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GM 533 Week 2 Lecture, Applied Managerial Statistics

GM 533 Week 2 Lecture, Applied Managerial Statistics

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Gm533 week 2_live_lecture Gm533 week 2_live_lecture Presentation Transcript

  • Welcome! To the Week 2Live Lecture/Discussion Applied Managerial Statistics (GM533) Lecturer – Prof. Brent Heard Please note that I borrowed these charts from Joni Bynum and the textbook publisher. Thanks Joni! I will put my touch on them (in blue) as we go along. 4-1 1
  • Week 2: Probability - Overview• Week 2 Terminal Course Objectives (TCOs)• The basic idea of probability• The four main problem types we’ll cover 4-2 2
  • Week 2 TCOs• TCO B Probability Concepts and Distributions: Given a managerial problem, utilize basic probability concepts, and standard probability distributions, e.g., binomial, normal, as is appropriate, to formulate a course of action which addresses the problem. 4-3 3 View slide
  • Week 2 TCOs (cont’d)• TCO F Statistics Software Competency: Students should be able to perform the necessary calculations for objectives A through E using technology, whether that be a computer statistical package or the TI-83, and be able to use the output to address a problem at hand.(I assume this includes Excel. I show you examples in Excel because you have access to it in the workplace. If your instructor requires you to use Minitab, that is their decision. I am only the lecturer) 4-4 4 View slide
  • Probability: the Basic Idea “We use the concept of probability to deal with uncertainty. Intuitively, the probability of an event is a number (between zero and one) that measures the chance, or likelihood, that the event will occur.” The text book, p. 171.(Probabilities must be between 0 and 1 or 0 and 100%. Zero means it can’t happen, 1 means it will definitely. Everything else is in between) 4-5 5
  • The 4 Main Probability Problem Types1. Contingency Tables2. Expected Value3. The Binomial Distribution4. The Normal Distribution 4-6 6
  • 1. Example of Contingency Tables• Each month a brokerage house studies various companies and rates each company’s stock as being either “low risk” or “moderate to high risk.” In a recent report, the brokerage house summarized its findings about 15 aerospace companies and 25 food retailers in the following table: 4-7 7
  • Part A. 1. If we randomly select one of thetotal of 40 companies, find:Company Low Moderate Total • The probability thatType Risk to High Risk the company is a food retailerAerospace 6 9 15companyFood 15 10 25retailer 25 .625 40Total 21 19 40 4-8 8
  • Part A. 2. If we randomly select one of thetotal of 40 companies, find:Company Low Risk Moderate Total • The probability thatType to High Risk the company’s stock is “low risk”Aerospace 6 9 15company 21 .525 40Food 15 10 25retailerTotal 21 19 40 4-9 9
  • Part A. 4. If we randomly select one of thetotal of 40 companies, find:Company Low Risk Moderate Total • The probability thatType to High Risk the company is a food retailer AND has a stock that isAerospace 6 9 15 “low risk”company 15 .375 40Food 15 10 25retailer (See that this value comesTotal 21 19 40 from“within” the table.) 4-10 10
  • Part A. 5. If we randomly select one of thetotal of 40 companies, find:Company Low Risk Moderate Total • The probability thatType to High Risk the company is a food retailer OR has a stock that is “lowAerospace 6 9 15 risk”company 31 .775 40Food 15 10 25retailer (Mathematically this isTotal 21 19 40 (25+21- 15)/40. I had to subtract out the “overlap” where I counted it twice.) 4-11 11
  • Part B. 2. If we randomly select one of thetotal of 40 companies, find:Company Low Risk Moderate Total • The probability thatType to High Risk the company’s stock is moderate to high risk GIVENAerospace 6 9 15 THAT the firm is acompany food retailer. 10 .40 25Food 15 10 25retailer (When I heard “Given That” it should Alert me that the denominator hasTotal 21 19 40 changed. I’m only dealing with a subset of the data now.) 4-12 12
  • Part C If we randomly select one of the totalof 40 companies:Company Low Risk Moderate Total • Determine if the companyType to High type is independent of the Risk level of risk of the firm’s stock.Aerospace 6 9 15 P(Aero.) =? P(Aero.Low Risk)company 15 6 .375 .2857 40 21 .357 .2857Food 15 10 25retailer Since they are not equal, company type is notTotal 21 19 40 independent of level of risk, i.e., they are dependent. 4-13 13
  • 2. Example Expected Value Problem There are many investment options which offer prospective gains and losses with their associated probabilities. For example, an oil company is considering drilling a number of wells, and, based on geological data and operating costs, they determine the following outcomes and probabilities. 4-14 14
  • Example Expected Value Problem (cont’d)x = the outcome in $ p(x) x*p(x)- $40,000 (no oil) .25 (- $40,000)*.25 = - $10,000$10,000 (some oil) .70 ($10,000)*.7 = $7,000$70,000 (much oil) .05 ($70,000)*.05 = $3,500 1.00 -$10,000 + $7,000 + $3,500 = $500 (This is just the total of the probabilities, it should always be one. Don’t make this harder than what it is.) 4-15 15
  • The Binomial Distribution #1 The Binomial Experiment: 1. An experiment consists of n identical trials (e.g., we flip a coin n=4 times). 2. Each trial (e.g., an individual coin flip) results in either “success” or “failure” (e.g., a head or a tail). 3. Probability of success, p (e.g., p=.5), is constant from trial to trial 4. Trials are independent (e.g., what we get on one flip does not affect the next flip). Note: The probability of failure, q, is 1 – p and is constant from trial to trial If x is the total number of successes in n trials of a binomial experiment, then x is a binomial random variable (e.g., the number of heads out of 4 flips) 4-16 16
  • Binomial Distribution using Excel Template• Templates for Binomial• Found at: http://highered.mcgraw- hill.com/sites/0070620164/student_view0/excel_templates.htmlDownload Binomial Distribution file. When you first open it, go to the Review tab at the top and click on “Unprotect Sheet.” Save it and you have your own personal Binomial Calculator to use to check your Minitab answers and to use at work. 4-17 17
  • Binomial using Excel Template Example• We have a binomial experiment with p = .6 and n = 3. Set up the probability distribution and compute the mean, variance, and standard deviation.• Find P(x=2), P(x≤2) and P(x>1) 4-18 18
  • Binomial using Excel Template Example (cont.) All I did was enter 3 for n and 0.6 for p. The template does the rest. It gives me the mean, variance and standard deviation. 4-19 19
  • Binomial using Excel Template Example (cont.)P(x=2) = 0.4320 P(x≤2) = 0.7840 P(x>1) = 0.6480 Note (x>1) is the same as saying “at least 2” 4-20 20
  • 3.Example of a Binomial ProbabilityProblem Consider the following scenario: historically, thirty percent of all customers who enter a particular store make a purchase. Suppose that six customers enter the store in a given time period, and that these customers make independent purchase decisions. Each individual customer either does or does not make a purchase. This is what makes this a binomial situation. 4-21 21
  • Typical Questions for the BinomialProbability Distribution• What is the probability that at least three customers make a purchase?• What is the probability that two or fewer customers make a purchase?• What is the probability that at least one customer makes a purchase? 4-22 22
  • Answers using Excel Template (Do in Minitab also) At least 1 would be 0.88242 or fewer would be 0.7443 At least 3 would be 0.2557 4-23 23
  • Example Binomial Problem – Mini TabSolution looks like this Probability Density • Cumulative Distribution Function Function• Binomial with n = 6 and p = • Binomial with n = 6 and p = 0.3 0.3• x P( X = x ) • x P( X <= x )• 0 0.117649 • 0 0.11765• 1 0.302526 • 1 0.42017• 2 0.324135 • 2 0.74431• 3 0.185220 • 3 0.92953• 4 0.059535 • 4 0.98906• 5 0.010206 • 5 0.99927• 6 0.000729 • 6 1.00000 4-24 24
  • Binomial Probabilities Explanations P( x 3) P( x 3) P( x 4) P( x 5) P( x 6) 0.18522 0.05954 0.01021 0.00073 0.25569 P( x 3) 1 P( x 2) 1.00000 0.74431 0.25569 P(x 2) 0.74431 P( x 1) 1.00000 P( x 0) 1.00000 0.11765 0.88235 4-25 25
  • The Normal Probability DistributionThe normal curve issymmetrical and bell-shaped• The normal is symmetrical about its mean • The mean is in the middle under the curve • So is also the median• The normal is tallest over its mean• The area under the entire normal curve is 1 • The area under either half of the curve is 0.5 5-26 26
  • The Position and Shapeof the Normal Curve (a) The mean positions the peak of the normal curve over the real axis (b) The variance 2 measures the width or spread of the normal curve 5-27 27
  • The Standard Normal Distribution #1 If x is normally distributed with mean and standard deviation , then the random variable z x z is normally distributed with mean 0 and standard deviation 1; this normal is called the standard normal distribution 5-28 28
  • The Standard Normal Distribution #2 z measures the number of standard deviations that x is from the mean • The algebraic sign on z indicates on which side of is x • z is positive if x > (x is to the right of on the number line) • z is negative if x < (x is to the left of on the number line) 5-29 29
  • Cumulative Areas under the StandardNormal Curve 4-30 30
  • Cumulative Areas under the StandardNormal Curve (cont’d) 4-31 31
  • Finding Areas under the StandardNormal Curve 4-32 32
  • Finding Areas under the StandardNormal Curve (cont’d) 4-33 33
  • Normal Distribution using an Excel Template• You can use the Normal Distribution Template from the same site I noted to do these.• Just go to the site, download the template, open it, click “Unprotect Sheet.” Save it to your computer and you have a “Normal Distribution Calculator” to check your Minitab and hand calculations.• When using for Standard Normal Distribution calculations, enter 0 for the mean and 1 for the standard deviation 4-34 34
  • Standard Normal Distribution with ExcelTemplate 4-35 35
  • Finding Areas under the StandardNormal Curve (cont’d) 4-36 36
  • Finding Z Points ona Standard Normal Curve 4-37 37
  • Brent’s Example of the Normal Distribution• Let’s say that failures of a particular component are normally distributed with a mean of 2.1 failures per year and a standard deviation of .6.• Find the probability of having less than two failures in a year.• Find the probability of having more than 3 failures in a year.• Find the probability of having between 1 and 3 failures in a year. 4-38 38
  • Brent’s Example of the Normal Distribution (cont.) I put in mean and std. dev. Probability of having between one and three failures in a year = 0.8998 Probability of having moreProbability of having less than than three failures in a year =two failures in a year = 0.4338 0.0668 4-39 39
  • Good day/night• See you next time 4-40 40