假设我有一个像{1,2,3,...,23}这样的人口,我想生成一个样本,以便样本的平均值等于6 .
我尝试使用 sample
函数,使用自定义概率向量,但它不起作用:
population <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23)
mean(population)
minimum <- min(population)
maximum <- max(population)
amplitude <- maximum - minimum
expected <- 6
n <- length(population)
prob.vector = rep(expected, each=n)
for(i in seq(1, n)) {
if(expected > population[i]) {
prob.vector[i] <- (i - minimum) / (expected - minimum)
} else {
prob.vector[i] <- (maximum - i) / (maximum - expected)
}
}
sample.size <- 5
sample <- sample(population, sample.size, prob = prob.vector)
mean(sample)
样本的平均值大约是人口的平均值(振荡大约12),我希望它大约是6 .
一个很好的样本是:
-
{3,5,6,8,9},平均值= 6.2
-
{2,3,4,8,9},平均值= 5.6
问题与sample integer values in R with specific mean不同,因为我有一个特定的人口,我不能只生成任意实数,他们必须在人口中 .
概率向量图:
1 回答
你可以试试这个:
分解: