#opgave med kast af 5 terninger maxantoejne=rep(0,100000) for (i in 1:100000){ x=sample(c(1:6),5,replace=T) maxantoejne[i]=max(x) } z=maxantoejne<6 I=as.numeric(z) p=mean(I) #0.3933 lower=p-3*sd(I)/sqrt(100000) upper=p+3*sd(I)/sqrt(100000) lower #0.397628 upper #0.406932 upper-lower #0.009303932 #opgave med Poissonfordeling exp(-3) exp(-3)*3^1/1 exp(-3)*3^2/2 dpois(c(0:2),3) x=rpois(10000,3) I=x==0 mean(I) I=x==1 mean(I) I=x==2 mean(I) var(x) #2.94 #quite close to 3 z=x^2 mean(z) #12.0608 mean(z)-2*sd(z)/sqrt(length(z)) mean(z)+2*sd(z)/sqrt(length(z)) #11.71957 12.23883