Question

The Oklahoma Pipeline Company projects the following pattern of inflows from an investment. The inflows are spread over time to reflect delayed benefits. Each year is independent of the others. Year 1 Year 5 Year 10 Cash Inflow Probability Cash Inflow Probability Cash Inflow Probability $ 60 0.40 $ 50 0.30 $ 40 0.40 80 0.20 80 0.40 80 0.20 100 0.40 110 0.30 120 0.40 The expected value for all three years is $80. Compute the standard deviation for each of the three years. (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Answer 1

Expected value = Year 1 Cash flow (3) Probability P) D: (x-expected value) Dx 2 P*Dx 2 600.4 -20 400 160 80 0.21 10 이 이 100 0.41 20 400 160 320 Standard deviation= sqrt(P*Dx 2) = 17.88854382 Year 2 Cash flow Probability Dx (x-expected value) Dx 2P*Dx 2 500.3 -30 900 270 801 0.4 0 0 0 1101 0.3 30 9001 270| 5:40 Standard deviation = sqrt(SPDx 2) = 23.23790008 Year 3 Cash flow Probability Dx (x-expected value) Dx 2 P*Dx 2 400.4 -40 1600 640 80 0.21 10 이 이 120 4 0. 4 0 1600 640 1280 Standard deviation = sqrt(SPDx 2) = 35.77708764