概率学,比较有难度,请对自身有信心的能人进来帮帮我.题目是英文,请见图片,我大致翻译下...某公司生

概率学,比较有难度,请对自身有信心的能人进来帮帮我.题目是英文,请见图片,我大致翻译下...某公司生

(a) g(x)=x, for x<L g(x)=x-L for x≥LSee the attached figure for the plot assuming L=2E[y]=μy=∫(0,+∞)g(x)f(x)dx =∫(0,L)(xf(x)dx+∫(L,+∞( (x-L)f(x)dx =∫(0,+∞)xf(x)dx-∫(L,+∞)Lf(x)dx =μx-L+L∫(0,L)f(x)dx (b)Suppose x follows a normal distribution with mean μx and variance (σx)^2,we want to show that there exists a value μ0 of μx that minimizes μyμy=μx-L+L∫(0,L)f(x)dx for normal distribution, f(x)={1/[(2π(σx)^2)^(1/2)]}*exp{-(L-μx)^2/[2(σx)^2]To minimize μx, we set d μy/ d μx=0=1-L f(x) to solve for μx=μ0, so that there exists a value of μx=μ0 that minimizes μy(c)L=2m, σx=0.02m,0=1-2*{1/[(2π(0.02)^2)^(1/2)]}*exp{-(2-μx)^2/[2(0.02)^2]=1-100*exp{-(2-μx)^2/[2(0.02)^2]value of μx that minimizes the amount of lost material is μx=μ0=2±0.0543

哦，回答的不错