设随机变量X的概率密度为f(x)

Question

如题所述

Answer 1

(1)
f(x)
=Ax ; 0≤x<1
=2-x ; 1≤x<2
=0 ; elsewhere
∫(0->1) Ax dx +∫(1->2) (2-x)dx =1
A/2 -(1/2) [(2-x)^2]|(1->2) =1
A/2 + 1/2 =1
A= 1
(2)
P(0.5≤X≤1.5)
=P(X≤1.5) -P(X<0.5)
=1/2+∫(1->1.5) (2-x) dx - ∫(0->0.5) x dx
=1/2-(1/2)[(2-x)^2]|(1->1.5) - 1/8
=1/2 -(1/2)( 1/4- 1 ) -1/8
=1/2 + 3/8 -1/8
=3/4
(2)
E(X)
=∫(0->1) x^2 dx +∫(1->2) x(2-x)dx
=(1/3)[x^3]|(0->1) + [x^2 - (1/3)x^3]|(1->2)
=1/3 + [(4-8/3)-(1-1/3)]
=1/3 + 2/3
=1
(3)
E(X^2)
=∫(0->1) x^3 dx +∫(1->2) x^2.(2-x)dx
=(1/4)[x^4]|(0->1) + [ (2/3)x^3 - (1/4)x^4]|(1->2)
=1/4 + [(16/3-4) -(2/3-1/4)]
=1/4 + ( 4/3 - 5/12)
=1/4 + 11/12
=14/12
=7/6
D(X)=E(X^2)-[E(X)]^2 =7/6 - 1= 1/6