Question

PRACTICE 4 DMUR only You are considering two investment projects each having the same cost. Each project is facing the following events, probabilities and net profits: ALTERNATIVES: EVENTS: NET PROFITS: PROBABILITIE: aj: Newspaper ei e2 e3 4000 6000 9000 .3 .SO 2 a: Pamphlet ei e2 e3 3000 7000 8000 .30 SO .20 1. Construct a decision tree and show which project you would chose by using the expected value method )? 2. Calculate the coefficient of variation of each project, and determine which one should you chose accordingly? 3. Use the Z-table, and show the likelihood that Project and Project 2 will yield a net profit between $7000 and $9000. 4. Assume that the first investment project will cost you S2000 less than the second one Support how that would change your decision, if at all. DMUR: Results Xbar Sigma COV Normal Dist% A B C Your Be Sure To Support Answers (Give Reasons!) Use the table provided above for a RECAP of your results. The table has extra rows - You will need them for practice #5. Be sure to show your work, state a clear conclusion in each step and always support your answers.

Answer 1

Answer:

Given that,

$\rightarrow$ You are considering two investment projects each having the same cost. Each project is facing the following events, probabilities and net profits:

Alternatives	a1:Newspaper	a2:Pamphlet
Events	e1	e2	e3	e1	e2	e3
Net Profits	4000	6000	9000	3000	7000	8000
Probabilitie	0.3	0.50	0.2	0.30	0.50	0.20

Newspaper:

Event	Prob(P)	NP=x	P(x)	(x- T )	P$\times$(x- T )^2
C1	0.3	4000	1200	-2000	1,200,000
C2	0.50	6000	3000	0	0
C3	0.2	9000	1800	3000	1,800,000
			=6000		=3,000,000

Pamphlet:

Event	Prob(P)	NP=x	P(x)	(x- T )	P$\times$(x- T )^2
C1	0.3	3000	900	-3000	2,700,000
C2	0.5	7000	3500	1000	500,000
C3	0.2	8000	1600	2000	800,000
			=6000		=4,000,000

Newspaper N.P P=0.3 19-4000 + 1900 P=0.50 → Cy = 6000 →3000 p=02 C3 = 9000 1800 6000 X-6000 Pamphlet Net profit P 20.3 → 9 z 3000 7900 P=0.5 → C2=7000 3500 to C928000 + 1600 6000 X = 6000 120.2

Both are better

(2).

Newspaper:

X = ΣΡ(x) = 6000

$\sigma _{x}=\sqrt{P\times (x-\bar{x})^{2}}=\sqrt{4,000,000}=2000$

The coefficient of variation (COV):

$Cov=\frac{\sigma _{x}}{\bar{x}}$

=2000/6000

=33.33%

Similarly,

Pamphlet:

X = ΣΡ(x) = 6000

$\sigma _{x}=\sqrt{P\times (x-\bar{x})^{2}}=\sqrt{3,000,000}=1732.051$

The coefficient of variation (COV):

$Cov=\frac{\sigma _{x}}{\bar{x}}$

=1732.051/6000

=28.87 %

Lower the better = Newspaper

(3).

For Newspaper:

$Z_1=\frac{x-\mu }{\sigma }=\frac{7000-6000}{1732.051}$

=1000/1732.051

=0.58

And,

$Z_2=\frac{x-\mu }{\sigma }=\frac{9000-6000}{1732.051}$

=3000/1732.501

=1.73

The are between Z of 7000-Z of 9000 is, 0.2392.

And,

For pamphlet:

$Z_1=\frac{x-\mu }{\sigma }=\frac{7000-6000}{2000}$

=1000/2000

=0.5

And,

$Z_2=\frac{x-\mu }{\sigma }=\frac{9000-6000}{2000}$

=3000/2000

=1.5

The are between Z of 7000-Z of 9000 is, 0.2417.

ju: 6000 H peo dezo u=6dco ad MO 770 3:5 Pamphlet 2:05 32143 Newspapers

(4).

$\rightarrow$If first investment project costs $2000 less than second.

Then, expected returns.

Newspaper=6000+2000=8000

Pamphlet=6000+0=6000

No, decision should not changes we still recommed newspaper project.