Probability Concepts and Applications

Transcription

1 Chapter 2 Probability Concepts and Applications To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson Learning Objectives After completing this chapter, students will be able to: 1. Understand the basic foundations of probability analysis. 2. Describe statistically dependent and independent events. 3. Describe and provide examples of both discrete and continuous random variables. 4. Explain the difference between discrete and continuous probability distributions. 5. Calculate expected values and variances and use the normal table. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢ ١

2 Chapter Outline 2.1 Introduction 2.2 Fundamental Concepts 2.8 Random Variables 2.9 Probability Distributions 2.10 The Binomial Distribution 2.11 The Normal Distribution 2.12 The F Distribution 2.13 The Exponential Distribution 2.14 The Poisson Distribution Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣ Introduction Life is uncertain; we are not sure what the future will bring. Probability is a numerical statement about the likelihood that an event will occur. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤ ٢

3 Fundamental Concepts 1. The probability, P, of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1. That is: 0 P (event) 1 2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٥ Random Variables A random variable assigns a real number to every possible outcome or event in an experiment. X = number of refrigerators sold during the day Discrete random variables can assume only a finite or limited set of values. Continuous random variables can assume any one of an infinite set of values. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٦ ٣

4 Random Variables Numbers EXPERIMENT Stock 50 Christmas trees Inspect 600 items Send out 5,000 sales letters Build an apartment building Test the lifetime of a lightbulb (minutes) Table 2.4 OUTCOME Number of Christmas trees sold Number of acceptable items Number of people responding to the letters Percent of building completed after 4 months Length of time the bulb lasts up to 80,000 minutes RANDOM VARIABLES RANGE OF RANDOM VARIABLES X 0, 1, 2,, 50 Y 0, 1, 2,, 600 Z 0, 1, 2,, 5,000 R 0 R 100 S 0 S 80,000 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٧ Random Variables Not Numbers EXPERIMENT OUTCOME RANDOM VARIABLES Students respond to a questionnaire Strongly agree (SA) Agree (A) Neutral (N) Disagree (D) Strongly disagree (SD) 5 if SA 4 if A.. X = 3 if N.. 2 if D.. 1 if SD RANGE OF RANDOM VARIABLES 1, 2, 3, 4, 5 One machine is inspected Defective Not defective Y = 0 if defective 1 if not defective 0, 1 Consumers respond to how they like a product Good Average Poor 3 if good. Z = 2 if average 1 if poor.. 1, 2, 3 Table 2.5 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٨ ٤

5 Probability Distribution of a Discrete Random Variable For discrete random variables a probability is assigned to each event. The students in Pat Shannon s statistics class have just completed a quiz of five algebra problems. The distribution of correct scores is given in the following table: Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٩ Probability Distribution of a Discrete Random Variable RANDOM VARIABLE (X Score) NUMBER RESPONDING PROBABILITY P (X) 5 AP = 10/ = 20/ = 30/ = 30/ = 10/100 Total = 100/100 Table 2.6 The Probability Distribution follows all three rules: 1. Events are mutually exclusive and collectively exhaustive. 2. Individual probability values are between 0 and Total of all probability values equals 1. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٠ ٥

6 Slide 10 AP6 11e Table 2.6 does not have the Outcome column because it's about quiz scores Annie Puciloski; 01/02/2011

7 Expected Value of a Discrete Probability Distribution The expected value is a measure of the central tendency of the distribution and is a weighted average of the values of the random variable. where P E X i ( X i ) n i= 1 ) E(X n ( X ) = X P( ) i= 1 i X i ( X ) + X P( X ) X P( X ) = X P n n = random variable s possible values = probability of each of the random variable s possible values = summation sign indicating we are adding all n possible values = expected value or mean of the random sample AP7 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١١ Expected Value of a Discrete Probability Distribution For Dr. Shannon s class: E n ( X ) = X P( ) i= 1 i X i = 5(0.1) + 4(0.2) + 3(0.3) + 2(0.3) + 1(0.1) = = 2.9 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٢ ٦

8 Slide 11 AP7 Text has random "variable" not "sample"; see p. 35 Annie Puciloski; 01/02/2011

9 Variance of a Discrete Probability Distribution For a discrete probability distribution the variance can be computed by σ 2 = Variance = [ X n i= 1 2 E( X )] P( i X i ) where X i = random variable s possible values E(X ) = expected value of the random variable [ X i E( X )] = difference between each value of the random variable and the expected mean P( X i ) = probability of each possible value of the random variable Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٣ Variance of a Discrete Probability Distribution For Dr. Shannon s class: variance = 5 i= 1 [ X 2 E( X )] P( i X i 2 2 variance = ( ) ( 0. 1) + ( ) ( 0. 2) ( ) ( 0. 3) + ( ) ( 0. 3) + 2 ( ) ( 0. 1) = = ) Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٤ ٧

10 Variance of a Discrete Probability Distribution A related measure of dispersion is the standard deviation. σ = Variance = 2 σ where σ = square root = standard deviation Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٥ Variance of a Discrete Probability Distribution A related measure of dispersion is the standard deviation. σ = Variance = 2 σ where σ = square root = standard deviation For Dr. Shannon s class: σ = Variance = = Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٦ ٨

11 Probability Distribution of a Continuous Random Variable Since random variables can take on an infinite number of values, the fundamental rules for continuous random variables must be modified. The sum of the probability values must still equal 1. The probability of each individual value of the random variable occurring must equal 0 or the sum would be infinitely large. The probability distribution is defined by a continuous mathematical function called the probability density function or just the probability function. This is represented by f (X). Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٧ Probability Distribution of a Continuous Random Variable Probability Weight (grams) Figure 2.6 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٨ ٩

12 The Binomial Distribution Many business experiments can be characterized by the Bernoulli process. The Bernoulli process is described by the binomial probability distribution. 1. Each trial has only two possible outcomes. 2. The probability of each outcome stays the same from one trial to the next. 3. The trials are statistically independent. 4. The number of trials is a positive integer. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-١٩ The Binomial Distribution The binomial distribution is used to find the probability of a specific number of successes in n trials. We need to know: n = number of trials p = the probability of success on any single trial We let r = number of successes q = 1 p = the probability of a failure Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٠ ١٠

13 The Binomial Distribution The binomial formula is: n! Probability of r successes in n trials = p r!( n r)! r r q n The symbol! means factorial, and n! = n(n 1)(n 2) (1) For example 4! = (4)(3)(2)(1) = 24 By definition 1! = 1 and 0! = 1 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢١ Table 2.7 The Binomial Distribution Binomial Distribution for n = 5 and p = NUMBER OF HEADS (r) Probability = 5! r!(5 r)! (0.5) r (0.5) 5 r = 5! 0!(5 0)! (0.5) 0 (0.5) = 5! 1!(5 1)! (0.5) 1 (0.5) = 5! 2!(5 2)! (0.5) 2 (0.5) = 5! 3!(5 3)! (0.5) 3 (0.5) = 5! 4!(5 4)! (0.5) 4 (0.5) = 5! 5!(5 5)! (0.5) 5 (0.5) 5 5 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٢ ١١

14 Solving Problems with the Binomial Formula We want to find the probability of 4 heads in 5 tosses. n = 5, r = 4, p = 0.5, and q = = 0.5 Thus 5! 4 P = (4 successes in 5 trials) = !(5 4)! 5 4 5( 4)( 3)( 2)( 1) = ( )( 0. 5) = ( 3)( 2)( 1)( 1!) Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٣ Solving Problems with Binomial Tables MSA Electronics is experimenting with the manufacture of a new transistor. Every hour a random sample of 5 transistors is taken. The probability of one transistor being defective is What is the probability of finding 3, 4, or 5 defective? So n = 5, p = 0.15, and r = 3, 4, or 5 We could use the formula to solve this problem, but using the table is easier. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٤ ١٢

15 Solving Problems with Binomial Tables n r Table 2.8 (partial) We find the three probabilities in the table for n = 5, p = 0.15, and r = 3, 4, and 5 and add them together. P Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٥ Table 2.8 (partial) Solving Problems with Binomial Tables P ( 3 or more defects) = P( 3) + P( 4) + P( 5) n r We find the three probabilities in the table for n = 5, p = 0.15, and r = 3, 4, and 5 and add them together P = = Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٦ ١٣

16 Solving Problems with Binomial Tables It is easy to find the expected value (or mean) and variance of a binomial distribution. Expected value (mean) = np Variance = np(1 p) For the MSA example: Expected value = np = 5( 0. 15) = Variance = np( 1 p) = 5( 0. 15)( 0. 85) = Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٧ The Normal Distribution The normal distribution is the one of the most popular and useful continuous probability distributions. The formula for the probability density function is rather complex: ( x µ ) 1 2 2σ f ( X ) = e σ 2π 2 The normal distribution is specified completely when we know the mean, µ, and the standard deviation, σ. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٨ ١٤

17 The Normal Distribution The normal distribution is symmetrical, with the midpoint representing the mean. Shifting the mean does not change the shape of the distribution. Values on the X axis are measured in the number of standard deviations away from the mean. As the standard deviation becomes larger, the curve flattens. As the standard deviation becomes smaller, the curve becomes steeper. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٢٩ The Normal Distribution 40 µ = Smaller µ, same σ µ = Larger µ, same σ Figure µ = 60 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٠ ١٥

18 The Normal Distribution Same µ, smaller σ Same µ, larger σ Figure 2.9 µ Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣١ The Empirical Rule For a normally distributed random variable with mean µ and standard deviation σ, then 1. About 68% of values will be within 1σ of the mean. 2. About 95.4% of values will be within 2σ of the mean. 3. About 99.7% of values will be within 3σ of the mean. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٢ ١٦

19 The Empirical Rule 68% 16% 16% 1σ +1σ a µ b 95.4% 2.3% 2.3% 2σ +2σ a µ b 99.7% 0.15% 0.15% Figure σ +3σ a µ b Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٣ The F Distribution It is a continuous probability distribution. The F statistic is the ratio of two sample variances. F distributions have two sets of degrees of freedom. Degrees of freedom are based on sample size and used to calculate the numerator and denominator of the ratio. df 1 = degrees of freedom for the numerator df 2 = degrees of freedom for the denominator The probabilities of large values of F are very small. Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٤ ١٧

20 The F Distribution Figure 2.15 F α Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٥ The F Distribution Consider the example: df 1 = 5 df 2 = 6 α = 0.05 From Appendix D, we get F α, df1, df = F , 5, 6 = 4.39 This means P(F > 4.39) = 0.05 The probability is only 0.05 F will exceed Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٦ ١٨

21 The F Distribution F value for 0.05 probability with 5 and 6 degrees of freedom 0.05 Figure 2.16 F = 4.39 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٧ The Exponential Distribution The exponential distribution (also called the negative exponential distribution) is a continuous distribution often used in queuing models to describe the time required to service a customer. Its probability function is given by: where x f ( X ) = µ e µ X = random variable (service times) µ = average number of units the service facility can handle in a specific period of time e = (the base of natural logarithms) Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٨ ١٩

22 The Exponential Distribution 1 Expected value = = Average service time µ 1 Variance = 2 µ f(x) X Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٣٩ Figure 2.17 Arnold s Muffler Shop Arnold s Muffler Shop installs new mufflers on automobiles and small trucks. The mechanic can install 3 new mufflers per hour. Service time is exponentially distributed. What is the probability that the time to install a new muffler would be ½ hour or less? Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٠ ٢٠

23 Arnold s Muffler Shop Here: X = Exponentially distributed service time µ = average number of units the served per time period = 3 per hour t = ½ hour = 0.5hour P(X 0.5) = 1 e -3(0.5) = 1 e -1.5 = = Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤١ Arnold s Muffler Shop Note also that if: P(X 0.5) = 1 e -3(0.5) = 1 e -1.5 = = Then it must be the case that: P(X>0.5) = = Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٢ ٢١

24 Arnold s Muffler Shop Probability That the Mechanic Will Install a Muffler in 0.5 Hour Figure 2.18 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٣ The Poisson Distribution The Poisson distribution is a discrete distribution that is often used in queuing models to describe arrival rates over time. Its probability function is given by: where x λ e P( X ) = X! P(X) = probability of exactly X arrivals or occurrences λ = average number of arrivals per unit of time (the mean arrival rate) e = 2.718, the base of natural logarithms X = specific value (0, 1, 2, 3, ) of the random variable λ Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٤ ٢٢

25 The Poisson Distribution The mean and variance of the distribution are both λ. Expected value = λ Variance = λ Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٥ Poisson Distribution We can use Appendix C to find Poisson probabilities. Suppose that λ = 2. Some probability calculations are: x λ λ e P( X ) = X! e 1(0.1353) P(0) = = = ! e 2(0.1353) P(1) = = = ! e 4(0.1353) P(2) = = = ! 2 Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٦ ٢٣

26 Exponential and Poisson Together If the number of occurrences per time period follows a Poisson distribution, then the time between occurrences follows an exponential distribution: Suppose the number of phone calls at a service center followed a Poisson distribution with a mean of 10 calls per hour. Then the time between each phone call would be exponentially distributed with a mean time between calls of 6 minutes (1/10 hour). Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall 2-٤٧ ٢٤