Distribution function

X 2 3 4
P 0.35 0.35 0.3

The data in this table do I calculate the distribution function F(x) and then probability p(2.5 <ξ <3.25) p(2.8 <ξ) and p(3.25> ξ)

Correct result:

p1 =  0.35
p2 =  0.3
p3 =  0.65

Solution:

F2=0.3 F3=F2+0.35=0.3+0.35=1320=0.65 F4=F3+0.35=0.65+0.35=1 p1=0.35=720F_{2}=0.3 \ \\ F_{3}=F_{2}+0.35=0.3+0.35=\dfrac{ 13 }{ 20 }=0.65 \ \\ F_{4}=F_{3}+0.35=0.65+0.35=1 \ \\ p_{1}=0.35=\dfrac{ 7 }{ 20 }
p2=0.3=310p_{2}=0.3=\dfrac{ 3 }{ 10 }
p3=0.3+0.35=1320=0.65p_{3}=0.3+0.35=\dfrac{ 13 }{ 20 }=0.65



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