Question

The following is the joint probability distribution of number of car crashes (C) and car make (M).

C = 0                          C = 1 C = 2 C = 3 C = 4

TOYOTA (M = 0)            0.35                          0.065                       0.05                          0.025                       0.01

OTHER (M = 1)              0.45                          0.035                       0.01                          0.005                       0.00

Question 1 Part A: What is the variance of the number of crashes?

Question 1 Part B: Calculate the covariance between C and M.

Question 1 Part C: What is the average number of crashes if the car make is Toyota?

Question 1 Part D:

Suppose car manufacturers are penalized (P) on the basis of the following formula

P = 60,000 + 6C – 2M

Caculate: The Average Penalty (P) and the Variance of Penalty (P)

Answer 1

following informtio has been regenerated

				C
		0	1	2	3	4	total
M	0	0.35	0.065	0.05	0.025	0.01	0.5
	1	0.45	0.035	0.01	0.005	0	0.5
	total	0.8	0.1	0.06	0.03	0.01	1

E(C)=sum(c*P(C))=0*0.8+1*0.1+2*0.06+3*0.03+4*0.01= 0.35

E(C²)=sum(c*c*P(C))=0*0*0.8+1*1*0.1+2*2*0.06+3*3*0.03+4*4*0.01= 0.77

E(M)=0*0.5+1*0.5=0.5

E(M²)=0*0*0.5+1*1*0.5=0.5

E(CM)=0*0*0.35+0*1*0.065+0.*2*0.05+0*3*0.025+0*4*0.01+ 1*0*0.35+1*1*0.065+1*2*0.05+1*3*0.025+1*4*0.01=0.28

Var(M)=E(M2)-E(M)*E(M)=0.5-0.5*0.5=0.25

(part A) Variance(C)=E(C²)- E(C)*E(C)=0.77-0.35*0.35=0.6475

(Part B) Cov(C,M)=E(CM)-E(C)E(M)=0.28-0.35*0.5=0.105

(part C) E(C|M=0)=sum(c*P(C|M=0))=0*0.35+1*0.065+2*0.05+3*0.025+4*0.01=0.28

(part D)P = 60,000 + 6C – 2M

E(P)=E( 60,000 + 6C – 2M)=60000+6*0.35-2*0.5=60001.1

Var(P)=Var(60000+6C-M)=36Var(C)+4Var(M)-24Cov(M,C)=36*0.6475+4*0.25-12*0.105=23.05