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
$	50			0.40			$	30			0.35			$	20			0.30
	70			0.20				70			0.30				70			0.40
	90			0.40				110			0.35				120			0.30

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.)

Answer 1

>>>>

Variance of Year 1 = Σ (X - Expected Return)^2 * Probability

=  [[(50 - 70)^2*0.40] + [(70-70)^2*0.20] + [(90 - 70)^2*0.40]]

= [160 + 0+160]

= 320

Standard Deviation of Year 1 = Square root of Variance

= (320)^(1/2)

= 17.88854

Therefore, Standard Deviation of Year 1 is 17.89

>>>>

Variance of Year 5 = Σ (X - Expected Return)^2 * Probability

=  [[(30 - 70)^2*0.35] + [(70-70)^2*0.30] + [(110 - 70)^2*0.35]]

= [560 + 0 + 560]

= 1120

Standard Deviation of Year 5 = Square root of Variance

= (1120)^(1/2)

= 33.4664

Therefore, Standard Deviation of Year 5 is 33.47

>>>>

Variance of Year 10 = Σ (X - Expected Return)^2 * Probability

=  [[(20 - 70)^2*0.30] + [(70-70)^2*0.40] + [(120 - 70)^2*0.30]]

= [750 + 0 + 750]

= 1500

Standard Deviation of Year 10 = Square root of Variance

= (1500)^(1/2)

= 38.72983

Therefore, Standard Deviation of Year 10 is 38.73