已知连续型随机变量X具有概率密度为f(x)={？
f(x)=Ax(1-x)^3 ; 0≤x≤1=0 ; elsewhere(1)∫(0->1) f(x) dx =1∫(0->1) Ax(1-x)^3 dx =1A∫(0->1) [-(1-x)+1](1-x)^3 dx =1A∫(0->1) [(1-x)^3 -(1-x)^4] dx =1A[ (1-x)^5/5 - (1-x)^4/4]|(0->1) =1A( 1/4-1/5)=1A=20(2)0≤x≤1F(x)=∫(0->x) f(t) dt=20∫(0->x) t(1-t)^3 dt=20∫(0->x) [(1-t)^3 -(1-t)^4] dt=20[ (1-t)^5/5 - (1...全部
f(x)=Ax(1-x)^3 ; 0≤x≤1=0 ; elsewhere(1)∫(0->1) f(x) dx =1∫(0->1) Ax(1-x)^3 dx =1A∫(0->1) [-(1-x)+1](1-x)^3 dx =1A∫(0->1) [(1-x)^3 -(1-x)^4] dx =1A[ (1-x)^5/5 - (1-x)^4/4]|(0->1) =1A( 1/4-1/5)=1A=20(2)0≤x≤1F(x)=∫(0->x) f(t) dt=20∫(0->x) t(1-t)^3 dt=20∫(0->x) [(1-t)^3 -(1-t)^4] dt=20[ (1-t)^5/5 - (1-t)^4/4]|(0->x)=20[ (1-x)^5/5 - (1-x)^4/4 - ( 1/5-1/4) ]=20[ (1-x)^5/5 - (1-x)^4/4 +1/20 ]=4(1-x)^5 - 5(1-x)^4 +1ieF(x)=0 ; x=4(1-x)^5 - 5(1-x)^4 +1 ; 0≤x≤1=1 ; x>1(3)P(0=F(1/2)=4(1/2)^5 - 5(1/2)^4 +1=1/8-5/16 +1=(18-5)/16=13/16。
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