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 < 1
0.33 1 < x < 3
0.44 3 < x < 4
0.48 4 < x < 6
0.86 6 < x < 12
1 12 < x
(a) What is the pmf of X?
x 1 3 4 6 12
p(x)

(b) Using just the cdf, compute P(3 = X = 6) and P(4 = X).
P(3 < X < 6) =
P(4 <

X) =

a)

The pmf of x is

x     p(x)

1      0.33

3     0.11

4     0.04

6     0.38

12    0.14

b)

P(3≤X≤6)=0.53

P(X≥4)=0.56

Step-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), so

f(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 that

f(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 that

F(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=0.33

f(3)=F(3)-F(2)=0.44-0.33=0.11

f(4)=F(4)-F(3)=0.48-0.44=0.04

f(6)=F(6)-F(5)=0.86-0.48=0.38

f(12)=F(12)-F(11)=1-0.86=0.14

The pmf of x is

x     p(x)

1      0.33

3     0.11

4     0.04

6     0.38

12    0.14

b)

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=0.53

P(3≤X≤6)=0.53

P(X≥4)=?

P(X≥4)=1-P(X<4)=1-P(X≤3)=1-F(3)=1-0.44=0.56

P(X≥4)=0.56

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>