Chapter 2 Probability Concepts and Applications Prepared by

Chapter 2 Probability Concepts and Applications Prepared by Lee Revere and John Large To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -1 1

Learning Objectives Students will be able to: 1. Understand the basic foundations of probability analysis. 2. Describe statistically dependent and independent events. 3. Use Bayes’ theorem to establish posterior probabilities. 4. Describe and provide examples of both discrete and continuous random variables. 5. Explain the difference between discrete and continuous probability distributions. 6. Calculate expected values and variances and use the Normal table. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -2 2

Chapter Outline 2. 1 2. 2 2. 3 2. 4 2. 5 2. 6 2. 7 Introduction Fundamental Concepts Mutually Exclusive and Collectively Exhaustive Events Statistically Independent Events Statistically Dependent Events Revising Probabilities with Bayes’ Theorem Further Probability Revisions To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -3 3

Chapter Outline continued 2. 8 2. 9 2. 10 2. 11 2. 12 Random Variables Probability Distributions The Binomial Distribution The Normal Distribution The Exponential Distribution 2. 13 The Poisson Distribution To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -4 4

Introduction § Life is uncertain! § We must deal with risk! § A probability is a numerical statement about the likelihood that an event will occur. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -5 5

Basic Statements about Probability 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. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -6 6

Diversey Paint Example Demand for white latex paint at Diversey Paint and Supply has always been either 0, 1, 2, 3, or 4 gallons per day. Over the past 200 days, the frequencies of demand are represented in the following table: Qty Demanded 0 1 2 3 4 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna No. of Days 40 80 50 20 10 Total 200 2 -7 7

Diversey Paint Example (continued) Probabilities of Demand Probability Quantity Freq. (Relative Freq) Demand (days) (40/200) = 0. 20 0 40 (80/200) = 0. 40 1 80 (50/200) = 0. 25 2 50 (20/200) = 0. 10 3 20 (10/200) = 0. 05 4 10 Total Prob =1. 00 Total days = 200 Note: 0 P(event) 1 and P(event) = 1 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -8 8

Types of Probability Objective probability is based on logical observations: P ( event ) = Number of times event occurs Total number of outcomes or occurrences Determined by: § Relative frequency – Obtained using historical data (Diversey Paint) § Classical method – Known probability for each outcome (tossing a coin) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -9 9

Types of Probability Subjective probability is based on personal experiences. Determined by: § Judgment of experts § Opinion polls § Delphi method § Others To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -10 10

Mutually Exclusive Events § Events are said to be mutually exclusive if only one of the events can occur on any one trial. Example: a fair coin toss results in either a head or a tail. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -11 11

Collectively Exhaustive Events § Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome. § Heads and tails as possible outcomes of coin flip. Example: a collectively exhaustive list of possible outcomes for a fair coin toss includes heads and tails. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -12 12

Die Roll Example Outcome Probability of Roll 1 1/6 2 1/6 This is a collectively exhaustive list of 3 1/6 potential outcomes 4 1/6 for a single die roll. 5 1/6 6 1/6 Total = 1 The outcome is a mutually exclusive event because only one event can occur (a 1, 2, 3, 4, 5, or 6) on any single roll. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -13 13

Twin Birth Example A woman is pregnant with nonidentical twins. Following is a list of collectively exhaustive, mutually exclusive possible outcomes: Outcome of Birth Probability Boy/Boy Boy/Girl/Boy ¼ ¼ What is the probability that both babies will be girls? / boys? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -14 14

In-Class Practice Assuming a traditional 52 -card deck, can you identify if these outcomes are mutually exclusive and/or collectively exhaustive ? ? § § § Draw a spade and a club Draw a face card and a number card Draw an ace and a 3 Draw a club and a nonclub Draw a 5 and a diamond Draw a red card and a diamond To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -15 15

Law of Addition: Mutually Exclusive P (event A or event B) = P (event A) + P (event B) or: P (A or B) = P (A) + P (B) Example: P (spade or club) = P (spade) + P (club) = 13/52 + 13/52 = 26/52 = 1/2 = 50% To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -16 16

Law of Addition: not Mutually Exclusive P(event A or event B) = P(event A) + P(event B) P(event A and event B both occurring) or P(A or B) = P(A)+P(B) - P(A and B) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -17 17

Venn Diagram P(A and To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna B) P(A) 2 -18 P(B) 18

Venn Diagram P(A or B) - + P(A) P(B) = To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna P(A and B) P(A or B) 2 -19 19

In-Class Example: Specialized University offers four different graduate degrees: business, education, accounting, and science. Enrollment figures show 25% of their graduate students are in each specialty. Although 50% of the students are female, only 15% are female business majors. If a student is randomly selected from the University’s registration database: § What is the probability the student is a business or education major? § What is the probability the student is a female or a business major? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -20 20

Specialized University Solution The probability that the student is a business or education major is mutually exclusive event. Thus: P(Bus or Edu) = P(Bus) + P(Edu) =. 25 +. 25 =. 50 or 50% The probability that the student is a female or a business major is not mutually exclusive because the student could be a female business major. Thus: P(Fem or Bus) = P(Fem) + P(Bus) – P(Fem and Bus) =. 50 +. 25 -. 15 =. 60 or 60% To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -21 21

Statistical Dependence § Events are either Ø statistically independent (the occurrence of one event has no effect on the probability of occurrence of the other), or Østatistically dependent (the occurrence of one event gives information about the occurrence of the other). To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -22 22

Which Are Independent? (a) Your education (b) Your income level (a) Draw a jack of hearts from a full 52 -card deck (b) Draw a jack of clubs from a full 52 -card deck (a) Chicago Cubs win the National League pennant (b) Chicago Cubs win the World Series To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -23 23

Probabilities: Independent Events § Marginal probability: the probability of an event occurring: P(A) § Joint probability: the probability of multiple, independent events, occurring at the same time: P(AB) = P(A)*P(B) § Conditional probability (for independent events): Ø the probability of event B given that event A has occurred: P(B|A) = P(B) Ø or, the probability of event A given that event B has occurred: P(A|B) = P(A) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -24 24

Venn Diagram: P(A|B) P(A) P(B|A) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna P(A|B) 2 -25 25

Independent Events Example 1. P(black ball drawn on first draw) A bucket • P(B) = 0. 30 contains 3 black (marginal balls and 7 probability) green balls. We 2. P(two green balls draw a ball drawn) from the bucket, • P(GG) = replace it, and P(G)*P(G) = draw a second ball. 0. 70*0. 70 = 0. 49 (joint probability for two independent events) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -26 26

Independent Events Example continued 1. P(black ball drawn on second draw, first draw was green) ØP(B|G) = P(B) = 0. 30 (conditional probability) 2. P(green ball drawn on second draw, first draw was green) ØP(G|G) = 0. 70 (conditional probability) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -27 27

Probabilities: Dependent Events § Marginal probability: probability of an event occurring: P(A) § Conditional probability (for dependent events): Ø The probability of event B given that event A has occurred: P(B|A) = P(AB)/P(A) Ø The probability of event A given that event B has occurred: P(A|B) = P(AB)/P(B) § Joint probability: The probability of multiple events occurring at the same time: P(AB) = P(B|A)*P(A) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -28 28

Venn Diagram: P(B|A) P(A P(B) and P(A) B) / P(B) P(A) P(B|A) = P(AB)/P(A) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -29 29

Venn Diagram: P(A|B) P(A P(B) and P(A) B) / P(A) P(AB) P(A|B) = P(AB)/P(B) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -30 30

Dependent Events Example Then: Assume that we § P(WL) = 4/10 = 0. 40 have an urn § P(WN) = 2/10 = 0. 20 containing 10 balls of the following § P(W) = 6/10 = 0. 60 descriptions: § P(YL) = 3/10 = 0. 3 § 4 are white (W) and lettered (L) § 2 are white (W) and numbered (N) § 3 are yellow (Y) and lettered (L) § 1 is yellow (Y) and numbered (N) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna § P(YN) = 1/10 = 0. 1 § P(Y) = 4/10 = 0. 4 2 -31 31

Dependent Events Example Frequencies in table format L N Total W 4 2 6 Y 3 1 4 Total 7 3 10 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -32 32

Dependent Events Example Probabilities in table format L N Total W 0. 4 0. 2 0. 6 Y 0. 3 0. 1 0. 4 Total 0. 7 0. 3 1. 0 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -33 33

Dependent Events Example L Marginal N W P(WL) P(WN) P(W) Y P(YL) P(YN) P(Y) Marginal P(L) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna P(N) 2 -34 34

Dependent Events Example (continued) Then: § P(Y) =. 4 - marginal probability § P(L|Y) = P(YL)/P(Y) = 0. 3/0. 4 = 0. 75 - conditional probability § P(W|L) = P(WL)/P(L) = 0. 4/0. 7 = 0. 57 - conditional probability To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -35 35

Dependent Events: Joint Probability Example Your stockbroker informs you that if the stock market reaches the 10, 500 point level by January, there is a 70% probability that Tubeless Electronics will go up in value. Your own feeling is that there is only a 40% chance of the market reaching 10, 500 by January. What is the probability that both the stock market will reach 10, 500 points, and the price of Tubeless will go up in value? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -36 36

Dependent Events: Joint Probabilities Solution Let M represent the event of the Then: stock market P(MT) =P(T|M)P(M) reaching the = (0. 70)(0. 40) 10, 500 point = 0. 28 level, and T represent the event that Tubeless goes up. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -37 37

Revising Probabilities: Bayes’ Theorem Bayes’ theorem can be used to calculate revised or posterior probabilities. Prior Probabilities Bayes’ Process Posterior Probabilities New Information To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -38 38

General Form of Bayes’ Theorem To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -39 39

Posterior Probabilities Example A cup contains two dice identical in appearance. One, however, is fair (unbiased), the other is loaded (biased). The probability of rolling a 3 on the fair die is 1/6 or 0. 166. The probability of tossing the same number on the loaded die is 0. 60. We have no idea which die is which, but we select one by chance, and toss it. The result is a 3. What is the probability that the die rolled was fair? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -40 40

Posterior Probabilities Example (continued) § We know that: P(fair) = 0. 50 P(3|fair) = 0. 166 P(loaded) = 0. 50 P(3|loaded) = 0. 60 - marginal probability § Then: P(3 and fair) = P(3|fair)P(fair) = (0. 166)(0. 50) = 0. 083 P(3 and loaded) = P(3|loaded)P(loaded) = (0. 60)(0. 50) = 0. 300 - joint probability To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -41 41

Posterior Probabilities Example continued A 3 can occur in combination with the state “fair die” or in combination with the state ”loaded die. ” The sum of their probabilities gives the marginal probability of a 3 on a toss: P(3) = 0. 083 + 0. 030 = 0. 383 - marginal probability Then, the probability that the die rolled was the fair one is given by: - conditional probability To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -42 42

Further Probability Revisions To obtain further information as to whether the die just rolled is fair or loaded, let’s roll it again…. Again we get a 3. Given that we have now rolled two 3 s, what is the probability that the die rolled is fair? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -43 43

Further Probability Revisions continued § We know from before that: P(fair) = 0. 50, P(loaded) = 0. 50 § Then: P(3, 3|fair) = P(3|fair)*P(3|fair) = (0. 166) = 0. 027 P(3, 3|loaded) = P(3|loaded)*P(3|loaded) = (0. 60) = 0. 360 § So: P(3, 3 and fair) = P(3, 3|fair)*P(fair) = (0. 027)(0. 50) = 0. 013 P(3, 3 and loaded) = P(3, 3|loaded)P(loaded) = (0. 36)(0. 50) = 0. 180 Thus, the probability of getting two 3 s is a marginal probability obtained from the sum of the probability of two joint probabilities: P(3, 3) = 0. 013 + 0. 180 = 0. 193 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -44 44

Further Probability Revisions continued § Using the probabilities from the previous slide: To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -45 45

Further Probability Revisions continued To give the final comparison: P(fair|3) = 0. 22 P(loaded|3) = 0. 78 P(fair|3, 3) = 0. 067 P(loaded|3, 3) = 0. 933 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -46 46

Random Variables § Discrete random variable - can assume only a finite or limited set of values - i. e. , the number of automobiles sold in a year. § Continuous random variable can assume any one of an infinite set of values - i. e. , temperature, product lifetime. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -47 47

Random Variables (Numeric) Random Variable Number of trees sold Number acceptable Number of people responding % completed after 4 months Time bulb lasts - up to 80, 000 minutes X = number of trees sold Y = number acceptable Z = number of people responding To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna R = % completed after 4 months 0, 1, 2, …, 600 0, 1, 2, …, 5, 000 0 R 100 0 S 80, 000 S = time bulb burns 2 -48 Range of Random Variable 0, 1, 2, , 50 48 Discrete Stock 50 Xmas trees Inspect 600 items Send out 5, 000 sales letters Build an apartment building Test the lifetime of a light bulb (minutes) Outcome Continuous Experiment

Random Variables (Non-numeric) Experiment Outcome Random Variable Range of Random Variable Students Strongly agree (SA) X = 5 if SA 1, 2, 3, 4, 5 respond to a Agree (A) 4 if A questionnaire Neutral (N) 3 if N Disagree (D) 2 if D Strongly Disagree (SD) 1 if SD One machine is Defective Y = 0 if defective 0, 1 inspected Not defective 1 if not defective Consumers Good Z = 3 if good 1, 2, 3 respond to how Average 2 if average they like a Poor 1 if poor product To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -49 49

Probability Distributions §Probability distribution – the set of all possible values of a random variable and their associated probabilities. In a discrete probability distribution a probability between 0 and 1 is assigned to each discrete variable. The sum of the probabilities sum to 1. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -50 50

Discrete Probability Distribution Example Dr. Shannon asked students to respond to the statement, “The textbook was well written and helped me acquire the necessary information. ” Shown below is the discrete probability distribution of the respondents. Outcome SA A N D SD X # Responding 5 10 4 20 3 30 2 30 1 10 P(X) 0. 10 0. 20 0. 30 0. 10 Sum of P(X) = 1. 0 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -51 51

Discrete Probability Distribution Graph A graphical display of a probability distribution yields information about its shape, the central tendency (expected values) and the spread of the data (variance). Below is a graphical depiction of Dr. Shannon’s student responses. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -52 52

Expected Value of a Discrete Prob. Distribution § The expected value of a discrete probability distribution is: n E( X ) = å X i P( X i ) =1 i For Dr. Shannon’s class: Thus, the mean response to Dr. Shannon’s question is between disagree (2) and neutral (3), with the average being closer to neutral. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -53 53

Variance of a Discrete Probability Distribution § The variance of a discrete probability distribution is: n 2 s = å[X i - E(X )] P(X i ) 2 i= 1 For Dr. Shannon’s class: Thus, the standard deviation for Dr. Shannon’s question is 1. 29 = 1. 136 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -54 54

Variance of a Discrete Probability Distribution § The standard deviation is a measure of the dispersion or spread of the data that is related to the variance. The formula for the standard deviation of all probability functions is: s= s 2 For Dr. Shannon’s class: To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -55 55

Binomial Distribution § The binomial distribution is a probability distribution with: Øtrials that follow a Bernoulli process and have two possible outcomes. Øprobabilities that stay the same from one trial to the next. Øtrials that are statistically independent. Øa positive integer number of trials. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -56 56

Binomial Formulas § The binomial formula can be used to determine the probability of r successes in n trials. n! r n-r = pq r!(n - r)! Where, n = number of trials r = number of successes p = probability of success q = probability of failure (1 -p) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -57 57

Binomial Formulas (continued) § For a binomial distribution, the expected value, or mean, is: m = np § The variance is: s 2 = np (1 - p) § The standard deviation is: s= s 2 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -58 58

Soda Selection: Binomial Example § Suppose 50% of your friends prefer diet soda to regular soda. § You decide to practice your new binomial skills while studying with five friends. § You bring both diet and regular soda to your next study session and offer one to each of your friends. Ø What is the probability that only one of your friends will select a diet soda? Ø What is the probability that three of your friends will select the diet soda? Ø What is the expected value, variance, and standard deviation of your experiment? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -59 59

Soda Selection Solution What is the probability that only one of your friends will select a diet soda? =. 1563 What is the probability that three of your friends will select the diet soda? =. 3125 These questions can be answered using the binomial formula, where n = 5, r = 1 then 3, p =. 5 and q =. 5. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -60 60

Soda Selection Solution Below is a graphical depiction of the answers. P(r=3) P(r=1) 0 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 1 2 2 -61 3 4 61 5

Soda Selection Solution – Binomial Table Now let’s answer these same questions using the binomial table: p= 50% success n r 0. 05 0. 10 0. 15 0. 20 0. 25 0. 30 0. 35 0. 40 0. 45 0. 50 1 0 0. 9500 0. 9000 0. 8500 0. 8000 0. 7500 0. 7000 0. 6500 0. 6000 0. 5500 0. 5000 1 0. 0500 0. 1000 0. 1500 0. 2000 0. 2500 0. 3000 0. 3500 0. 4000 0. 4500 0. 5000 0 0. 9025 0. 8100 0. 7225 0. 6400 0. 5625 0. 4900 0. 4225 0. 3600 0. 3025 0. 2500 1 0. 0950 0. 1800 0. 2550 0. 3200 0. 3750 0. 4200 0. 4550 0. 4800 0. 4950 0. 5000 2 0. 0025 0. 0100 0. 0225 0. 0400 0. 0625 0. 0900 0. 1225 0. 1600 0. 2025 0. 2500 0 0. 8574 0. 7290 0. 6141 0. 5120 0. 4219 0. 3430 0. 2746 0. 2160 0. 1664 0. 1250 1 0. 1354 0. 2430 0. 3251 0. 3840 0. 4219 0. 4410 0. 4436 0. 4320 0. 4084 0. 3750 2 0. 0071 0. 0270 0. 0574 0. 0960 0. 1406 0. 1890 0. 2389 0. 2880 0. 3341 0. 3750 3 0. 0001 0. 0010 0. 0034 0. 0080 0. 0156 0. 0270 0. 0429 0. 0640 0. 0911 0. 1250 0 0. 8145 0. 6561 0. 5220 0. 4096 0. 3164 0. 2401 0. 1785 0. 1296 0. 0915 0. 0625 1 0. 1715 0. 2916 0. 3685 0. 4096 0. 4219 0. 4116 0. 3845 0. 3456 0. 2995 0. 2500 2 0. 0135 n= 5 friends 0. 0486 0. 0975 0. 1536 0. 2109 0. 2646 0. 3105 0. 3456 0. 3675 0. 3750 3 0. 0005 0. 0036 0. 0115 0. 0256 0. 0469 0. 0756 0. 1115 0. 1536 0. 2005 0. 2500 4 0. 0000 0. 0001 0. 0005 0. 0016 0. 0039 0. 0081 0. 0150 0. 0256 0. 0410 0. 0625 0 0. 7738 0. 5905 0. 4437 0. 3277 0. 2373 0. 1681 0. 1160 0. 0778 0. 0503 0. 0313 1 0. 2036 0. 3281 0. 3915 0. 4096 0. 3955 0. 3602 0. 3124 0. 2592 0. 2059 0. 1563 2 0. 0214 0. 0729 0. 1382 0. 2048 0. 2637 0. 3087 0. 3364 0. 3456 0. 3369 0. 3125 3 0. 0011 0. 0081 0. 0244 0. 0512 0. 0879 0. 1323 0. 1811 0. 2304 0. 2757 0. 3125 4 0. 0000 0. 0005 0. 0022 0. 0064 0. 0146 0. 0284 0. 0488 0. 0768 0. 1128 0. 1563 5 0. 0000 0. 0001 0. 0003 0. 0010 0. 0024 0. 0053 0. 0102 0. 0185 0. 0313 0 0. 7351 0. 5314 0. 3771 0. 2621 0. 1780 0. 1176 0. 0754 0. 0467 0. 0277 0. 0156 1 0. 2321 0. 3543 0. 3993 0. 3932 0. 3560 0. 3025 0. 2437 0. 1866 0. 1359 0. 0938 2 0. 0305 0. 0984 0. 1762 0. 2458 0. 2966 0. 3241 0. 3280 0. 3110 0. 2780 0. 2344 3 0. 0021 0. 0146 0. 0415 0. 0819 0. 1318 0. 1852 0. 2355 0. 2765 0. 3032 0. 3125 0. 0154 2 -620. 0330 62 0. 0595 0. 0951 0. 1382 0. 1861 0. 2344 0. 0015 0. 0044 0. 0102 0. 0205 0. 0369 0. 0609 0. 0938 0. 0001 0. 0002 0. 0007 0. 0018 0. 0041 0. 0083 0. 0156 2 3 4 5 6 To accompany Quantitative Analysis 4 0. 0001 0. 0012 0. 0055 for Management, 9 e 5 0. 0000 0. 0001 0. 0004 by Render/Stair/Hanna 6 0. 0000

Soda Selection Solution continued What is the expected value, variance, and standard deviation of your experiment? expected value (μ) = np = 5(. 5) = 2. 5 variance = np(1 -p) = 5(. 5) = 1. 25 standard deviation = = = 1. 118 So, on average half of your five friends will select diet soda – this is intuitive because 50% of your friends prefer diet soda. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -63 63

Normal Distribution § The normal distribution is the most popular and useful continuous probability distribution. ØSpecified completely by the mean and standard deviation ØSymmetrical, with the midpoint representing the mean ØValues on the X axis are measured in the number of standard deviations away from the mean. ØAs the standard deviation becomes larger the curve flattens. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -64 64

Normal Distribution continued Normal Distribution Probability density function - f(X) To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -65 65

Normal Distribution for Different Values of Different values of the mean shift the curve, but do not affect the shape of the distribution. =40 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna =50 2 -66 =60 66

Normal Distribution for Different Values of Different values of the standard deviation will flatten the curve, but do not affect the mean. =1 =0. 3 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna =0. 2 2 -67 67

Three Common Areas under the Curve Three commonly used areas under the normal curve are +/- 1, 2 and 3 standard deviations. To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -68 68

The Relationship Between Z and X =100 =15 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -69 69

Haynes Construction Company Example Haynes Construction Company builds primarily triplex and quadraplex apartment buildings for investors, and it is believed that the total construction time follows a normal distribution. The mean time to construct a triplex is 100 days, and the standard deviation is 20 days. Failure to complete the construction in 125 days results in penalty fees. However, early completion of 75 days or less results in a bonus. What is the probability that Haynes will pay a penalty fee? What is the probability that Haynes will receive a bonus? To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -70 70

Haynes Construction Company Solution What is the probability Haynes will pay a penalty fee? = (125 – 100) / 20 = 1. 25 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -71 71

The Standard Normal Table There is an 89. 44% chance the construction will be complete in 125 days or less, thus, there is a 1 -. 89435, or a 10. 57% chance the construction will take longer causing Haynes to pay a penalty! Z 0 0. 01 0. 02 0. 03 0. 04 0. 05 0. 06 0. 0 0. 50000 0. 50399 0. 50798 0. 51197 0. 51595 0. 51994 0. 52392 0. 1 0. 53983 0. 54380 0. 54776 0. 55172 0. 55567 0. 55962 0. 56356 0. 2 0. 57926 0. 58317 0. 58706 0. 59095 0. 59483 0. 59871 0. 60257 0. 3 0. 61791 0. 62172 0. 62552 0. 62930 0. 63307 0. 63683 0. 64058 0. 4 0. 65542 0. 65910 0. 66276 0. 66640 0. 67003 0. 67364 0. 67724 0. 5 0. 69146 0. 69497 0. 69847 0. 70194 0. 70540 0. 70884 0. 71226 0. 72575 0. 72907 0. 73237 0. 73565 0. 73891 0. 74215 0. 74537 0. 75804 0. 76115 0. 76424 0. 76730 0. 77035 0. 77337 0. 77637 0. 8 0. 78814 0. 79103 0. 79389 0. 79673 0. 79955 0. 80234 0. 80511 0. 9 0. 81594 0. 81859 0. 82121 0. 82381 0. 82639 0. 82894 0. 83147 1. 0 0. 84134 0. 84375 0. 84614 0. 84849 0. 85083 0. 85314 0. 85543 1. 1 0. 86433 0. 86650 0. 86864 0. 87076 0. 87286 0. 87493 0. 87698 1. 2 0. 88493 0. 88686 0. 88877 0. 89065 0. 89251 0. 89435 0. 89617 1. 3 0. 90320 0. 90490 0. 90658 0. 90824 0. 90988 0. 91149 0. 91308 1. 4 0. 91924 0. 92073 0. 92220 0. 92364 0. 92507 0. 92647 0. 92785 0. 935742 -72 0. 93699 0. 93822 0. 93943 0. 94062 To accompany Quantitative Analysis 0. 93319 0. 93448 for 1. 5 Management, 9 e by Render/Stair/Hanna 72

Haynes Construction Company Solution What is the probability Haynes will pay a penalty fee? = (75 – 100) / 20 = -1. 25 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -73 73

The Standard Normal Table The standard normal table does not have negative value, so we must look up the positive value and subtract from one ~ this works because of the symmetrical property of the normal. Thus, there is a 1 -. 89435, or a 10. 57% chance the construction will take less than 75 days and Haynes will get a bonus! Z 0 0. 01 0. 02 0. 03 0. 04 0. 05 0. 06 0. 0 0. 50000 0. 50399 0. 50798 0. 51197 0. 51595 0. 51994 0. 52392 0. 1 0. 53983 0. 54380 0. 54776 0. 55172 0. 55567 0. 55962 0. 56356 0. 2 0. 57926 0. 58317 0. 58706 0. 59095 0. 59483 0. 59871 0. 60257 0. 3 0. 61791 0. 62172 0. 62552 0. 62930 0. 63307 0. 63683 0. 64058 0. 4 0. 65542 0. 65910 0. 66276 0. 66640 0. 67003 0. 67364 0. 67724 0. 5 0. 69146 0. 69497 0. 69847 0. 70194 0. 70540 0. 70884 0. 71226 0. 72575 0. 72907 0. 73237 0. 73565 0. 73891 0. 74215 0. 74537 0. 75804 0. 76115 0. 76424 0. 76730 0. 77035 0. 77337 0. 77637 0. 8 0. 78814 0. 79103 0. 79389 0. 79673 0. 79955 0. 80234 0. 80511 0. 9 0. 81594 0. 81859 0. 82121 0. 82381 0. 82639 0. 82894 0. 83147 1. 0 0. 84134 0. 84375 0. 84614 0. 84849 0. 85083 0. 85314 0. 85543 1. 1 0. 86433 0. 86650 0. 86864 0. 87076 0. 87286 0. 87493 0. 87698 1. 2 0. 88493 0. 88686 0. 88877 0. 89065 0. 89251 0. 89435 0. 89617 1. 3 0. 90320 0. 90490 0. 90658 0. 90824 0. 90988 0. 91149 0. 91308 1. 4 0. 91924 0. 92073 0. 92220 0. 92364 0. 92507 0. 92647 0. 92785 0. 935742 -74 0. 93699 0. 93822 0. 93943 0. 94062 To accompany Quantitative Analysis 0. 93319 0. 93448 for 1. 5 Management, 9 e by Render/Stair/Hanna 74

Haynes Construction Company Example Other questions, such as the probability that the construction will be completed between 110 and 125 days can also be answered. = (125 – 100) / 20 = 1. 25 and = (110 – 100) / 20 = 0. 5 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -75 75

Haynes Construction Company Example The area under the curve must be found for both z values and subtracted from one another. Z (. 5) =. 69146 Z (1. 25) =. 89435 P(. 5 < Z < 1. 25) =. 89435 -. 69146 =. 20289 or 20. 29% To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -76 76

The Negative Exponential Distribution § The negative exponential distribution is a continuous distribution that is often used to describe the time required to service a customer. =5 To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna Expected value = 1/ Variance = 1/ 2 2 -77 77

The Poisson Distribution § The Poisson distribution is a discrete distribution that is often used to describe arrival rates. =2 Expected value = Variance = To accompany Quantitative Analysis for Management, 9 e by Render/Stair/Hanna 2 -78 78