Question

Given here is the joint probability function associated with data obtained in a study of the auto-mobile accident in which a child (under age 5 years) was in the car and at least one fatality occurred. Especially, the study focused on whether or not the child survived and what type of seatbelt (if any) he or she used, yı = 0, if the child survived, yi = 1 if not and y2 = 0, if no belt used y2 = 1, if adult belt used y2 = 2, if car - seat belt used yi is the number of fatalities per child and, since children's car seats usually utilize two belts, Y2 is the number of seatbelts in use at the time of the accident,

Answer 1

No. of Fatalities per Child If child If child doesn't survive survives No. of seat belts in use Y1=0 Y=1 No seat belt used Y2=0 0.38 0.17 0.55 Adult seat belt used Y2=1 0.14 0.02 0.16 Car seat belt used Y=2 0.24 0.05 0.29 0.76 0.24 1.00

(a). Therefore, Marginal probability distribution for Fatalities :-

If child survives    => P (Y1 = 0) = 0.76

If child doesn't survive => P (Y1 = 1) = 0.24

(b). And,  Marginal probability distribution for No. of seat belts in use :-

If no seat belt used => P (Y2 = 0) = 0.55

If adult seat belt used => P (Y2 = 1) = 0.16

If car seat belt used => P (Y2 = 2) = 0.29

(c). Now, P [ (Y2 =1) / (Y1 = 0) ] = 0.14 / 0.76 = 0.184

(d). and, E [ Y2 / (Y1 = 0)] = [ (0 x 0.38 / 0.76 ) + (1 x 0.14 / 0.76 ) + (2 x 0.24 / 0.76 ) ]

= 0 + 0.184 + 0.632 = 0.816

(e). Expected value of no. of seat belts in use :-

E (Y2) = [(0 x 0.55) + (1 x 0.16) + (2 x 0.29)] = 0 + 0.16 + 0.58 = 0.74

(f). Expected value of no. of fatalities per child :-

E (Y1) = [(0 x 0.76) + (1 x 0.24)] = 0 + 0.24 = 0.24