An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the past 3 years. The pmf of Y is: y 0 1 2 3
P(y) 0.50 0.25 0.20 0.05

Required:
a. Compute E(Y) and V(3Y−10)
b. Suppose an individual with Y violations incurs a surcharge of $(95Y2−25). Calculate the expected amount of the surcharge.
c. What is the standard deviation of surcharges?

a.

E(y)=0.8

V(3y-10)=7.74

b.

The expected amount of surcharge is 117.5.

c.

The standard deviation of surcharges is 217.6723.

Step-by-step explanation:

a.

y          0      1      2    3

p(y)    0.5 0.25 0.2 0.05

y*p(y)  0    0.25 0.4 0.15

E(y)=sum(y*p(y))=0+0.25+0.4+0.15

E(y)=0.8.

y²          0      1      4    9

y²p(y)   0 0.25 0.8 0.45

V(y)=E(y²)-[E(y)]²

E(y²)=sum(y²*p(y))=0+0.25+0.8+0.45=1.5

V(y)=E(y²)-[E(y)]²

V(y)=1.5-[0.8]²

V(y)=1.5-0.64

V(y)=0.86

V(3y-10)=9V(y)=9*0.86=7.74

V(3y-10)=7.74

b.

E(95Y²-25)=95E(Y²)-25=95*1.5-25=142.5-25=117.5

E(95Y²-25)=117.5

The expected amount of surcharge is 117.5.

c.

SD(95Y²-25)=?

SD(95Y²-25)=√V(95Y²-25)

V(95Y²-25)=95²*V(Y²)

V(95Y²-25)=9025*V(Y²)

V(Y²)=E[(Y²)²]-[E(Y²)]²

y^4          0      1      16    81

y^4*p(y)   0  0.25   3.2 4.05

E[(Y²)²]=sum(y^4*p(y))=7.5

V(Y²)=E[(Y²)²]-[E(Y²)]²

V(Y²)=7.5-[1.5]²

V(Y²)=7.5-2.25

V(Y²)=5.25

V(95Y²-25)=9025*V(Y²)

V(95Y²-25)=9025*5.25

V(95Y²-25)=47381.25

SD(95Y²-25)=√47381.25

SD(95Y²-25)=217.6723

The standard deviation of surcharges is 217.6723.

6 meters

step-by-step explanation: