Assignment 2 Part 2 - ADM 2303 Business Statistics

Assignment 2 – Part 2
1. Project Success
a.
Using the general multiplication rule:
Psolar (success) = P (success | solar) x P (solar)
Psolar (success) = 0.75 x 0.70 = 0.525
Pwind (success) = P (success | wind) x P (wind)
Pwind (success) = 0.65 x 0.30 = 0.195
Psolar + Pwind = 0.525 + 0.195 = 0.72
Therefore, the probability of success is 0.72 or 72%.
b.
Using Bayes’ theorem
𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠|𝑠𝑜𝑙𝑎𝑟|)∙𝑃(𝑠𝑜𝑙𝑎𝑟)
P (solar | success) = 𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠)

(0.75 𝑥 0.70)
P (solar | success) = 0.72

P (solar | success) = 0.7292
Therefore, the probability that the company chooses solar is 0.7292 or 72.9%
2. Financial Evaluation
a.
Project solar:
Expected value = (0.30 x 5) + (0.50 x 12) + (0.20 x 18)
Expected value = 1.5 + 6 + 3.6 = 11.1
Variance = (5^2 x 0.30) + (12^2 x 0.50) + (18^2 x 0.20)
Variance = 7.5 + 72 + 64.8 = 144.3
Variance = 144.3 - 11.1^2
Variance = 21.09
Project Wind:
Expected value = (8 x 0.40) + (14 x 0.35) + (20 x 0.25)

, Expected value = 3.2 + 4.9 + 5 = 13.1
Variance = (8^2 x 0.40) + (14^2 x 0.35) + (20^2 x 0.25)
Variance = 25.6 + 68.6 + 100 = 194.2
Variance = 194.2 - 13.1^2
Variance = 22.59
b.
Project solar:
√21.09
CV = 11.1

CV = 0.414
CV = 41.4%
Project wind:
√22.59
CV = 13.1

CV = 0.363
CV = 36.3%
c. Project solar has a higher financial risk because of its higher CV.
d.
Expected value (total) = 11.1 + 13.1 = 24.2
Variance (total) = 21.09 + 22.59 = 43.68
3. Operational Disruptions
a.
P (D ≥ 2) = 1 – P (D = 0) - P (D = 1)
𝑒 −3 30
P (D = 0) = = 0.0498
0!

𝑒 −3 31
P (D = 1) = = 0.1494
1!

P (D ≥ 2) = 1 – 0.0498 - 0.1494
P (D ≥ 2) = 0.8008
Therefore, the probability of at least 2 disruptions in a month is 0.8008 or 80.08%.
b.