An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows: F(x) =0 - x < 10.33 - 1 < x < 30.44 - 3 < x < 40.48 - 4 < x < 60.86 - 6 < x < 121 - 12 < xa. What is the pmf of X? x 1 3 4 6 12p(x)b. Using just the cdf, compute P(3 ≤ X ≤ 6) and P(4 ≤ X).P(3 < X < 6) =P(4 < X) =

Answer:a)The pmf of x isx     p(x)1      0.333     0.114     0.046     0.3812    0.14b)P(3≤X≤6)=0.53P(X≥4)=0.56Step-by-step explanation:a)We have to find pmf of x.We know that F(x)=P(X≤ x)F(0)=P(X≤ 0)=P(X=0)=f(0)F(1)=P(X≤ 1)=P(X=0)+P(X=1)=f(0)+f(1)As, f(0)=F(0), sof(1)=F(1)-F(0)F(2)=P(X≤ 2)=P(X=0)+P(X=1)+P(X=2)=F(1)+f(2)f(2)=F(2)-F(1)So, we can say thatf(3)=F(3)-F(2)f(4)=F(4)-F(3)f(6)=F(6)-F(5)f(12)=F(12)-F(11).We are given thatF(0)=0 ,F(1)=0.33 ,F(2)=0.33 ,F(3)=0.44, F(4)=0.48, F(5)=0.48, F(6)=0.86, F(7)=0.86, F(8)=0.86, F(9)=0.86, F(10)=0.86, F(11)=0.86, F(12)=1.We have to find f(1), f(3), f(4), f(6) and f(12).f(1)=F(1)-F(0)=0.33-0=0.33f(3)=F(3)-F(2)=0.44-0.33=0.11f(4)=F(4)-F(3)=0.48-0.44=0.04f(6)=F(6)-F(5)=0.86-0.48=0.38f(12)=F(12)-F(11)=1-0.86=0.14The pmf of x isx     p(x)1      0.333     0.114     0.046     0.3812    0.14b)P(3≤X≤6)=?We know that P(a<X≤b)=F(b)-F(a)P(3≤X≤6)=P(2<X≤6)=F(6)-F(2)=0.86-0.33=0.53P(3≤X≤6)=0.53P(X≥4)=?P(X≥4)=1-P(X<4)=1-P(X≤3)=1-F(3)=1-0.44=0.56P(X≥4)=0.56