View the step-by-step solution to:

# CHAPTER 9 Theoretical Probability Models Notes Chapter 9 is a straightforward treatment of probability modeling using five standard distributions....

Dear,
How You draw "decision tree"? In a Word or Ppt, or maybe in some particular program/software?
TY
Jasmin

118 CHAPTER 9 Theoretical Probability Models Notes Chapter 9 is a straightforward treatment of probability modeling using five standard distributions. The five distributions (binomial, Poisson, exponential, normal, beta) were chosen because they cover a variety of different probability-modeling situations. Tables of probabilities for four of the distributions (all except the exponential, probabilities for which are easily calculated) are provided in the appendices. In addition to the five distributions treated in the main part of the chapter, the uniform distribution is developed in problems 9.27 - 9.29, the triangular distribution in 9.30, and the lognormal distribution in problem 9.36 and the Municipal Solid Waste case study. Depending on the nature of the course and the level of the students, instructors may wish to introduce other distributions. One note of caution: Chapter 9 provides an introduction to probability distributions that are used in Chapter 10 (fitting model parameters and natural conjugate priors) as well as Chapter 11 (creating random variates in a simulation). In particular, if the course is intended to move on to Chapter 11, it is important to expose students to the uniform distribution, the triangular distribution, and some of the other distributions as well. In addition to using the tables provided in the appendices, RISKview can be used for computing and seeing distributions. Step-by-step instructions for viewing theoretical distributions with RISKview are provided in the chapter. RISKview is a pop-up dialog box within @RISK that is activated using the Define Distribution button. The desired probabilities can be determined by sliding the delimiters (marked by inverted triangles) or by entering Left and Right X and P values in the statistics grid. Topical cross-reference for problems Bayes’ theorem 9.17, 9.21 Beta distribution 9.4, 9.5, 9.22, 9.24 Binomial distribution 9.1, 9.5, 9.15 - 9.19, 9.29, 9.31, 9.34, 9.35, Overbooking Central limit theorem 9.36, Municipal Solid Waste Empirical rule 9.11 Exponential distribution 9.5, 9.8, 9.12, 9.14, 9.15, Earthquake Prediction Linear transformations 9.14, 9.25, 9.26 Lognormal distribution 9.36, Municipal Solid Waste Memoryless property 9.12, 9.13, 9.28 Normal distribution 9.2, 9.5 - 9.7, 9.11, 9.25, 9.26, 9.31, 9.36, Municipal Solid Waste Pascal sampling 9.18 Poisson distribution 9.3, 9.5, 9.9, 9.10, 9.13, 9.15, 9.19 - 9.21, 9.32 - 9.35, Earthquake Prediction PrecisionTree 9.35 Requisite models 9.23 RISKview 9.1-9.15, 9.18, 9.19, 9.22, 9.24-9.34 Sensitivity analysis 9.17, 9.24, Earthquake Prediction Triangular distribution 9.30 Uniform distribution 9.27 - 9.29
119 Solutions 9.1. Find P(Gone 6 or more weekends out of 12) = P B ( R 6 | n = 12, p = 0.65) = P B ( R ' 6 | n = 12, p = 0.35) = 0.915. Being gone 6 or more weekends out of 12 is the same as staying home on 6 or fewer. Using RISKview, start RISKview. Select a function as the distribution source, the binomial as the distribution type, 12 for the n parameter, and 0.65 for the p parameter. In the statistics grid, set the Left X value to 5.5, and the Right X value to 12. The Difference P value then shows the desired probability: 91.54%. This distribution is saved as a RISKview file titled “Problem 9.1.rvp”. 9.2. P(Loss) = P N ( X < 0 | μ = 2000, σ = 1500) = P( Z < 0 - 2000 1500 ) = P( Z < -1.33) = 0.0918. P(Gain greater than 4000) = P N ( X > 4000 | μ = 2000, σ = 1500) = P( Z > 4000 - 2000 1500 ) = P( Z > 1.33) = 1 - P( Z 1.33) = 1 - 0.9082 = 0.0918. Note that P( Z -1.33) = P( Z 1.33) because of symmetry of the normal distribution. Using RISKview, select a function as the distribution source, the normal as the distribution type, 2000 for the µ parameter, and 1500 for the σ parameter. To determine the probability of a loss, in the statistics grid, set the Left X value to 4000, and the Right X value to 0. The Difference P value then shows the desired probability: 9.12%. To determine the probability that the return will be greater than 4000, set the Left X value to 4000 and the Right X value to 8000. The Difference P value then shows the desired probability: 9.12%. This distribution is saved as a RISKview file titled “Problem 9.2.rvp”. 9.3. P(No chocolate chips) = P P ( X = 0 | m = 3.6) = 0.027 from Appendix C. P(Fewer than 5 chocolate chips) = P P ( X < 5 | m = 3.6) = P P ( X 4 | m = 3.6) = 0.706 from Appendix D. P(More than 10 chocolate chips) = P P ( X > 10 | m = 3.6) = 1 - P P ( X 10 | m = 3.6) = 1 - 0.999 = 0.001. Using RISKview, select a function as the distribution source, the Poisson as the distribution type, and 3.6 for the λ parameter. To determine the probability of no chips, in the statistics grid, set the Left X value to -0.5, and the Right X value to 0.5. The Difference P value then shows the desired probability: 2.73%. To determine the probability that there are fewer than 5 chips, set the Left X value to -0.5 and the Right X value to 4.5. The Difference P value then shows the desired probability: 70.64%. To determine the probability that there are more than 10 chips, set the Left X value to 10.5 and the Right X value to 15. The Difference P value then shows the desired probability: 0.13%. This distribution is saved as a RISKview file titled “Problem 9.3.rvp”.
Show entire document

### Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

### -

Educational Resources
• ### -

Study Documents

Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

Browse Documents