Question

Leann, with a $300,000 bequest from her father and a business degree from Athabasca University, is considering opening a gift shop in North Edmonton. If her shop is highly successful, she expects an annual net profit of $220,000. If the business is moderately successful, she expects $130,000. If not successful, she expects to have zero net profit. Under any circumstances, she is not contemplating any loss. Her anticipated probabilities of these three options are: 0.5, 0.3 and 0.2, respectively.

a) Calculate her expected net profit. Also calculate the standard deviation of her profit.

b)The business requires a $300,000 investment. If she has a 20% opportunity cost on invested funds of similar riskiness, should the project be undertaken?

c) Suppose Leann considers two alternative investment options instead of opening a gift shop. She has the option to buy a risk free asset that will pay 10%, or she can invest in a stock that has a 0.3 chance of paying 10%, a 0.2 chance of paying 22%, and a 0.5 chance of providing a 20% return. If she invests $160,000 in the stock and $140,000 in the risk free asset, determine the expected percentage return on the stock and the standard deviation.

Answer 1

(a) Making a probability density function table

Profit (x)	Probability	X * P(X = x)
If highly successful 220000	0.5	110000
If moderately successful 130000	0.3	39000
If unsuccessful 0	0.2	.0

Expected net profit = $\sum x*P(X=x)$

Expected net profit = 149000

Profit	probability	$x^{2}*P(X=x)$
220000	0.5	24200000000
130000	0.3	5070000000
0	0.2	0

Variance = $\sum x^{2}*P(X=x)-(E(x))^{2}$

= 29270000000 - $149000^{2}$

= 7069000000

SD = $\sqrt{Var}$

= 84077.3453

(b)

20% opportunity cost of the investment of $300000 = 20% * 300000

= $ 60000

This is less than the expected net profit in (a) which was $149000.

Therefore he should undertake the project

(c) This can be done similarly as the (a)

I am using absolute values for easy calculations

Return(%)	Pobability	x* P(X=x)
10	0.3	3
22	0.2	4.4
20	0.5	10

Expected % return = $\sum x*P(X=x)$

Expected % return = 17.4%

Variance = $\sum x^{2}*P(X=x)-(E(x))^{2}$   

= 326.8 - $17.4^{2}$

= 24.04 % %

SD = $\sqrt{Var}$

= 4.903%