------------------------------------------------------------------------ [软体程式类别]: R [程式问题]: 绘图 [软体熟悉度]: 中(3个月到1年) [问题叙述]: 我画了一个标准常态分布图，想要在正负一个标准差间的面积部份，画上斜线 标示出这68%的面积，请问要如何做?谢谢 [程式范例]: rm(list=ls()) x <- seq(-4,4,length=1000) f.x <- dnorm(x, mean=0, sd=1) plot(x, f.x, type = "l", lty= 1, axes = F, lwd = 1, xlab = "z value", ylab="") axis(1, at=c(-4,-3,-2,-1,0,+1,+2,+3,+4),tick=T,pos=0, label=c("-4","-3","-2","-1","mu","+1","+2","+3","+4")) lines(c(-1,-1),c(0, dnorm(-1)),lty=3) lines(c(1,1),c(0, dnorm(1)),lty=3) lines(c(-2,-2),c(0, dnorm(-2)),lty=1) lines(c(2,2),c(0, dnorm(2)),lty=1) #points(-1,dnorm(-1),pch=19) #points(1,dnorm(1),pch=19) ----------------------------------------------------------------------------- --

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