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)

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=0.65=1320p_{ 3 } = 0.3+0.35 = 0.65 = \dfrac{ 13 }{ 20 }



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