Capítulo 8 Pi04
8.1 Cargamos
# load("C:/Users/uo287610/Downloads/Framingham3.RData")
cor(Framingham[,c("DBP", "SBP")], use="complete")## DBP SBP
## DBP 1.000000 0.792146
## SBP 0.792146 1.0000008.2 Representación
Línea de regresión relaciona dos vaiables linealmente y=a+bx, a termino independiente y b coef regr: DBP = intercept (30.7)+ (0.40)* SBP. La aprox es buena por r=0.79. Al trabajar con dos variables=> R^2 = r^2.
plot(Framingham$DBP, Framingham$SBP) # falta reg line
# lm(Framingham$DBP, Framingham$SBP)
# x sobre y y viceversa seran iguales a sin r=1 8.3
cor(Framingham[,c("FRW", "DBP")], use="complete")## FRW DBP
## FRW 1.0000 0.3388
## DBP 0.3388 1.0000plot(Framingham$FRW, Framingham$DBP) # a donde corte ejey # flta reg line y lo tro
8.4 Comparar distribuciones discreta binomial
8.4.1 Distribución binomial B(10, 0.4) B(n, pi)
# rep indepe de experimento de bernu exto pi y 1-p fracaso
plot(0:10, dbinom(0:10, size=10, prob=.4)) # falta acumulado
# y tabal de probabilidaes# Calcular binomial
pbinom(c(7), size=10, prob=0.4, lower.tail=TRUE)## [1] 0.9877054# P(2<X<=5)
pbinom(c(5,2), size=10, prob=0.4, lower.tail=TRUE) # osumar 3 ## [1] 0.8337614 0.1672898# P(x>4) = 1-p(x<4) = 1-p(x<=3)
1-pbinom(c(3), size=10, prob=0.4, lower.tail=TRUE)## [1] 0.6177194# P(X>3)
pbinom(c(3), size=10, prob=0.4, lower.tail=F)## [1] 0.6177194# Cuantiles asociados a las probabilidades 0.1,0.5 y 0.75
qbinom(c(.1, .5, .75), size=10, prob=0.4, lower.tail=T)## [1] 2 4 5#
qbinom(c(.9, .5, .25), size=10, prob=0.4, lower.tail=F) #.25 a la dcha## [1] 2 4 5# Extraer muestras aleatorias
MBinom01 <- as.data.frame(matrix(rbinom(1*100, size=10, prob=0.4), ncol=100))
MBinom02 <- as.data.frame(matrix(rbinom(1*20, size=10, prob=0.4), ncol=20))8.5 Erjercicio gripe
# Tenemos una distribución binomial B(15, 0.3)
# P(x=0)
pbinom(c(0), size=15, prob=0.3)## [1] 0.004747562# P(x>2)
pbinom(c(2), size=15, prob=0.3, lower.tail=F)## [1] 0.8731723# P(3<=x<=5) = p(x<=5)-p(x<3)
# = p(x<=2)
t<-pbinom(c(5,2), size=15, prob=0.3)
t[1]-t[2]## [1] 0.5947937# Muestra 20, n=10, pi=0.2
MBinom03 <- as.data.frame(matrix(rbinom(1*20, size=10, prob=0.2), ncol=20))
# ^- máx 10 éxitos8.6 Distribución contínua
# Y -> N(20,2)
local({
.x <- seq(13.419, 26.581, length.out=1000)
plotDistr(.x, dnorm(.x, mean=20, sd=2), cdf=FALSE, xlab="x", ylab="Density",
main=paste("Normal Distribution: Mean=20, Standard deviation=2"))
}) # 99% - 3veces deviacion tipica hasta +3 deviaciom tipica
local({
.x <- seq(13.419, 26.581, length.out=1000)
plotDistr(.x, pnorm(.x, mean=20, sd=2), cdf=TRUE, xlab="x",
ylab="Cumulative Probability",
main=paste("Normal Distribution: Mean=20, Standard deviation=2"))
})# P(Y<=12)
pnorm(c(12), mean=20, sd=2)
# P(18<Y<20)
t<- pnorm(c(20,18), mean=20, sd=2)
t[1]-t[2]
# P(x>21)
1-pnorm(c(21), mean=20, sd=2) # mayor a algo = cola derecha?
pnorm(c(21), mean=20, sd=2, lower.tail = F)
# cuartiles de la distribución
qnorm(c(.25,.5,.75), mean=20, sd=2)
# muestar aleatorias
NS1 <- as.data.frame(matrix(rnorm(2*12, mean=20, sd=2), ncol=12))
rownames(NS0) <- paste("sample", 1:2, sep="")
colnames(NS0) <- paste("obs", 1:12, sep="")
NS0 <- within(NS0, {
mean <- rowMeans(NS0[,1:12])
})# N(106, 8)
# Inferior a 120
pnorm(c(120), mean=106, sd=8)## [1] 0.9599408# Porcentaje
(pnorm(c(130), mean=106, sd=8)-pnorm(c(90), mean=106, sd=8))*100## [1] 97.59# 1er cuartil q1=c0.25
qnorm(c(.25), mean=106, sd=8) # 255 diabeticos por debajo del resultado## [1] 100.6041# n=12, media5 y sd 3
NS2 <- as.data.frame(matrix(rnorm(1*12, mean=5, sd=3), ncol=12))
NS2<-as.data.frame(t(NS2))
summary(NS2)## V1
## Min. :-1.858
## 1st Qu.: 3.518
## Median : 5.629
## Mean : 5.270
## 3rd Qu.: 6.929
## Max. :12.702NS2 <- as.data.frame(matrix(rnorm(10^3, mean=5, sd=3), ncol=10^3))
NS2<-as.data.frame(t(NS2))
summary(NS2)## V1
## Min. :-4.286
## 1st Qu.: 2.866
## Median : 4.849
## Mean : 4.889
## 3rd Qu.: 7.013
## Max. :15.011