Question

概率论，求大神解答！！！

Answer 1

Answer 2

f(x) = kx, ; 0≤x≤3
= 2 - x/2 ; 3<x≤4
=0 ; elsewhere

x≤3
F(x) = ∫(0->x) kt dt
= (1/2)kx^2
F(3) = (9/2)k

3<x≤4
F(x) = F(3) +∫(3->x) ( 2- t/2) dt
= (9/2)k + [ 2t -(1/4)t^2 ]|(3->x)
= (9/2)k + 2x -(1/4)x^2 - 6 + 9/4
= (9/2)k + 2x -(1/4)x^2 - 15/4

F(4) =1
(9/2)k + 2(4) -(1/4)4^2 - 15/4 =1
(9/2)k + 8 -4 - 15/4 =1
(9/2) k = 3/4
k = 1/6

F(x) = 0 ; x<0
= (1/12)x^2 ; 0≤x≤3
= -3 +2x -(1/4)x^2 ; 3<x<4
=1 ; x≥4

P(1≤X≤7/2)
= F(7/2) -F(1)
= (1/12) [ (7/12)^2 -1 ]
= 1/36