'a plus d'intérêt)
  il s'agit d'une hypothèse.\\
  Cet intervalle de fluctuation servira donc surtout à vérifier ...
  %la crédibilité d'une hypothèse sur $p$.
}

\vspace{1cm}

\begin{enumerate}
 \item Déterminer l'intervalle de fluctuation au seuil de $95\%$ du critère \og gaucher \fg dans la
 population des éléphants pratiquant le saut en parachute.\\\\\\

 \item Peut-on ou non réfuter l'hypothèse faite sur les gauchers ? Argumenter.
 \vspace{2.5cm}
 
 
 \item Et pour le critère \og daltonien \fg ?

\Cadre[Critère de décision]{
  Pour une étude statistique d'un caractère connu de la population 
  (âge, sexe, taille, etc), on considère qu'une hypothèse sur une proportion $p$ d'apparition de ce caractère
  dans la population est crédible (ou qu'un échantillon est \textbf{représentatif} lorsqu'on connaît $p$), si 
  \vspace{3cm}
  }

\end{enumerate}

\Rqs{
\item Si la fréquence observée n'appartient pas à l'intervalle de fluctuation au seuil de $95\%$, on
considérera que l'hypothèse de départ sur $p$ est ...\\\\
%\textbf{peu crédible} et on la rejettera.\\
Dans ce cas, on a un risque de se tromper dans $5\%$, puisque $5\%$ des fréquences établies dans ces conditions 
ne sont pas dans cet intervalle. On peut donc refuser l'hypothèse pour $5\%$ d'échantillons \og en trop \fg. \\
Par contre, si $f$ est dans l'intervalle de fluctuation au seuil de $95\%$, on considérera 
l'hypothèse sur $p$ est ... \\\\
%\textbf{crédible}, 
Dans ce cas, on ne connait pas le risque d'erreur, ie le risque d'accepter une hypothèse \og en trop \fg (à partir 
d'échantillons établis dans d'autres conditions).\\
Dans tous les cas, \textbf{on est sûr de rien} ! Donc on évitera le vocabulaire \og vrai \fg ou \og faux \fg.
\item Tout est une question d'équilibre : \\
Si l'on veut diminuer l'erreur de rejet, par exemple en prenant un intervalle de fluctuation au seuil de $100\%$, 
on ne rejettera donc aucune hypothèse \og en trop \fg, par contre, on les acceptera tous, donc évidemment, 
beaucoup trop. Toute hypothèse sur $p$ semblera crédible, ce qui n'a aucun intérêt.\\
Ainsi, dans la pratique, on utilise surtout les seuils de $95\%$ et de $99\%$. 
}

%\newpage

\partieBac{Un nouvel intervalle de fluctuation} 
 \begin{enumerate}
  \item Si l’on choisit au hasard un échantillon de 324 éléphants pratiquant le saut en parachute, 
  quelle loi suit la variable aléatoire $X$ qui dénombre le nombre d’éléphants daltoniens ?
  \textit{Justifier.}\\\\\\\\
  \item Sur tableur, construire :
  \begin{enumerate}
  \item Dans la colonne A la listes des valeurs $k$ possibles pour $X$.\\
  \item Dans la colonne B la liste des probabilités $P(X=k)$ correspondantes. \\
  \textit{Utiliser} \fbox{\texttt{=LOI.BINOMIALE(~k~;~n~;~p~;~0~)}} \, 
  \textit{où $k$ est la valeur d'une cellule.}\\
  \item Le diagramme en bâton correspondant aux colonnes A et B.\\
  \textit{Dans \og Plage de données \fg , cocher les deux cases}.\\
 \item Dans la colonne C les probabilités cumulées correspondantes.
 \hfill \fbox{\textbf{Appeler le professeur pour valider !}}
 \end{enumerate}
 \item En observant le diagramme, proposer une \textbf{méthode} pour établir un nouvel intervalle de fluctuation à 
 $95\%$ des fréquences $\dfrac{X}{n}$. \\\\\\
 \item Grâce à la colonne C, déterminer précisément cet intervalle. \\\\
 \item Interpréter cet intervalle et conclure.\\\\
 \item A partir de combien de daltoniens peut-on considérer que la situation dans le club est anormale avec
 cette méthode de prise de décision ? \\\\
 \item Avec cette méthode, donner l'intervalle de fluctuation au seuil de $95\%$ de la fréquence des gauchers.
 \textit{Préciser la loi de la variable aléatoire $Y$ qui compte le nombre de gauchers.}\\\\
\item Commenter.\\\\\\\\
\end{enumerate}

\Cadre[Pour résumer]{
  On appelle $X$ la variable aléatoire qui compte le nombre de personnes d'un échantillon de taille $n$,
  possèdant un certain caratère, de probabilité d'apparition $p$. On a donc $X\hookrightarrow B(n,p)$\\
  On note $f=\dfrac{X}{n}$ la fréquence d'apparition de caractère.\\
  La fréquence $f$ se situe dans au moins $95\%$ des cas dans l'intervalle 
  $\Big[ \ldots\ldots\ldots , \ldots\ldots\ldots \Big]$ ,
  où :\\
  \begin{itemize}
   \item $a$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse strictement $\ldots\ldots\ldots$.
   \item $b$ désigne le plus petit entier $k$ à partir duquel $P(X\leq k)$ dépasse $\ldots\ldots\ldots$.
  \end{itemize}
  \begin{itemize}
  \item[\textbullet] \textbf{Avantages} :\\\\
  \textit{Sans parler du fait que cet intervalle
  est justifié par vos connaissances sur la loi binomiale ...}
  \item[\textbullet] \textbf{Inconvénients} : \\\\
   \end{itemize}
}

\vspace{1cm}

Ici $X \hookrightarrow B(400 ; 0.18)$.\\
%\begin{center}
\begin{tabular}{*{2}{|>{\centering}m{1.5cm}}|}\hline
 $k$ & $P(X\leq k)$\tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
 54 & 0.00953 \tabularnewline\hline
 55 & 0.01375 \tabularnewline\hline
 56 & 0.01944 \tabularnewline\hline
 57 & 0.02699 \tabularnewline\hline
 58 & 0.03678 \tabularnewline\hline
 59 & 0.04924  \tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
\end{tabular} 
\qquad 
\begin{tabular}{*{2}{|>{\centering}m{1.5cm}}|}\hline
 $k$ & $P(X\leq k)$\tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
 83 & 0.94587 \tabularnewline\hline
 84 & 0.94587 \tabularnewline\hline
 85 & 0.95825 \tabularnewline\hline
 86 & 0.96821 \tabularnewline\hline
 87 & 0.97609 \tabularnewline\hline
 88 & 0.98225 \tabularnewline\hline
 $\ldots$ & \ldots \tabularnewline\hline
\end{tabular} 
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\psline[linewidth=1.2pt,linecolor=red](56.5,-0.0053)(56.5,0.06)
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\psline{->}(48,0.008)(54,0.0005)
\psline{->}(99,0.008)(91,0.0005)
\rput[bl](40,0.053){\red{$a=\ldots\ldots$}}
\rput[bl](90,0.053){\red{$b=\ldots\ldots$}}
\begin{scriptsize}
\rput[bl](38,0.01){\red{Moins de $2.5\%$}}
\rput[bl](90,0.01){\red{Moins de $2.5\%$}}
\rput[bl](65,0.055){\blue{Plus de $95\%$}}
\end{scriptsize}
\end{pspicture*}
\end{minipage}

%\end{center}
\end{act}

\newpage


\Expl{
  On admet que dans la population d’enfants de 11 à 14 ans d’un département français le pourcentage
d’enfants ayant déjà eu une crise d’asthme dans leur vie est de $13\%$.\\
Un médecin d’une ville de ce département est surpris du nombre important d’enfants le consultant
ayant des crises d’asthme et en informe les services sanitaires. Ceux–ci décident d’entreprendre une
étude et d’évaluer la proportion d’enfants de 11 à 14 ans ayant déjà eu des crises d’asthme.\\
Ils sélectionnent de manière aléatoire 100 jeunes de 11 à 14 ans de la ville.\\
La règle de décision prise est la suivante : si la proportion observée est supérieure à la borne
supérieure de l’intervalle de fluctuation asymptotique au seuil de $95\%$ alors une investigation plus
complète sera mise en place afin de rechercher les facteurs de risque pouvant expliquer cette
proportion élevée.
\begin{enumerate}
 \item Déterminer l’intervalle de fluctuation asymptotique au seuil de $95\%$ de la proportion de jeunes de
11 à 14 ans ayant eu une crise d’asthme dans un échantillon de taille 100. 
\item  L’étude réalisée auprès des 100 personnes a dénombré 19 jeunes ayant déjà eu des crises d’asthme.\\
Que pouvez-vous conclure ?\\\\\\\\\\
\item  Le médecin n’est pas convaincu par cette conclusion et déclare que le nombre de personnes
interrogées était insuffisant pour mettre en évidence qu’il y avait plus de jeunes ayant eu des crises
d’asthme que dans le reste du département.\\\\\\\\\\
Combien faudrait-il prendre de sujets pour qu’une proportion observée de $19\%$ soit en dehors de
l’intervalle de fluctuation asymptotique ?\\\\\\\\
\end{enumerate}
}

\vspace{1cm}

\Cadre[Résumé]{
\begin{tabular}{|>{\centering}m{2cm}|>{\centering}m{8cm}|>{\centering}m{5cm}|}\hline
 & Intervalle de \textbf{fluctuation}\\ Au seuil 0.95\\ ($p$ connue)
 & Intervalle de \textbf{confiance} \\ Au seuil 0.95 \\ ($p$ inconnue) \tabularnewline\hline
 \textbf{SECONDE} & $n\geq 25$ , $0.2 \leq p \leq 0.8$ \\  $I=\left[ p-\dfrac{1}{\sqrt{n}} , p +\dfrac{1}{\sqrt{n}}\right]$ \\ & Sensibilisation\tabularnewline\hline
 \textbf{PREMIERE} & Avec la loi binomiale \\ $I=\left[\dfrac{a}{n} ; \dfrac{b}n \right]$\\ 
 & \cellcolor{gray} \tabularnewline\hline
 %\textbf{TERMINALE} &$n\geq 30$ , $np \leq 5$ et $n(1-p)  \leq 5$ \\ Asymptotique \\ $I_n=\left[p-1.96 \dfrac{\sqrt{p(1-p)}}{\sqrt{n}} ; p+1.96 \dfrac{\sqrt{p(1-p)}}{\sqrt{n}}\right]$\\
 %& $J = \left[ f-\dfrac{1}{\sqrt{n}} , f +\dfrac{1}{\sqrt{n}}\right]$ \tabularnewline\hline
\end{tabular} 
}
 \newpage
 
Voici un algorithme qui permet de retrouver le $a$ et le $b$ cherchés ci-dessus, ainsi que sa traduction 
pour algobox et pour chacun des modèles de vos TI.\hfill 
Ayez l'initiative de le recopier, cela sera bien pratique en exercice !

\begin{minipage}{8cm}
 \Algo{8cm}{
  \Entree{
    $n$ est un entier naturel\\
    $p$ est un réel de l'intervalle $[0;1]$
  }
  \Var{
    $s$ est un nombre réel de l'intervalle $[0,1]$\\
    $k$ est un entier naturel
  }
  \Deb{
    $k:=0$\\
    $s:=\binom{n}{0}p^0(1-p)^n$\\
    \While{$s\leq 0.025$}{
    $k := k + 1$\\
    $s:=s + \binom{n}{k}p^k(1-p)^{n-k}$
    }
    Afficher \og $a=$ \fg et la valeur $k$\\
    \While{$s< 0.975$}{
    $k := k + 1$\\
    $s:=s + \binom{n}{k}p^k(1-p)^{n-k}$
    }
    Afficher \og $b=$ \fg et la valeur $k$
    }
 }
\end{minipage}
\quad
\begin{minipage}{10.5cm}
 \begin{cadrecode}
\begin{small}
\begin{verbatim}
1   VARIABLES
2     n EST_DU_TYPE NOMBRE
3     p EST_DU_TYPE NOMBRE
4     s EST_DU_TYPE NOMBRE
5     k EST_DU_TYPE NOMBRE
6   DEBUT_ALGORITHME
7     LIRE n
8     LIRE p
9     k PREND_LA_VALEUR 0
10    s PREND_LA_VALEUR ALGOBOX_LOI_BINOMIALE(n,p,0)
11    TANT_QUE (s<=0.025) FAIRE
12      DEBUT_TANT_QUE
13      k PREND_LA_VALEUR k+1
14      s PREND_LA_VALEUR s+ALGOBOX_LOI_BINOMIALE(n,p,k)
15      FIN_TANT_QUE
16    AFFICHER "a ="
17    AFFICHER k
18    TANT_QUE (s<0.975) FAIRE
19      DEBUT_TANT_QUE
20      k PREND_LA_VALEUR k+1
21      s PREND_LA_VALEUR s+ALGOBOX_LOI_BINOMIALE(n,p,k)
22      FIN_TANT_QUE
23    AFFICHER "b ="
24    AFFICHER k
25  FIN_ALGORITHME

\end{verbatim}
\end{small}
\end{cadrecode}

\end{minipage}


\begin{minipage}{5.8cm} 
 \Algo[TI 82 à 84]{5.8cm}{
  \texttt{PROGRAM:INTERV\\
  :Prompt N\\
  :Prompt P\\
  :0 \textrightarrow \, K\\
  :BinomFdp(N,P,0) \textrightarrow \, S\\
  :While S<=0.025\\
  :K+1 \textrightarrow \, K\\
  :S+BinomFdp(N,P,K) \textrightarrow \, S\\
  :End\\
  :Disp  \textquotedbl A=\textquotedbl ,K\\
  :While S<=0.975\\
  :K+1 \textrightarrow \, K\\
  :S+BinomFdp(N,P,K) \textrightarrow \, S\\
  :End\\
  :Disp \textquotedbl B=\textquotedbl ,K
  }  
 }
 \textbf{BinomFdp} : \\
 \fbox{2nde} + \fbox{Vars} + \fbox{A : binomFdp}
\end{minipage}
\begin{minipage}{7cm} 
 \Algo[TI 89]{7cm}{
  \texttt{:interv(n,p)\\
  :Prgm
  :0 \textrightarrow \, k\\
  :tistat.binomddp(n,p,0) \textrightarrow \, s\\
  :While s<=0.025\\
  :k+1 \textrightarrow \, k\\
  :s+tistat.binomddp(n,p,k) \textrightarrow \, s\\
  :EndWhile\\
  :Disp  \textquotedbl a= \textquotedbl , k\\
  :While s<=0.975\\
  :k+1 \textrightarrow \, k\\
  :s+tistat.binomddp(n,p,k) \textrightarrow \, s\\
  :EndWhile\\
  :Disp \textquotedbl b= \textquotedbl , k\\
  :EndPrgm
    }  
 }
 \textbf{tistat.binomddp} (binomDdP) : \\
 \fbox{CATALOG} \, + \, \fbox{F3}
\end{minipage}
\begin{minipage}{6.3cm} 
 \Algo[TI Nspire]{6.3cm}{
  Define LibPub \textbf{intervfluc}$(n,p)=$\\
  Prgm\\
  Local $s$, $k$\\
  $k:=0$\\
  $s:=$ binomPdf($n$,$p$,0)\\
  While $s\leq 0.025$\\
  $k:=k+1$\\
  $s:=s+$ binomPdf($n$,$p$,$k$)\\
  EndWhile\\
  Disp  \textquotedbl a= \textquotedbl , $k$\\
  While $s\leq 0.975$\\
  $k:=k+1$\\
  $s:=s+$ binomPdf($n$,$p$,$k$)\\
  EndWhile\\
  Disp \textquotedbl b= \textquotedbl , $k$\\
  EndPrgm
   }
   \textbf{binomPdf} (Binomiale Ddp) : \\
    \fbox{catalogue} \, + \, \fbox{2} \\\\
    Ouvrir Probabilités puis Distributions
\end{minipage}


\end{document}