Question

he 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 In

The expected value for all three years is $70.

Compute the standard deviation for each of the three years. (Do not round intermediate calculations. Round your answer to 2 decimal places.)
  

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Final answer

Standard deviation
Year - 1 13.42
Year - 5 23.24
Year - 10 44.72

Explanation

Year - 1

Let X represents cash flow variables of Year - 1

Dx = Deviation from expected value = X - 70 ( note : expected value is given as 70 )

Dx2 = Dx * Dx

P = Probability

X Dx Dx2 P P * Dx2
55 -15 225 0.4 90
70 0 0 0.2 0
85 15 225 0.4 90
Sum of P * Dx2 = 180
Standard deviation 13.42

Standard deviation = Square Root of ( 180 ) = 13.42 ........... for Year - 1

Year - 5

X Dx Dx2 P P * Dx2
40 -30 900 0.3 270
70 0 0 0.4 0
100 30 900 0.3 270
Sum of P * Dx2 = 540
Standard deviation 23.24

Year - 10

X Dx Dx2 P P * Dx2
20 -50 2500 0.4 1000
70 0 0 0.2 0
120 50 2500 0.4 1000
Sum of P * Dx2 = 2000
Standard deviation 44.72
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