chapitre 1ere
authorDenise sur Lya <sekhmet@lya>
Tue, 31 May 2016 17:23:19 +0000 (19:23 +0200)
committerDenise sur Lya <sekhmet@lya>
Tue, 31 May 2016 17:23:19 +0000 (19:23 +0200)
DST corrigé 2nde

14 files changed:
1ere/chapitre8_loibino/chap8.pdf
1ere/chapitre8_loibino/chap8.tex
1ere/chapitre8_loibino/exemple.ods [new file with mode: 0644]
1ere/chapitre8_loibino/graphique.pdf [new file with mode: 0644]
1ere/chapitre8_loibino/graphique_exemple.pdf [new file with mode: 0644]
2nde/DST_8/dst8.tex
2nde/DST_8/dst8_A_corrige.pdf
2nde/DST_8/dst8_A_enonce.pdf
2nde/DST_8/dst8_B_corrige.pdf
2nde/DST_8/dst8_B_enonce.pdf
2nde/DST_8/fond_graphe_exoA.pdf [new file with mode: 0644]
2nde/DST_8/fond_graphe_exoA.svg [new file with mode: 0644]
2nde/DST_8/fond_graphe_exoB.pdf [new file with mode: 0644]
2nde/DST_8/fond_graphe_exoB.svg [new file with mode: 0644]

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Binary files a/1ere/chapitre8_loibino/chap8.pdf and b/1ere/chapitre8_loibino/chap8.pdf differ
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@@ -14,8 +14,8 @@
 On considère une expérience aléatoire qui n'a que deux issues : succès ($S$) et échec ($\bar{S}$). On note $p$ la probabilité de succès et $q = \ldots \ldots \ldots \ldots $ la probabilité d'échec.
 \begin{itemize2}
   \item Cette expérience s'appelle \souligne{épreuve de Bernoulli de paramètre $p$}
-  \item La variable aléatoire qui prend la valeur $1$ en cas de succès et $0$ en cas d'échec s'appelle \souligne{variable aléatoire de Bernoulli}.
-  \item La loi de probabilité de cette variable aléatoire (tableau ci-dessous) est appelée \souligne{loi de Bernoulli de paramètre $p$}.
+  \item La variable aléatoire qui prend la valeur $1$ en cas de succès et $0$ en cas d'échec s'appelle \souligne{variable aléatoire} \souligne{de Bernoulli}.
+  \item La loi de probabilité de cette variable aléatoire (tableau ci-dessous) est appelée \souligne{loi de Bernoulli de} \souligne{paramètre $p$}.
 \end{itemize2}
 
 \begin{tabular}{|c|p{2cm}|p{2cm}|}
@@ -167,6 +167,62 @@ $P(X\geq 3) = \dotfill$
 %%%%%%%%%%%
 \newpage
 \section{Intervalle de fluctuation}
+\begin{prop}[Rappel]
+Si $p$ est la proportion d'un caractère dans une population, avec $0,2 \leq p \leq 0,8$, alors si on prend un échantillon de taille $n \geq 25$, la fréquence $f$ observée du caractère dans cet échantillon appartient à l'intervalle de fluctuation :
 
+\[ IF = \hspace{6cm}\]
+
+avec une probabilité d'au moins \upc{0,95}.
+\end{prop}
+
+Grâce à la loi binômiale, on peut améliorer ce résultat.
+
+\subsection*{Intervalle de fluctuation avec la loi binômiale}
+La proportion de personnes ayant les yeux marron en France est de $0,34$. On choisit au hasard 100 personnes, et on souhaite savoir combien de personnes aux yeux marrons on peut trouver dans cet échantillon, avec une probabilité de $0,95$.
+
+On définit $X$ la variable aléatoire qui compte le nombre de personnes aux yeux marrons. \dotfill
+
+\lignepoint
+
+$X$ prend les valeurs \dotfill
+
+\bigskip
+
+On cherche ensuite deux bornes $a$ et $b$, de telle sorte que :
+\begin{itemize2}
+\item $P(a \leq X \leq b) \geq \upc{95}$
+\item \dotfill %$P(X\leq a-1) \leq \upc{2,5}$
+\item \dotfill %$P(X \geq b+1) \leq .... \equi P(X \leq b) \geq 97,5 pc$
+\end{itemize2}
+Autrement dit, $a$ est \dotfill
+
+$b$ est \dotfill
+
+\bigskip
+
+\includegraphics[width=0.7\textwidth]{graphique_exemple.pdf}
+
+%$P(X \leq 24) = 0,020$
+%$P(X\leq 25) = 0,034$
+%a = 25
+
+%$P(X \leq 42) = 0.962$
+%$P(X \leq 43) = 0.976$
+
+%$b = 43$
+
+On a donc $P(X \in \ldots \ldots \ldots) \geq \upc{95}$, donc la fréquence $f$ appartient à l'intervalle \ldots\ldots\ldots\ldots\ldots\ldots avec une probabilité au moins égale à \upc{95}.
+
+\subsection*{Intervalle de fluctuation}
+\begin{prop}
+Si une variable aléatoire $X$ suit une loi binômiale de paramètres $n$ et $p$, alors l'intervalle de fluctuation de la fréquence observée au seuil de \upc{95} est 
+\vspace{0.5cm}
+
+où 
+\begin{itemize2}
+\item \dotfill
+\item \dotfill
+\end{itemize2} 
+\end{prop}
 
 \end{document}
diff --git a/1ere/chapitre8_loibino/exemple.ods b/1ere/chapitre8_loibino/exemple.ods
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diff --git a/1ere/chapitre8_loibino/graphique.pdf b/1ere/chapitre8_loibino/graphique.pdf
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diff --git a/1ere/chapitre8_loibino/graphique_exemple.pdf b/1ere/chapitre8_loibino/graphique_exemple.pdf
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@@ -266,13 +266,32 @@ $f$ est une fonction dont on donne le tableau de variations :
 f(x) & \h{4} & \d & \b{-2} & \c & \h{0} & \d & \b{-1} \\
 \fin
 \end{minipage}
-\begin{minipage}{8cm}\includegraphics[width=8cm]{fond_graphe.pdf}\end{minipage}
+\begin{minipage}{8cm}
+\corrige
+  \version{
+  \includegraphics[width=8cm]{fond_graphe_exoA.pdf}}{
+  \includegraphics[width=8cm]{fond_graphe_exoB.pdf}}
+\else
+\includegraphics[width=8cm]{fond_graphe.pdf}
+\fi
+\end{minipage}
 
 \begin{enumerate}
 \item Peut-on dire que $f$ est décroissante sur \version{$[-2; 7]$}{$[-4; 4]$} ? Justifier brièvement. \bareme{0.5}
+\reponse{Non, car $f$ n'est pas décroissante sur tout cet intervalle : elle est croissante sur \version{$[3; 4]$}{$[1; 2]$}. }
 \item Quel est le maximum de $f$ sur son ensemble de définition, et où est-il atteint ? \bareme{0.5}
+\reponse{Le maximum de $f$ sur son ensemble de définition est $4$, atteint en $\version{-2}{-4}$}
 \item Quel est le minimum de $f$ sur $\left[-1; \frac{\version{7}{3}}{2} \right]$, et où est-il atteint ? \bareme{0.5}
+\reponse{Le minimum de $f$ sur $\left[-1; \frac{\version{7}{3}}{2} \right]$ est $-2$, atteint en $\version{3}{1}$.}
 \item On apprend que $f(0) = 0$. Étudier le signe de $f$ (sous forme de tableau ou avec des phrases). \bareme{1}
+\reponse{ Sous forme de tableau :
+
+\variations
+\version{x & -2 & & 0 & & 4 & & 7 \\}
+{x & -4 & & 0 & & 2 & & 4 \\}
+f(x) & \ga+ & \z & - & \z & \dr- \\
+\fin
+}
 \item Dessiner ci-dessus un graphique possible pour $f$. \bareme{1}
 
 
@@ -286,18 +305,30 @@ $ABCDEFGH$ est un cube de côté \ucm{\version{5}{4}}. On place les points $A'$,
 \end{minipage}
 \begin{enumerate}
 \item Dans quel intervalle peut varier $x$ ? \bareme{0,5}
+\reponse{
+$x$ doit être dans l'intervalle $[0; \version{5}{4} ]$.}
 \item Calculer l'aire lorsque $x=0$, et $x=\version{5}{4}$. 
 
 \emph{Aide : vous pouvez utiliser la propriété suivante : l'aire d'un triangle équilatéral de côté $c$ est $ \frac{c^2\sqrt{3}}{4}$.} \bareme{1,5}
 \reponse{
+Lorsque $x=0$, la figure n'est pas coupée, donc l'aire est égale à celle du cube :
+\[A = 6 \times \version{5}{4}^2 = \version{150}{96} \]
+
+Lorsque $x= \version{5}{4}$ :
+
 La surface de la figure est composée de 3 carrés \og entiers \fg, trois triangles rectangles isocèles, plus le triangle $A'C'F'$.
 
-L'aire des trois carrés est égale à \[ A_1 = 3 \times \version{5}{4}^2\]
+L'aire des trois carrés est égale à \[ A_1 = 3 \times \version{5}{4}^2 = \version{75}{48}\]
+
+L'aire des trois triangles rectangles isocèles est égale à \[ A_2 = 3 \times \frac{b \times h}{2} = 3 \frac{\version{5}{4}^2}{2} = \version{37,5}{24}\]
+
+L'aire du triangle équilatéral est égale à \[ A_3 = A'C'^2 \frac{\sqrt 3}{4}\]
+Dans le triangle rectangle $A'C'B$ rectangle en $B$, d'après le théorème de Pythagore, 
+\[A'C'^2 = A'B^2 + C'B^2 = \version{5}{4}^2 + \version{5}{4}^2 = \version{50}{32}\]
 
-L'aire des trois carrés coupés est égale à \[ A_2 = \frac{b \times h}{2} = \frac{\version{5}{4}^2}{2} = \]
+Donc \[ A_3 =  \version{50}{32} \times \frac{\sqrt 3}{4} = \version{\frac{25}{2}}{8} \sqrt 3\]
 
-L'aire du triangle équilatéral est égale à \[ A_3 = A'C' \frac{\sqrt 3}{4}\]
-Dans le triangle rectangle $A'C'B$ rectangle en $B$, d'après le théorème de Pythagore
+On a donc \[A =  \version{75}{48} + \version{37,5}{24} + \version{\frac{25}{2}}{8} \sqrt 3 \simeq \version{134,2}{85,9}\]
 }
 \item Montrer que 
 \[ A(x) = \version{150}{96} - \frac{3x^2}{2} + \frac{x^2\sqrt 3}{2}\]
@@ -305,18 +336,27 @@ Dans le triangle rectangle $A'C'B$ rectangle en $B$, d'après le théorème de P
 \reponse{
 La surface de la figure est composée de 3 carrés \og entiers \fg, trois carrés \og coupés \fg (un triangle rectangle isocèle en moins), plus le triangle $A'C'F'$.
 
-L'aire des trois carrés est égale à \[ A_1 = 3 \times \version{5}{4}^2\]
+L'aire des trois carrés (voir question précédente) est égale à \[ A_1 = \version{75}{48}\]
 
-L'aire des trois carrés coupés est égale à \[ A_2 = 3 \version{5}{4}^2 - 3 \frac{b \times h}{2} = 3 \version{5}{4}^2  - 3 \frac{x^2}{2} \]
+L'aire des trois carrés coupés est égale à \[ A_2 = 3 \version{5}{4}^2 - 3 \frac{b \times h}{2} = \version{75}{48}  - 3 \frac{x^2}{2} \]
 
 L'aire du triangle équilatéral est égale à \[ A_3 = A'C'^2 \frac{\sqrt 3}{4}\]
-Dans le triangle rectangle $A'C'B$ rectangle en $B$, d'après le théorème de Pythagore, $A'C'^2 = A'B^2 + C'B^2 = 2x^2$. D'où $A'C' = x \sqrt{2}$
+Dans le triangle rectangle $A'C'B$ rectangle en $B$, d'après le théorème de Pythagore, $A'C'^2 = A'B^2 + C'B^2 = 2x^2$.
+
 Donc \[ A_3 = 2x^2 \frac{\sqrt{3}}{4} = x^2 \frac{\sqrt{3}}{2} \]
 
-D'où \[ A(x) = 6 \version{5}{4}^2 - \frac{x^2}{2} (3 - \sqrt 3)\]
+D'où \[ A(x) =  \version{150}{96} - \frac{x^2}{2} (3 - \sqrt 3)\]
 
 }
 \item À l'aide de la calculatrice, dresser le tableau de variations de $A$ sur son ensemble de définition. \bareme{0,5}
+\reponse{
+
+\variations
+x & 0 & & \version{5}{4} \\
+A(x) & \h{ \version{150}{96}} & \d & \b{\version{134,2}{85,9}} \\
+\fin
+
+}
 \end{enumerate}
 
 
@@ -333,4 +373,13 @@ Le second coffre a pour inscription~: «~Il y a un trésor dans l'autre coffre~
 \smallskip
 
 Faut-il ouvrir un coffre (lequel) ? Les deux ? Aucun ?
+
+\reponse{Il faut considérer les deux cas : soit les deux coffres disent la vérité, soit les deux mentent.
+
+Si les deux mentent, alors : 
+\begin{enumerate}[Coffre 1 : ]
+\item il n'y a pas de piège ici ET pas de trésor dans l'autre coffre
+\item il n'y a pas de trésor dans le coffre 1
+\end{enumerate}
+On voit que cela se contredit... ce n'est donc pas cette option. Les deux coffres disent donc la vérité... raisonnement à continuer.}
 \end{document}
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