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 $ 120 0.40 $ 110 0.35 $ 100 0.30 130 0.20 130 0.30 130 0.40 140 0.40 150 0.35 160 0.30 The expected value for all three years is $130. Compute the standard deviation for each of the three years. (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Answer 1

Year 1:

Expected Value = 130

Variance = 0.40 * (120 - 130)^2 + 0.20 * (130 - 130)^2 + 0.40 * (140 - 130)^2
Variance = 80

Standard Deviation = (80)^(1/2)
Standard Deviation = 8.94

Year 5:

Expected Value = 130

Variance = 0.35 * (110 - 130)^2 + 0.30 * (130 - 130)^2 + 0.35 * (150 - 130)^2
Variance = 280

Standard Deviation = (280)^(1/2)
Standard Deviation = 16.3

Year 10:

Expected Value = 130

Variance = 0.30 * (100 - 130)^2 + 0.40 * (130 - 130)^2 + 0.30 * (160 - 130)^2
Variance = 540

Standard Deviation = (540)^(1/2)
Standard Deviation = 23.24