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
$	70			0.40			$	60			0.35			$	50			0.40
	90			0.20				90			0.30				90			0.40
	110			0.40				120			0.35				130			0.20

The expected value for all three years is $90.

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

Answer 1

Year 1:

Expected Value = $90

Variance = 0.40 * (70 - 90)^2 + 0.20 * (90 - 90)^2 + 0.40 * (110 - 90)^2
Variance = 320

Standard Deviation = (320)^(1/2)
Standard Deviation = $17.89

Year 5:

Expected Value = $90

Variance = 0.35 * (60 - 90)^2 + 0.30 * (90 - 90)^2 + 0.35 * (120 - 90)^2
Variance = 630

Standard Deviation = (630)^(1/2)
Standard Deviation = $25.10

Year 10:

Expected Value = $90

Variance = 0.40 * (50 - 90)^2 + 0.40 * (90 - 90)^2 + 0.20 * (130 - 90)^2
Variance = 960

Standard Deviation = (960)^(1/2)
Standard Deviation = $30.98