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 | ||||||||||||||
$ | 90 | 0.30 | $ | 80 | 0.30 | $ | 60 | 0.40 | |||||||||||
100 | 0.40 | 100 | 0.40 | 100 | 0.20 | ||||||||||||||
110 | 0.30 | 120 | 0.30 | 140 | 0.40 | ||||||||||||||
The expected value for
all three years is $100.
Compute the standard deviation for each of the three years.
(Do not round intermediate calculations. Round your answer
to 2 decimal places.)
Standard Deviation | |
Year 1 | |
Year 5 | |
Year 10 |
Year 1:
Expected Value = $100
Variance = 0.30 * (90 - 100)^2 + 0.40 * (100 - 100)^2 + 0.30 *
(110 - 100)^2
Variance = 60
Standard Deviation = (60)^(1/2)
Standard Deviation = $7.75
Year 5:
Expected Value = $100
Variance = 0.30 * (80 - 100)^2 + 0.40 * (100 - 100)^2 + 0.30 *
(120 - 100)^2
Variance = 240
Standard Deviation = (240)^(1/2)
Standard Deviation = $15.49
Year 10:
Expected Value = $100
Variance = 0.40 * (60 - 100)^2 + 0.20 * (100 - 100)^2 + 0.40 *
(140 - 100)^2
Variance = 1,280
Standard Deviation = (1,280)^(1/2)
Standard Deviation = $35.78
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