In: Statistics and Probability
PRACTICE 3 DMUR
Hudson Realty is considering a boost in advertising in order to reduce a large inventory of unsold houses. The management plans to make its media decision using the following data on the expected success of newspaper versus pamphlet promotions. Each promotion strategy requires the sane amount of capital:
ALTERNATIVES: a1:Newspaper a2: Pamphlet
EVENTS: e1 e2 e3 e1 e2 e3
NET PROFITS: 3000 7000 11000 5000 7000 9000
PROBABILITIE: .25 .50 .25 .25 .50 .25
1. Construct a decision tree and show which promotion alternative you would chose by using the expected value method ( )?
2. Calculate the coefficient of variation (CoV) of each alternative, and determine which one should be chosen accordingly?
3. Use the Z-table,and show the likelihood that Alternative1 as well as Alternative 2 will yield a net profit between $7000 and $9000.
ßAlternatives/Events Þ |
e1 |
e2 |
e3 |
a1: Newspaper |
$ 3,000 |
$ 7,000 |
$ 11,000 |
a2: Pamphlet |
$ 5,000 |
$ 7,000 |
$ 9,000 |
fi : probabilities |
.25 |
.5 |
.25 |
PRACTICE 4DMUR
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: .25 .50 .25 .30 .50 .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 (cov) 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.
ßAlternatives/Events Þ |
e1 |
e2 |
e3 |
a1: Newspaper |
$ 4,000 |
$ 6,000 |
$ 9,000 |
a1: fi : probabilities |
.25 |
.5 |
.25 |
a2: Pamphlet |
$ 3,000 |
$ 7,000 |
$ 8,000 |
a2: fi : probabilities |
.3 |
.5 |
.2 |
IF YOU HAVE ANY DOUBTS COMMENT BELOW I WILL BE TTHERE TO HELP YOU..ALL THE BEST..
1)Expected value for alternative a1
=3000*0.25+7000*0.5+11000*0.25
=7000
Expected value for alternative a2
=5000*0.25+7000*0.5+9000*0.25
=7000
Both alternative have same expected value.
2)first calculate standard deviation for both alternatives
For a1,
=2828.43
For a2,
=1414.21
COV1=2828.43/7000=0.4041
COV2=1414.21/7000=0.2020
COV2<COV1,so alternative2 is better.
3)alternative 1
P(7000<x<9000)
z =0 when x=7000
z=(9000-7000)/2828.43=0.707
P(0<=z<=0.707)=P(z<=0.707)-P(z<=0)=0.7602-0.5=0.2602
For alternative 2
z=(9000-7000)/1414.21=1.414
P(0<=z<=1.414)=P(z<=1.414)-P(z<=0)=0.9213-0.5=0.4213
I HOPE YOU UNDERSTAND..
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