The continuous r.v. X has the p.d.f:f(x)={( 1/2 x 0< X≤1 1/4(3-x) 1<X≤2 1/4 3<X≤4 0 otherwise)┤Find: P(1<X<3)
P(1<X<3)=∫_1^3〖f(x)〗 dx=1/4 ∫_1^2〖(3-x)〗 dx+∫_2^31/4 dx=1/4 [3x-1/2 x^2 ]_1^2+1/4==1/4+1/4 [6-2-3+1/2]=1/4 [5/2]=5/8
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P(1<X<3)=∫_1^3〖f(x)〗 dx=1/4 ∫_1^2〖(3-x)〗 dx+∫_2^31/4 dx
need an explanation for this answer? contact us directly to get an explanation for this answer=1/4 [3x-1/2 x^2 ]_1^2+1/4=
=1/4+1/4 [6-2-3+1/2]
=1/4 [5/2]=5/8