书中给出的程序为:
limite.central <- function (r=runif, distpar=c(0,1), m=.5,
s=1/sqrt(12),
n=c(1,3,10,30), N=1000) {
for (i in n) {
if (length(distpar)==2){
x <- matrix(r(i*N, distpar[1],distpar[2]),nc=i)
}
else {
x <- matrix(r(i*N, distpar), nc=i)
}
x <- (apply(x, 1, sum) - i*m )/(sqrt(i)*s)
hist(x,col='light blue',probability=T,main=paste("n=",i),
ylim=c(0,max(.4, density(x)$y)))
lines(density(x), col='red', lwd=3)
curve(dnorm(x), col='blue', lwd=3, lty=3, add=T)
if( N>100 ) {
rug(sample(x,100))
}
else {
rug(x)}}}
为了在验证当n足够大时,beta(0.5,0.5)分布的均值近似服从正态分布,我写下了
op <- par(mfrow=c(2,2))
limite.central(rbeta, distpar=c(0.5,0.5), m=0.5, s=sqrt(0.125), n=c(3, 10, 30 ,50))
par(op)
最后得到图像为
请问所出的问题在哪里?