The following is the joint probability distribution of number of car crashes (C)

ID: 3071893 • Letter: T

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)

following informtio has been regenerated

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

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

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