Question

I want question 8 answered question 7 is posted because data from that question is required to answer 8

7. 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 A. Report the marginal probability distribution for C B. What is the average number of car crash? C. What is the variance of the number of crashes? D. Calculate σCM and ρCM.

8. Suppose car manufacturers are penalized (P) on the basis of the following formula P = 60,000 + 6C – 2M Using your answers for Question 7, calculate the following A. The average penalty (P) B. The variance of penalty (P)

Answer 1

7.

STEP 1

C	C=0	C=1	C=2	C=3	C=4	total
M=0	0.85	0.065	0.05	0.025	0.01	0.50
M=1	0.45	0.035	0.01	0.005	0.00	0.50
Total	0.80	0.10	0.06	0.03	0.01	1

STEP 2

A) probability distribution of C is

C_i P(C_i=C)

0 0.80

1 0.10

  2 0.06

  3 0.03

  4 0.01

B) average number of car crash

= 0* 0.8 + 1* 0.1 + 2* 0.06 + 3 * 0.03 + 4 * 0.01

= 0.35

C) variance of the number of crashes

V(C) = E(C²) - E²(C)

   E(C²)= 0² * 0.08+ !²* 0.1 + 2² * 0.06 +3² * 0.03 +4² * 0.01

= 0.77

V(C) = 0.77 - (0.35)² = 0.6475

D) Cov (C,M) = E (CM) - E(C) .E(M)

E(CM) = 0*0*0.35 + 0*1*0.065 + ..............+ 1*4*0.00

=0.07

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

= 0.5

cov(C,M) = 0.07 - 0.35 * 0.5

= - 0.105

8. STEP 3

suppose car manufacture are penalized (P) on the basis of the fallowing formula

P = 60,000 + 6 C - 2 M

A) the average penalty (P)

E(P) = E( 60000+ 6C-2M)

= 60000 + 6(0.35) - 2(0.5)

= 60001.1

B) the variance of penalty (P)

V(P) = V ( 60000 +6C-M)

=36 V (6) + 4 V (M) - 24 cov (M,C)

= 36* 0.6475 +4* 0.25 - 12 * 0.105

= 23.05