为了获得总体平均值的正态概率分布,我是否使用rnorm函数并仅给出标准误差而不是标准偏差?
# probability distribution of a
# population mean of 100 with SD of 10, n = 50.
# convert to standard error
se <- 10/(50^.5)
# The rnorm function accepts sd, so I just use the
# se in its place to get the desired distribution
meanDist <- rnorm(1000,mean=100,sd=se)
hist(meanDist)
答案 0 :(得分:0)
是的,虽然hist
结果将显示平均而非人口价值分布的代表性抽样。