Parte 1: Modelo Teórico

media <- (9+11)/2
media
## [1] 10
varianza <- (11-9)^2/12
varianza
## [1] 0.3333333
p <- 1/2*(0.15)
p
## [1] 0.075

Parte II

unif1 <- runif(50,9,11)
unif2 <- runif(100,9,11)
unif3 <- runif(300,9,11)
unif4 <- runif(500,9,11)
unif5 <- runif(1000,9,11)
unif6 <- runif(8000,9,11)
par(mfrow=c(2,3))
hist(unif1,probability=TRUE)
hist(unif2,probability=TRUE)
hist(unif3,probability=TRUE)
hist(unif4,probability=TRUE)
hist(unif5,probability=TRUE)
hist(unif6,probability=TRUE)

  1. Vemos que el histograma para la muestra 6 se parece más al bosquejo.
mean(unif1)
## [1] 10.06801
mean(unif2)
## [1] 10.02805
mean(unif3)
## [1] 10.02453
mean(unif4)
## [1] 9.976133
mean(unif5)
## [1] 10.01163
mean(unif6)
## [1] 10.00416
  1. El que mas se acerca es el de la muestra 6
var(unif1)
## [1] 0.3086139
var(unif2)
## [1] 0.3392907
var(unif3)
## [1] 0.3379029
var(unif4)
## [1] 0.3350902
var(unif5)
## [1] 0.347345
var(unif6)
## [1] 0.3402359
  1. el que más se acerca es la varianza de la muestra 6
vector1 <- unif1 <= 9.15
vector2 <- unif2 <= 9.15
vector3 <- unif3 <= 9.15
vector4 <- unif4 <= 9.15
vector5 <- unif5 <= 9.15
vector6 <- unif6 <= 9.15

p1 <- sum(vector1)/length(vector1)
p2 <- sum(vector2)/length(vector2)
p3 <- sum(vector3)/length(vector3)
p4 <- sum(vector4)/length(vector4)
p5 <- sum(vector5)/length(vector5)
p6 <- sum(vector6)/length(vector6)

p1
## [1] 0.04
p2
## [1] 0.08
p3
## [1] 0.06333333
p4
## [1] 0.07
p5
## [1] 0.069
p6
## [1] 0.07575
  1. el que mas se acerca es la probabilidad de la muestra 6

problema 2

p7 <- 1-pnorm(224,200,15)
p8 <- pnorm(205,200,15)-pnorm(190,200,15)
vasos <- (1-pnorm(230,200,15))*1000
p7
## [1] 0.05479929
p8
## [1] 0.3780661
vasos
## [1] 22.75013

aproximadamente 23 vasos

problema 3

p9 <- pgamma(10, 3, rate = 0.5, lower.tail = F)
p9
## [1] 0.124652

problema 4

p10 <- pweibull(10, 2, scale = 10, lower.tail = F)
p10
## [1] 0.3678794

problema 5

p11 <- pexp(10,rate=1/6,lower.tail = FALSE)
p11
## [1] 0.1888756
p12 <- pexp(10,rate=1/6,lower.tail = TRUE)-pexp(5,rate=1/6,lower.tail = TRUE)
p12
## [1] 0.2457226