Exercice 1

Soit \(X_A\) (resp. \(X_B\)) le nombre de germinations de la variété A (resp. B).

1.1

On a \(X_A\sim\mathcal{B}(20,.2)\) et \(X_B\sim\mathcal{B}(40,.1)\).

1.2

\(\Pr(3\leq X_A\leq 5)\)

require(dplyr)
## Loading required package: dplyr
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
(pbinom(5,20,.2)-pbinom(2,20,.2))%>%round(4)
## [1] 0.5981

1.3

\(\Pr( X_B\geq 5)=\Pr( X_B> 4)\)

pbinom(4,40,.1,lower.tail = F)%>%round(4)
## [1] 0.371

1.4

\(\Pr( X_B= 6)=\) 0.1068.

dbinom(6,40,.1)%>%round(4)
## [1] 0.1068

1.5

On veut calculer \[ \Pr(X_A=X_B)=\sum_{x=0}^{20}\Pr(X_A=x)\Pr(X_B=x) \]

dbinom(0:20,20,0.2)%*%dbinom(0:20,40,0.1)%>%round(4)
##        [,1]
## [1,] 0.1537

1.6

On veut calculer \[ \Pr(X_A>X_B)=\sum_{x=1}^{20}\Pr(X_A=x)\Pr(X_B\leq x-1) \]

dbinom(1:20,20,.2)%*%pbinom(0:19,40,.1)%>%round(4)
##        [,1]
## [1,] 0.4266

Seconde méthode \[ \Pr(X_A>X_B)=\sum_{x=0}^{19}\Pr(X_A>x)\Pr(X_B= x) \]

pbinom(0:19,20,0.2,lower.tail = F)%*%dbinom(0:19,40,0.1)%>%round(4)
##        [,1]
## [1,] 0.4266

1.7

On veut calculer \[ \Pr(X_B>X_A)=\sum_{x=1}^{40}\Pr(X_A\leq x-1)\Pr(X_B= x) \]

pbinom(0:39,20,0.2)%*%dbinom(1:40,40,0.1)%>%round(4)
##        [,1]
## [1,] 0.4197

Seconde méthode \[ \Pr(X_B>X_A)=\sum_{x=0}^{20}\Pr(X_A= x)\Pr(X_B>x) \]

dbinom(0:20,20,0.2)%*%pbinom(0:20,40,0.1,lower.tail = F)%>%round(4)
##        [,1]
## [1,] 0.4197

1.8

\[ \Pr(X_B>10)\geq 0.99 \] On cherche par dichotomie

pbinom(10,197,0.1,lower.tail = F)
## [1] 0.9904094

On trouve \(197\) graines.

Exercice 2

\(D\sim \mathcal{N}(10000,600)\).

2.1

\(\Pr(D>11000)\)

pnorm(11000,10000,600,lower.tail = F)%>%round(4)
## [1] 0.0478

2.2

\(\Pr(9000<D<10400)\)

(pnorm(10400,10000,600)-pnorm(9000,10000,600))%>%round(4)
## [1] 0.6997

2.3

Soit \(N\) le nombre de machines qui durent plus longtemps que 10500. Elle suit une loi binomiale de paramètres \(n=10\) et de probabilité \(p=\Pr(D>10500)\). On cherche \(\Pr(N\geq 2)=\Pr(N> 1)\).

p=pnorm(10500,10000,600,lower.tail=F)
pbinom(1,10,p,lower.tail=F)%>%round(4)
## [1] 0.6312

2.4

\(\Pr(D>x)=0.1\)

qnorm(0.1,10000,600,lower.tail = F)%>%round()
## [1] 10769

2.5

\(\Pr(D<x)=0.1\)

qnorm(0.1,10000,600,lower.tail = T)%>%round()
## [1] 9231

Exercice 3

3.1

library(readxl)
Exercice3 <- read_excel("Exercice3.xlsx")
Exercice3$Cost=8*Exercice3$Temps
head(Exercice3)

3.2

hist(Exercice3$Cost)

3.3

Test<-t.test(Exercice3$Cost,conf.level = 0.95)
Test$conf.int
## [1] 54.85957 55.68976
## attr(,"conf.level")
## [1] 0.95

3.4

Hyptest<-t.test(Exercice3$Cost,mu=55,alternative = "greater")
Hyptest$p.value
## [1] 0.09570724

Comme la p-value est supérieure à 5%, on ne rejette par l’hypothèse nulle. Il n’y a pas suffisamment de preuve statistique pour conclure qu’en moyenne, le coût d’une livraison est supérieure à 55 euros.