Nicole Becker – Assignment 9 Complete the following textbook exercises: 2, 12 and 14 It is a decision under risk problem. Our state of nature is .7 and .3 and decision to make is which bottle to...

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Nicole Becker – Assignment 9
Complete the following textbook exercises:
2, 12 and 14
It is a decision under risk problem. Our state of nature is .7 and .3 and decision to make is which bottle to choose
Payoff table
State of nature
probability
Conditional payoff
returnable
Non returnable
Law passed
.7
80
25
Not passed
.3
40
60
68
25.5
As expected monetary payoff for returnable= .7*80+.3+40=68 is greater than EMP of non returnable bottles=.7*25+.3*60=25.5. so choosing returnable bottles is wise.
State of nature is good or bad weather with probabilities .4 and .6 respectively.
Total cost to travel by any means is= 75*time taken in journey+ cost of travel by that means
Payoff table
State of nature
Probability
(i)
Conditional probabilities in terms of monerty value $ (ii)
(i)*(ii)
Under perfect information
Car
Train
Plane
Car
Train
Plane
Minimum payoff(iii)
(i)*(iii)
Good weather
0.4
370
490
366.25
148
196
146.5
366.25
146.5
Bad weather
0.6
445
502.5
372.5
267
301.5
223.5
372.5
372.5
total
415
497.5
370
370
As EMV of traveling by plane is lowest this is the decision financial executive should choose.
So expected value of perfect information is =370-370=0.
C =20*0.5=10$ for all parts inspection
C=20*18*Sj= 360*Sj for none inspection
a)
State of
Probability of state
Inspecting (ii)
(i)*(ii)
Under perfect
nature
of nature (i)
information
All parts
None parts
All parts
None parts
Maximum payoff(iii)
(i)*(iv)
0.02
.7
10
7.2
7
5.04
10
7
0.04
.2
10
14.4
2
2.88
14.4
2.88
0.06
.1
10
21.6
1
2.16
21.6
2.16
10
10.08
12.04
b) Expected monetary value for all part inspection is less so this decision is better.
c) expected value of perfect information= 12.04-10= 2.02
Answered Same DayDec 23, 2021

Answer To: Nicole Becker – Assignment 9 Complete the following textbook exercises: 2, 12 and 14 It is a...

David answered on Dec 23 2021
124 Votes
Assignment # 1 Due September 22, 2003
Quantitative Methods II (QUA2321)
Final Exam “B”
Page 1 of 18
Quantitative Methods II (QUA2321)
Final Exam “B”
Page 18 of 18
EXAMINATION COVER PAGE
    COURSE CODE:
    QUA2321 - B
    COURSE TITLE:
    QUANTITATIVE METHODS II
    DELIVERY MODE:
    ON-LINE
Student Information
    STUDENT NAME:
(please print)
    
    DATE:
    
    INSTRUC
TOR NAME:
    
Exam Information
    Exam Duration:
    3 Hours
    Materials Required:
    
    Special Instructions:
    · One 8 ½ x 11 formula sheet permitted both sides - formulas only – must be handed in with exam
· Calculator Permitted
· Answer all questions in the space provided
· Student name to appear on all exam pages
· Return all exam materials including questions at end of exam
Question 1
The contingency table listed below show the frequency of persons injured by fireworks for the year 2001. The table describes the persons by age and sex.
    
    Age
    (in years)
    Sex
    Under 15
    15 or Older
    Male
    3357
    3852
    Female
    976
    2080
    Total
    4333
    5932
     
    Age
    (in years)
    
    Sex
    Under 15
    15 or Older
    Total
    Male
    3357
    3852
    7209
    Female
    976
    2080
    3056
    Total
    4333
    5932
    10265
a. What is the probability that the injured person was a male? (2 marks)
The probability that the injured person was a male is 7209/10265 = 0.702289
b. What is the probability that the injured person was a male under the age of 15? (2 marks)
The probability that the injured person was a male under the age of 15 is 3357/10265 = 0.327034
c. What is the probability that the injured person was either a male or under the age of 15? (2 marks)
The probability that the injured person was either a male or under the age of 15 is 7209/10265+4333/10265-3357/10265=0.79737
d. Given that a person selected at random was older than 15, what is the probability that they were female? (2 marks)
The probability that they were female given that a person selected at random was older than 15 is 2080/10265 = 0.20263
Question 2
The industry standards suggest 10 percent of new vehicles require warranty service within the first year. Jones Nissan sold 12 Nissans yesterday.
a. What is the probability that none of these vehicles requires warranty service? (2 marks)
Here we have
n =12 and p = 0.10
We need to find
P(X=0) = “=BINOMDIST(0,12,0.1,0)” = 0.28243
b. What is the probability exactly one of these vehicles requires warranty service? (2 marks)
Here we have
n =12 and p = 0.10
We need to find
P(X=1) = “=BINOMDIST(1,12,0.1,0)” = 0.376573
c. Determine the probability that exactly two of these vehicles require warranty service? (2 marks)
Here we have
n =12 and p = 0.10
We need to find
P(X=2) = “=BINOMDIST(2,12,0.1,0)” = 0.230128
d. Compute the mean and standard deviation of this probability distribution. (3 marks)
Mean = n*p = 12*0.10 = 1.2
Standard deviation = sqrt(n*p*(1-p) =sqrt(12*0.10*0.90) = 1.0392
Question 3
The average grade in a statistics course has been 69 with a standard deviation of 12.5. If a random sample of 50 is selected from this population, what is the probability that the average grade is more than 73? (5 marks)
Here we have
Mean = 69
Sd = 12.5
n = 50
We need to find
P(X>73) = P(Z > (73-69)/(12.5/sqrt(50))) = P(Z>2.26) = 0.5 – P(0Question 4
The commercial banks in the financial district are to be surveyed. Some of them are very large, with assets of more than $500 million; others are medium-sized, with assets between $100 million and $500 million; and the remaining banks have assets of less than $100 million. Explain how you would select a sample of these banks? (7 marks)
Since this is a...
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