SELECTED SIMULATION EXERCISES Exercise 4-1 People arrive at a newsstand at the rate of one every 10 ( 5 seconds. Most people buy only one paper but 20% buy two papers. It takes 5 ( 3 seconds to buy one paper and 7 ( 3 seconds to buy two papers. Simulate the sale of 100 papers, starting from the time the newsstand opens. Exercise 4-2 A series of moving stairways carry customers in an upward direction between four floors of a department store. People arrive at the foot of the stairs
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chips? If I had to guess, I would guess ‘5.’ Simply put, the odds I draw a ‘3’ from a hat with 5 chips are 1 in 5, the odds with 5 million chips is 1 in 2x10-7. It’s true we can’t know with complete certainty how many chips lay in the hat, but probability offers a charitable and rational answer. This method of investigation is present in Nick Bostrom’s essay: Are You Living In A Computer Simulation? In the essay Bostrom examines this question and reaches a conclusion
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Mathematics HL First examinations 2008 b DIPLOMA PROGRAMME MATHEMATICS HL First examinations 2008 International Baccalaureate Organization Buenos Aires Cardiff Geneva New York Singapore Diploma Programme Mathematics HL First published in September 2006 International Baccalaureate Organization Peterson House, Malthouse Avenue, Cardiff Gate Cardiff, Wales GB CF23 8GL United Kingdom Phone: + 44 29 2054 7777 Fax: + 44 29 2054 7778 Web site: www.ibo.org c International Baccalaureate
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Evaluating a Drug Licensing Opportunity Statement of the problem- Merck must determine whether or not to bid to license Davanrik My recommendation is that Merck must definitely make a competitive bid for Davanrik. The total expected value from the deal based numbers given in the Merck article is a healthy $14million so keeping a 20% incentive, Merck should bid no more than $11million for Davanrik as the initial licensing fee. Looking at the background of Merck, it is clear that it is a
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return Answer: ($0 $12,000 $10,000) $10,000 $2,000 $10,000 20% Logistics, Inc. doubled the annual rate of return predicted by the analyst. The negative net income is irrelevant to the problem. E8-2. Expected return Answer: Analyst 1 2 3 4 Total Probability 0.35 0.05 0.20 0.40 1.00 Return 5% 5% 10% 3% Expected return Weighted Value 1.75% 0.25% 2.0% 1.2% 4.70% E8-3. Comparing the risk of two investments Answer: CV1 0.10 0.15 0.6667 CV2 0.05 0.12 0.4167 Based solely on standard deviations
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Week 1 homework chapter 1 1. we have: Cf= 55000, cv= 8, p= 21; v= 10 000 a. Total cost TC = cf + v*cv TC= 55000+ 10000*8 TC= $135000 Total revenue TR= v*p TR= 10000*21 TR=$ 21000 Profit Profit= TR-TC Profit= TR-TC P= 21000-135000 P= $ -114 000 b.Break even volume, V= cf/ (p- cv) V= 55000/ (21- 8) V= 4230.77 recap tires, 2. monthly break even volume V= cf/ (p- cv)
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* * * Menu The math behind Six Sigma metrics Mon, 7 Jan 2008 By Valerie Bolhouse, Certified Six Sigma Blackbelt (This information supports an article appearing in the January 2008 issue of Vision Systems Design, "Quality Numbers: Six Sigma.") In nature and most manufacturing processes no two things are ever exactly the same. There exist small variations from part to part or measure to measure. If you were to acquire metrics on features of 100 "identical" parts and plot the values relative
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reliability are obtained for the circular connectedd (r,s)-out-of-(m,n):l? lattice system. FinalIy, it is proved that the reliability of the large system tmds to where p 1A and 7 are exp[-pArs] as n = , m Q - I 2 n -+ m if every component has failure probability ~ n - ' / / ~ , constant, ,LA, A > Ol 7 > r , ri rays Abstract I . INTRODUCTION The linear (circular) consecutive-k-out-of-n:F system has n llnewly (circularly) ordered components. Each compnent either functkms or fails. The systems fail
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Solutions Manual for Statistical Inference, Second Edition George Casella University of Florida Roger L. Berger North Carolina State University Damaris Santana University of Florida 0-2 Solutions Manual for Statistical Inference “When I hear you give your reasons,” I remarked, “the thing always appears to me to be so ridiculously simple that I could easily do it myself, though at each successive instance of your reasoning I am baffled until you explain your process.” Dr. Watson
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(y) = P (X ≤ y) = y i=1 f (x) dx. −∞ Sample covariance: g= 1 n−1 n Density function: 1 n−1 n (xi −x)(yi −y) = i=1 xi yi − nx y i=1 . f (x) = Evaluating probabilities: d F (x). dx Sample correlation: r= g . sx sy b P (a < X ≤ b) = a f (x) dx = F (b) − F (a). Probability Addition law: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). Multiplication law: P (A ∩ B) = P (A)P (B|A) = P (B)P (A|B). Partition law: For a partition B1 , B2 , . . . , Bk k k Expected
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