update
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parent
073cfab0aa
commit
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2
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7576
Excalidraw/révisions statistiques notes.excalidraw.md
Normal file
7576
Excalidraw/révisions statistiques notes.excalidraw.md
Normal file
File diff suppressed because it is too large
Load Diff
108
Statistiques révisions.md
Normal file
108
Statistiques révisions.md
Normal file
@ -0,0 +1,108 @@
|
||||
|
||||
# TD 0
|
||||
|
||||
## Exercice 1
|
||||
|
||||
$X \leadsto \mathcal{N}(0, 1)$
|
||||
|
||||
1. $P(-1.96 < X)$
|
||||
|
||||
Comme $X \leadsto \mathcal{N}(0, 1)$, on sait que $P(-1.96 < X) = P(X < 1.96) = \Pi(1.96) \approx 0.9750$
|
||||
|
||||
2. $P(-1.96 < X < 1.96)$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(-1.96 < X < 1.96) &= P(X < 1.96) - P(X < -1.96) \\
|
||||
&= P(X < 1.96) - 1 + P(X < 1.96) \\
|
||||
&= \Pi(1.96) + P(1.96) - 1 \\
|
||||
&\approx 2\times 0.9750 - 1 \\
|
||||
&\approx 0.95
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
## Exercice 2
|
||||
|
||||
$X \leadsto \mathcal{N}(3, 2)$
|
||||
|
||||
1. $P(1 \leq X \leq 2)$
|
||||
$$
|
||||
\begin{align}
|
||||
P(1 \leq X \leq 2) &= P(-2 \leq X - 3 \leq -1) \\
|
||||
&= P\left( -1 \leq \frac{X-3}{2} \leq -\frac{1}{2} \right) \\
|
||||
&= P\left( -1 \leq Y \leq -\frac{1}{2} \right) & \text{où } Y \leadsto \mathcal{N}(0, 1) \\
|
||||
&= P\left( \frac{1}{2} \leq Y \leq 1 \right) \\
|
||||
&= P(Y \leq 1) - P\left( \frac{1}{2} \leq Y \right) \\
|
||||
&= P(Y \leq 1) - 1 + P\left( Y \leq \frac{1}{2} \right) \\
|
||||
&= \Pi(1) + \Pi\left( \frac{1}{2} \right) - 1 \\
|
||||
&\approx 0.8413 + 0.6915 - 1 \\
|
||||
&\approx 0.5328
|
||||
\end{align}
|
||||
$$
|
||||
1. $P_{X<3} (1 \leq X \leq 4)$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P_{X<3} (1 \leq X \leq 4) &= P_{X < 3} (1 \leq X \leq 3) \\[1em]
|
||||
&= \frac{P(1 \leq X \leq 3)}{P(X < 3)} \\[1em]
|
||||
&= \frac{P\left( -\frac{1}{2} \leq \frac{X - 3}{2} \leq 0 \right) }{P\left( \frac{X-3}{2} \leq 0 \right)} \\[1em]
|
||||
&= \frac{P\left( -\frac{1}{2} \leq Y \leq 0 \right)}{P(Y \leq 0)} & \text{où } Y \leadsto \mathcal{N}(0, 1) \\[1em]
|
||||
&= \frac{P(Y \leq 0) - P\left( Y \leq -\frac{1}{2} \right)}{P(Y \leq 0)} \\[1em]
|
||||
&= \frac{P(Y \leq 0) - P\left( Y \geq \frac{1}{2} \right)}{P(Y \leq 0)} \\[1em]
|
||||
&= \frac{P(Y \leq 0) - 1 + P\left( Y \leq \frac{1}{2} \right)}{P(Y \leq 0)} \\[1em]
|
||||
&= \frac{P(Y \leq 0) + P\left( Y \leq \frac{1}{2} \right) - 1}{P(Y \leq 0)} \\[1em]
|
||||
&= \frac{\Pi(0) + \Pi\left( \frac{1}{2} \right) - 1}{\Pi(0)} \\[1em]
|
||||
&\approx \frac{0.5 + 0.6915 - 1}{0.5} \\
|
||||
&\approx 0.383
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
## Exercice 3
|
||||
|
||||
1. calculer $\mu$ et $\sigma$
|
||||
|
||||
$P(X \leq 165) = 0.58$
|
||||
$P(165 \leq X \leq 180) = 0.38$
|
||||
$P(180 \leq X) = 0.04$
|
||||
|
||||
$P\left( \frac{X-\sigma}{\mu} \leq \frac{165 - \sigma}{\mu} \right) = 0.58 \iff \Pi\left( \frac{165-\sigma}{\mu} \right) = 0.58$
|
||||
|
||||
$P\left( \frac{165 - \sigma}{\mu} \leq \frac{X-\sigma}{\mu} \leq \frac{180-\sigma}{\mu} \right) = 0.38 \iff \Pi\left( \frac{180-\sigma}{\mu} \right) + \Pi\left( \frac{165-\sigma}{\mu} \right) - 1 = 0.38$
|
||||
|
||||
$P\left( \frac{180-\sigma}{\mu} \leq \frac{X-\sigma}{\mu} \right) = 0.04 \iff \Pi\left( \frac{180-\sigma}{\mu} \right) = 0.96$
|
||||
|
||||
Donc, on a les égalités suivantes :
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
\begin{cases}
|
||||
\frac{165-\sigma}{\mu} = 0.21\\
|
||||
\frac{180-\sigma}{\mu} = 1.75
|
||||
\end{cases}
|
||||
&\iff
|
||||
\begin{cases}
|
||||
0.21\mu + \sigma = 165 \\
|
||||
1.75\mu + \sigma = 180
|
||||
\end{cases} \\
|
||||
&\iff
|
||||
\begin{cases}
|
||||
1.54\mu = 15 \\
|
||||
1.75\mu + \sigma = 180
|
||||
\end{cases} \\
|
||||
&\iff
|
||||
\begin{cases}
|
||||
\mu = 9.74 \\
|
||||
17.05 + \sigma = 180
|
||||
\end{cases} \\
|
||||
& \iff
|
||||
\begin{cases}
|
||||
\mu = 9.74 \\
|
||||
\sigma = 162.95
|
||||
\end{cases}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
|
||||
|
@ -14,6 +14,7 @@ kung_fu: 0
|
||||
> where file.mtime > date(this.file.name) and file.mtime < (date(this.file.name) + dur(1 day)) sort file.mtime asc
|
||||
> ```
|
||||
|
||||
|
||||
## Devoirs
|
||||
> [!smalltodo]+ Devoirs
|
||||
> ```dataview
|
||||
|
33
daily/2023-10-24.md
Normal file
33
daily/2023-10-24.md
Normal file
@ -0,0 +1,33 @@
|
||||
---
|
||||
spaced_repetition: 0
|
||||
kung_fu: 0
|
||||
---
|
||||
|
||||
## Todo
|
||||
- spaced repetition : `INPUT[toggle(onValue(1), offValue(0)):spaced_repetition]`
|
||||
- kung-fu : `INPUT[number:kung_fu]` minutes
|
||||
|
||||
|
||||
## I did
|
||||
> [!smallquery]- Modified files
|
||||
> ```dataview
|
||||
> LIST file.mtime
|
||||
> where file.mtime > date(this.file.name) and file.mtime < (date(this.file.name) + dur(1 day)) sort file.mtime asc
|
||||
> ```
|
||||
|
||||
## Devoirs
|
||||
> [!smalltodo]+ Devoirs
|
||||
> ```dataview
|
||||
> TABLE difficulty as "", due as "date", title as "description", file.etags as "tags"
|
||||
> FROM #devoir
|
||||
> WHERE contains(due, date(this.file.name))
|
||||
> ```
|
||||
> > [!done]- Devoirs faits
|
||||
> > ```dataview
|
||||
> > TABLE difficulty as "", due as "date", title as "description"
|
||||
> > FROM #devoir-fait
|
||||
> > WHERE contains(due, date(this.file.name))
|
||||
> > ```
|
||||
|
||||
## I am gratefull to
|
||||
|
@ -1,3 +1,81 @@
|
||||
#fac #PM
|
||||
|
||||
versa-tile
|
||||
|
||||
# Le business model : Elaboration du projet de création d’entreprise : Etapes à suivre
|
||||
|
||||
## 1. Définir le projet
|
||||
### Description de l'activité
|
||||
|
||||
- particularités
|
||||
- caractéristiques techniques, présentation, caractère innovant du produit
|
||||
- caractéristiques de l'offre sur le marché
|
||||
- marque
|
||||
- éléments recherchés par les clients
|
||||
- positionnement sur le marché
|
||||
- distinguer l'offre principale et les services associés
|
||||
|
||||
- caractéristiques techniques, présentation, particularités
|
||||
- ordinateurs portables
|
||||
- adaptables
|
||||
- s'adapter au besoin de chaque utilisateur
|
||||
- s'adapter à différents contextes d'utilisation pour un même utilisateur
|
||||
- entièrement remplacable
|
||||
- réparable façilement
|
||||
- durable (à vie)
|
||||
- l'utilisateur peut changer d'utilisation sans racheter un ordinateur
|
||||
- caractère innovant
|
||||
- remplace le modèle des ordinateurs portables qui ne sont pas réparables
|
||||
- permet une plus grande versatilité
|
||||
- l'ordinateur versa-tile peut s'adapter à chaque situation
|
||||
- plus durable que de racheter un ordinateur complet
|
||||
- caractéristiques de l'offre sur le marché
|
||||
- marque
|
||||
-
|
||||
|
||||
|
||||
## 2. L’étude de marché
|
||||
|
||||
|
||||
|
||||
Vous devez réaliser une étude de marché. Les sujets à étudier concernent :
|
||||
|
||||
- **La demande (clientèle potentielle)**
|
||||
- nombre de clients
|
||||
- distribution géographique des clients
|
||||
- fréquence de consommation des produits
|
||||
- budget consacré
|
||||
- **L’offre (la concurrence)**
|
||||
- concurrents directs
|
||||
- concurrents indirects
|
||||
- nombre de concurrents
|
||||
- **Les fournisseurs**
|
||||
- quels sont les fournisseurs potentiels
|
||||
- **La réglementation**
|
||||
- lois / normes qui régissent l'activité
|
||||
- **L’environnement (exploiter outil PESTEL)**
|
||||
- L’évolution des habitudes de consommations est-elle favorable au projet ? Pourquoi ?
|
||||
|
||||
|
||||
|
||||
## 3. Définir le mode de distribution / méthode de prospection
|
||||
|
||||
- Où les clients pourront-ils acheter votre produit ou faire appel à votre service ?
|
||||
- Quel canal de distribution avez-vous retenu (détaillant, grossistes, GMS commerces de proximité, réseau de franchisés, etc…)
|
||||
- Quels sont les différents moyens de communication que vous allez utiliser pour faire connaître votre entreprise (relations publiques, relations presse,, support de communication)
|
||||
|
||||
|
||||
|
||||
## 4. L’identification de votre entreprise
|
||||
|
||||
- Identifiez l'entreprise (et justifier cette identification)
|
||||
- un nom
|
||||
- une marque
|
||||
- un logo
|
||||
|
||||
Expliquez votre démarche pour la protection juridique de ce nom et de ce logo
|
||||
|
||||
## 5. La forme juridique de votre entreprise
|
||||
|
||||
- Quelle forme juridique choisie
|
||||
- Présentez-la et expliquez votre choix
|
||||
|
10
r.md
Normal file
10
r.md
Normal file
@ -0,0 +1,10 @@
|
||||
---
|
||||
|
||||
mindmap-plugin: rich
|
||||
|
||||
---
|
||||
|
||||
# statistiques
|
||||
``` json
|
||||
{"theme":"","mindData":[[{"id":"dac65e9c-d82f-61ad","text":"statistiques","isRoot":true,"main":true,"x":4000,"y":4000,"isExpand":true,"layout":{"layoutName":"multipleTree","direct":"right"},"stroke":""},{"id":"082b6c3f-738b-d8e3","text":"Loi normale","stroke":"darkorange","x":4104,"y":4084,"layout":null,"isExpand":true,"pid":"dac65e9c-d82f-61ad"},{"id":"6ba52f6b-6d5d-4e2d","text":"propriétés","stroke":"darkorange","x":4184,"y":4147,"layout":null,"isExpand":true,"pid":"082b6c3f-738b-d8e3"},{"id":"13e702d2-5e6d-46ad","text":"Pour une loi centrée normalisée, on peut utiliser $\\Pi$","stroke":"darkorange","x":4184,"y":4377,"layout":null,"isExpand":true,"pid":"082b6c3f-738b-d8e3"},{"id":"35ca61fd-e965-c461","text":"$P(X < - \\alpha) = P(X > \\alpha)$","stroke":"darkorange","x":4243.5,"y":4193,"layout":null,"isExpand":true,"pid":"6ba52f6b-6d5d-4e2d"},{"id":"d06929d7-1fb9-d945","text":"$P(X \\leq \\alpha ) = 1-P(X \\geq a)$","stroke":"darkorange","x":4243.5,"y":4285,"layout":null,"isExpand":true,"pid":"6ba52f6b-6d5d-4e2d"},{"id":"c0937930-10a7-8d57","text":"inverser le signe $\\iff$ inverser le sens du $>$","stroke":"darkorange","x":4354,"y":4239,"layout":null,"isExpand":true,"pid":"35ca61fd-e965-c461"},{"id":"7d17bad0-7ec9-b73f","text":"inverser le sens du $>$ $\\iff$ faire $1-P$","stroke":"darkorange","x":4360.5,"y":4331,"layout":null,"isExpand":true,"pid":"d06929d7-1fb9-d945"}]],"induceData":[],"wireFrameData":[],"relateLinkData":[],"calloutData":[]}
|
||||
```
|
BIN
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sources/cours/S5/L3_info_statistiques_ch0.pdf
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sources/cours/S5/L3_info_statistiques_tableau_loi_normale.pdf
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Loading…
x
Reference in New Issue
Block a user