864 words - 4 pages

Each unit sells for $ 5 Lot size = 1000 units Therefore the earning per lot = $ 5000 There are two states of nature: 1st state S1 = defective units per lot 15% 2nd state S2 = defective units per lot 35% If no inspection for S1 number of units defective = 15% x 1000 = 150 Cost of replacement of defective unit sent to the customer = $10 per unit Therefore the pay off = 5000 Â¡V (150 x 10) = $ 3500 If no inspection for S2 number of units defective = 35% x 1000 = 350 Cost of replacement of defective unit sent to the customer = $10 per unit Therefore the pay off = 5000 Â¡V (350 x 10) = $ 1500 If 100% inspection for S1 sampling cost = 1000 x 2= 2000 Replacement cost = $0.50 ...view middle of the document...

15 0.6 0.7225 0.4335 0.7195 S2: p = 0.35 0.4 0.4225 0.1690 0.2805 TOTAL 0.6025 1.000 P (X=1/ p=0.15) = (2/1) x (0.15)^1 x (0.85)^1 = 0.255 P (X=1/ p=0.35) = (2/1) x (0.35)^1 x (0.65)^1 = 0.455 X = 1 State Prior probability Conditional probability of sample outcome given state Intersection of state and sample outcome Posterior probability S1: p = 0.15 0.6 0.255 0.1530 0.4567 S2: p = 0.35 0.4 0.455 0.1820 0.5433 TOTAL 0.3350 1.000 P (X=2/ p=0.15) = (2/2) x (0.15)^2 x (0.85)^0 = 0.0225 P (X=2/ p=0.35) = (2/2) x (0.35)^2 x (0.65)^0 = 0.1225 X = 2 State Prior probability Conditional probability of sample outcome given state Intersection of state and sample outcome Posterior probability S1: p = 0.15 0.6 0.0225 0.0135 0.2160 S2: p = 0.35 0.4 0.1225 0.0490 0.7840 TOTAL 0.0625 1.000 Action a 1= 100% sampling Action a 2 = no sampling EP (a1: X=0) = 0.7195 x 2925 + 0.2805 x 2825 = 2896.95 EP (a2: X=0) = 0.7195 x 3500 + 0.2805 x 1500= 2939 Select a2 pay off = 2939 EP (a1: X=1) = 0.4567 x 2925 +...

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