Question

设随机变量X的概率密度为f(x)={x,0<=x<1;2-x,1<=x<2;0,其他}求P{x>=1/2}，P{1/2<X<3/2}步骤写清楚些，谢谢

Answer 1

(0≤x≤1)
F(x)=∫(-∞，+∞)f(x)dx
=∫[0,x]f(t)dt
=∫[0,x]tdt
=t^2/2[0,x]
=x^2/2
1≤x≤2时
F(x)=∫(-∞，+∞)f(x)dx
=∫[0,1]tdt +∫[1,x](2-t)dt
=1/2+(2t-t^2/2)[1,x]
=1/2+2x-x^2/2-2+1/2
=2x-x^2/2-1
x≥0时
F(x)=1

P{x>=1/2}=1-F(1/2)=1-(1/2)^2/2=7/8

P{1/2<X<3/2}=F(3/2)-F(1/2)=2*3/2-(3/2)^2/2-1-(1/2)^2/2=3/4

Answer 2

如题，f(x)={x,0<=x<1;2-x,1<=x<2;0,其他}
P{x>=1/2}=P{1>x>=1/2}+P{2>x>=1}2+P{x>=2}=∫ X从1/2到1+∫ （2-X）从1到2+0=7/8

P{1/2<X<3/2}=P{1>x>=1/2}+P{3/2>x>=1}=∫ X从1/2到1+∫ （2-X）从1到3/2=3/4