Ann completed her bachelors degree from Athabasca University

Ann completed her bachelor’s degree from Athabasca University while working as the manager of real estate investment firm in Edmonton. Ann must decide if her company should invest in a strip mall in northwest Edmonton. If the shopping center is highly successful, after tax profits will be $200,000 per year. Moderate success would yield an annual profit of $90,000, while the project will lose $25,000 per year if it is unsuccessful. Past experience suggests that there is a 30% chance that the project will be highly successful, a 50% chance of moderate success, and a 20% probability that the project will be unsuccessful.

a. Calculate the expected value and standard deviation of profit.

b.Suppose this project requires an $800,000 investment. If the above business has a 9% opportunity cost on invested funds in another investment that has no risk, should the project be undertaken? Explain why or why not.

c.Suppose that after completing her degree, Ann was offered a new job opportunity that is based on sales commission. In it, she can earn $350 a day with a 30% probability, $400 with a 40% probability, or $250 with a 30% probability. Calculate the expected value and variance of her earnings, and interpret the results.

Solution

Ans a)

Expeced Value of Project = Sum of (Probability of event* Outcome of event)

=30%*$200,000+50%*$90000+20%*(-$25000)=$60000+$45000-$5000=$100000

Standard Deviation = (Expected Value (Outcome of Project^2)-(Expected Value(Outcome of Project))^2

(Expected Value (Outcome of Project^2)=Sumof (Probability of outcome*Outcome^2)

=0.30*(200000^2)+0.5(90000^2)+0.2(-25000^2)=$15925000000

Standard Deviation=sqrt(15925000000-10000000000)=$76974.02

Ans b)

If Investment required for project is 800,000 then Return on Investment is (Expected Value of Profit each year /Investment)=(100000/800000)=12.5%

Hence Profit margin is higher than opportunity cost each year then we recommend to undertake this project

Ans c)

Expected Earning=($350*0.3+$400*0.4+$250*0.3)=$340

Variance of earnings=(Expected Earninf^2)-((Expected Earning)^2

Expected Earninf^2=0.3(350^2)+0.4(400^2)+0.3(250^2)=119500

Variance of earnings=119500-115600=$3900

We can say thta with 66% certainity earning can range from $119500+$3900=$123,400 to $119500-$3900=$115,600

We can say thta with 95% certainity earning can range from $119500+2*$3900=$127,300 to $119500-2*$3900=$111,700