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Statistics for Managers 5th Edition Chapter 5 Some Important Discrete Probability Distributions . Chap 5-1 Learning Objectives In this chapter, you learn: The properties of a probability distribution To calculate the expected value, variance, and standard deviation of a probability distribution To calculate probabilities from Binomial, Hypergeometric and Poisson distributions How to use the Binomial, Hypergeometric and Poisson distributions to solve business problems . Chap 5-2 Introduction to Probability Distributions Random Variable Represents a possible numerical value from an uncertain event Random Variables Ch. 5 Discrete Random Variable Continuous Random Variable Ch. 6 Chap 5-3 Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times) Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5) Chap 5-4 Discrete Probability Distribution Experiment: Toss 2 Coins. T T H H T H T H Probability Distribution X Value Probability 0 1/4 = 0.25 1 2/4 = 0.50 2 1/4 = 0.25 Probability 4 possible outcomes Let X = # heads. 0.50 0.25 0 . 1 2 X Chap 5-5 Discrete Random Variable Summary Measures Expected Value (or mean) of a discrete distribution (Weighted Average) N E(X) Xi P( Xi ) i1 Example: Toss 2 coins, X = # of heads, compute expected value of X: X P(X) 0 0.25 1 0.50 2 0.25 E(X) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0 Chap 5-6 Discrete Random Variable Summary Measures (continued) Variance of a discrete random variable N σ 2 [X i E(X)]2 P(Xi ) i1 Standard Deviation of a discrete random variable σ σ2 N 2 [X E(X)] P(Xi ) i i1 where: E(X) = Expected value of the discrete random variable X Xi = the ith outcome of X P(Xi) = Probability of the ith occurrence of X Chap 5-7 Discrete Random Variable Summary Measures (continued) Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(X) = 1) σ [X E(X)] P(X ) i i 2 σ (0 1)2 (0.25) (1 1)2 (0.50) (2 1)2 (0.25) 0.50 0.707 Possible number of heads = 0, 1, or 2 Chap 5-8 Example of Expected Value A company is considering a proposal to develop a new product. The initial cash outlay would be $1 million, and development time would be three years. If successful, the firm anticipates that net profit (revenue minus initial cash outlay) over the 5 year life cycle of the product will be $1.5 million. If moderately successful, net profit will reach $1.2 million. If unsuccessful, the firm anticipates zero cash inflows. The firms assigns the following probabilities to the 5-year prospects for this product: successful, .60; moderately successful, .30; and unsuccessful, .10. What is the expected net profit? Ignore time value of money. Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. Chap 5-9 Calculation of Expected Value To calculate the expected value E(X) = Σ [(X) P(X)] = (1.5)(.6) + (1.2)(.3) + (-1)(.1) = + 1.16 Since the expected value Is positive this project passes the initial approval process Chap 5-10 Investment Returns The Mean Consider the return per $1000 for two types of investments. Investment Economic P(XiYi) Condition Passive Fund X Aggressive Fund Y 0.2 Recession - $25 - $200 0.5 Stable Economy + $50 + $60 0.3 Expanding Economy + $100 + $350 Chap 5-11 Investment Returns The Mean E(X) = μX = (-25)(.2) +(50)(.5) + (100)(.3) = 50 E(Y) = μY = (-200)(.2) +(60)(.5) + (350)(.3) = 95 Interpretation: Fund X is averaging a $50.00 return and fund Y is averaging a $95.00 return per $1000 invested. Chap 5-12 Chap 5-12 Investment Returns Standard Deviation σ X (-25 50) 2 (.2) (50 50) 2 (.5) (100 50) 2 (.3) 43.30 σ Y (-200 95) 2 (.2) (60 95) 2 (.5) (350 95) 2 (.3) 193.71 Interpretation: Even though fund Y has a higher average return, it is subject to much more variability and the probability of loss is higher. Chap 5-13 Chap 5-13 Investment Returns Covariance σ XY (-25 50)(-200 95)(.2) (50 50)(60 95)(.5) (100 50)(350 95)(.3) 8250 Interpretation: Since the covariance is large and positive, there is a positive relationship between the two investment funds, meaning that they will likely rise and fall together. Chap 5-14 Chap 5-14 The Sum of Two Random Variables: Measures Expected Value: E ( X Y ) E ( X ) E (Y ) Variance: Var( X Y ) σ 2X Y σ 2X σY2 2σ XY Standard deviation: σ X Y σ 2X Y Chap 5-15 Chap 5-15 Portfolio Expected Return and Expected Risk Investment portfolios usually contain several different funds (random variables) The expected return and standard deviation of two funds together can now be calculated. Investment Objective: Maximize return (mean) while minimizing risk (standard deviation). Chap 5-16 Chap 5-16 Portfolio Expected Return and Expected Risk Portfolio expected return (weighted average return): E(P) w E ( X ) (1 w) E (Y ) Portfolio risk (weighted variability) σ P w 2σ 2X (1 w )2 σ 2Y 2w(1 - w)σ XY where w = portion of portfolio value in asset X (1 - w) = portion of portfolio value in asset Y Chap 5-17 Chap 5-17 Portfolio Expect Return and Expected Risk Recall: Investment X: Investment Y: E(X) = 50 σX = 43.30 E(Y) = 95 σY = 193.21 σXY = 8250 Suppose 40% of the portfolio is in Investment X and 60% is in Investment Y: E(P) .4 (50) (.6) (95) 77 σ P (.4) 2 (43.30) 2 (.6)2 (193.21) 2 2(.4)(.6)( 8250) 133.04 The portfolio return is between the values for investments X and Y considered individually. Chap 5-18 Chap 5-18 Binomial Probability Distribution A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse Two mutually exclusive and collectively exhaustive categories e.g., head or tail in each toss of a coin; defective or not defective light bulb Generally called “success” and “failure” Probability of success is p, probability of failure is 1 – p Constant probability for each observation e.g., Probability of getting a tail is the same each time we toss the coin Chap 5-19 Binomial Probability Distribution (continued) Observations are independent The outcome of one observation does not affect the outcome of the other Two sampling methods Infinite population without replacement Finite population with replacement Chap 5-20 Possible Binomial Distribution Settings A manufacturing plant labels items as either defective or acceptable A firm bidding for contracts will either get a contract or not A marketing research firm receives survey responses of “yes I will buy” or “no I will not” New job applicants either accept the offer or reject it Chap 5-21 Binomial Distribution Formula n! X nX P(X) p (1-p) X ! (n X)! P(X) = probability of X successes in n trials, with probability of success p on each trial X = number of ‘successes’ in sample, (X = 0, 1, 2, ..., n) n = sample size (number of trials or observations) p = probability of “success” Example: Flip a coin four times, let x = # heads: n=4 p = 0.5 1 - p = (1 - 0.5) = 0.5 X = 0, 1, 2, 3, 4 Chap 5-22 Example: Calculating a Binomial Probability What is the probability of one success in five observations if the probability of success is .1? X = 1, n = 5, and p = 0.1 n! P(X 1) p X (1 p)n X X!(n X)! 5! (0.1)1(1 0.1) 51 1! (5 1)! (5)(0.1)(0 .9) 4 0.32805 Chap 5-23 Binomial Distribution Characteristics Mean μ E(x) np Variance and Standard Deviation σ np(1 - p) 2 σ np(1 - p) Where n = sample size p = probability of success (1 – p) = probability of failure Chap 5-24 Binomial Example A doctor prescribes Nexium to 60% of his patients who have gastro-intestinal problems. What is the probability that out of the next 10 patients: 1. Six of them are given a prescription of Nexium 2. At least 6 are given a prescription of Nexium? Chap 5-25 Binomial Probabilities Table X P(X) 6 0.250822656 7 0.214990848 8 0.120932352 9 0.040310784 10 0.006046618 Data Sample size Probability of success 10 0.6 P(<=X) 0.617719398 0.832710246 0.953642598 0.993953382 1 P(<X) 0.366896742 0.617719398 0.832710246 0.953642598 0.993953382 P(>X) 0.382280602 0.167289754 0.046357402 0.006046618 0 P(>=X) 0.633103258 0.382280602 0.167289754 0.046357402 0.006046618 Chap 5-26 The Poisson Distribution It is appropriate to use the Poisson distribution when: You have an event that occurs randomly through time and space You know the average number of successes you expect to observe over a given time frame or in a given space Chap 5-27 Poisson Distribution Formula x e P( X) X! where: X = number of events in an area of opportunity = expected number of events e = base of the natural logarithm system (2.71828...) Chap 5-28 Poisson Example On average there are three thread defects in a 10 yard bolt of fine wool fabric. What is the probability of finding no more than two thread defects in a randomly chosen 10 bolt lot? Chap 5-29 Solution Poisson Probabilities Data Average/Expected number of successes: Poisson Probabilities Table X P(X) P(<=X) 0 0.049787 0.049787 1 0.149361 0.199148 2 0.224042 0.423190 3 0.224042 0.647232 4 0.168031 0.815263 P(<X) 0.000000 0.049787 0.199148 0.423190 0.647232 3 P(>X) 0.950213 0.800852 0.576810 0.352768 0.184737 P(>=X) 1.000000 0.950213 0.800852 0.576810 0.352768 Chap 5-30 Poisson Distribution Characteristics Mean μλ Variance and Standard Deviation σ λ 2 σ λ where = expected number of events Chap 5-31 The Hypergeometric Distribution “n” trials in a sample taken from a finite population of size N Sample taken without replacement Outcomes of trials are dependent Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population Chap 5-32 Hypergeometric Distribution Formula A N A X n X P( X) N n Where N = population size A = number of successes in the population N – A = number of failures in the population n = sample size X = number of successes in the sample n – X = number of failures in the sample Chap 5-33 Properties of the Hypergeometric Distribution The mean of the hypergeometric distribution is nA μ E(x) N The standard deviation is nA(N - A) N - n σ 2 N N -1 Where N-n is called the “Finite Population Correction Factor” N -1 from sampling without replacement from a finite population Chap 5-34 Using the Hypergeometric Distribution ■ Example: 3 different computers are checked from 10 in the department. 4 of the 10 computers have illegal software loaded. What is the probability that 2 of the 3 selected computers have illegal software loaded? N = 10 A=4 n=3 X=2 A N A 4 6 X n X 2 1 (6)(6) P(X 2) 0.3 120 N 10 n 3 The probability that 2 of the 3 selected computers have illegal software loaded is .30, or 30%. . Chap 5-35 Hypergeometric Distribution in PHStat (continued) Complete dialog box entries and get output … N = 10 A=4 n=3 X=2 P(X = 2) = 0.3 Chap 5-36 Chapter Summary Addressed the probability of a discrete random variable Discussed the Binomial distribution Discussed the Poisson distribution Discussed the Hypergeometric distribution Chap 5-37