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
$	35			0.40			$	20			0.35			$	10			0.40
	60			0.20				60			0.30				60			0.40
	85			0.40				100			0.35				110			0.20

The expected value for all three years is $60.

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

This is how the entire question is posed. No other data available.

Answer 1

The standard deviation can be calculated as given below

We first calculate the mean or expected value = Sum of all Pi x Ri

We then calculate the Devioation from the mean, substracting the mean from Ri

We then square the deviations from mean

WE then multiply the mean deviatin squares with the probability.

To calculate Variance, we add the above value for each probability

Standard Devbiation = (Varianvce)^1/2

For the year 1

Probability (Pi)	1 year Cash Flow (Ri)	Pi x Ri	Deviation from Mean(Ri- Rm)	Square of Deviation from Mean (Ri-Rm)^2	(Ri - Rm)^2 x Pi
0.40	35.00	14.00	(46.00)	2,116	846
0.20	60.00	12.00	(48.00)	2,304	461
0.40	85.00	34.00	(26.00)	676	270
Total		60.00		5,096	1,578
Mean Return (Rm)	= SUM(Pi x Ri)	60.00
Variance	(= Sum(Ri-Rm)^2 x Pi)	1,577.60
SD	(=Variance)^1/2	39.72

The mean or expected value = 60 &

Standard deviation = 39.72

For the year 5

Probability (Pi)	5th year Cash Flow (Ri)	Pi x Ri	Deviation from Mean(Ri- Rm)	Square of Deviation from Mean (Ri-Rm)^2	(Ri - Rm)^2 x Pi
0.35	20.00	7.00	(53.00)	2,809	983
0.30	60.00	18.00	(42.00)	1,764	529
0.35	100.00	35.00	(25.00)	625	219
Total		60.00		5,198	1,731
Mean Return (Rm)	= SUM(Pi x Ri)	60.00
Variance	(= Sum(Ri-Rm)^2 x Pi)	1,731.10
SD	(=Variance)^1/2	41.61

The mean is 60 and Standard deviation = 41.61

For the year 10

Probability (Pi)	10th year Cash Flow (Ri)	Pi x Ri	Deviation from Mean(Ri- Rm)	Square of Deviation from Mean (Ri-Rm)^2	(Ri - Rm)^2 x Pi
0.40	20.00	8.00	(46.00)	2,116	846
0.40	60.00	24.00	(30.00)	900	360
0.20	110.00	22.00	(32.00)	1,024	205
Total		54.00		4,040	1,411
Mean Return (Rm)	= SUM(Pi x Ri)	54.00
Variance	(= Sum(Ri-Rm)^2 x Pi)	1,411.20
SD	(=Variance)^1/2	37.57

Mean is 54 (This however does not match with the observation in question that all expected values are 60)

Standard Deviation = 37.57