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Chapter 3-PART I QUANTITATIVE METHODS FOR BUSINESS DISCREETE PROBABILITY DISTRIBUTION John Kooti Slide 1 Chapter 3, Part I Discrete Probability Distributions Random Variables Discrete Random Variables Expected Value and Variance Binomial Probability Distribution Poisson Probability Distribution Slide 2 Random Variables A random variable is a numerical description of the outcome of an experiment. A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals. Slide 3 Example: JSL Appliances Discrete random variable with a finite number of values: Let x = number of TV sets sold at the store in one day where x can take on 5 values (0, 1, 2, 3, 4) Discrete random variable with an infinite sequence of values: Let x = number of customers arriving in one day where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive. Slide 4 Discrete Probability Distributions The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable. The required conditions for a discrete probability function are: f(x) > 0 Sf(x) = 1 We can describe a discrete probability distribution with a table, graph, or equation. Slide 5 Example: JSL Appliances Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below right) was developed. Number Units Sold of Days x f(x) 0 80 0 .40 1 50 1 .25 2 40 2 .20 3 10 3 .05 4 20 4 .10 200 1.00 Slide 6 Example: JSL Appliances A graphical representation of the probability distribution for TV sales in one day .50 .40 Probability .30 .20 .10 0 1 2 3 4 Values of Random Variable x (TV sales) Slide 7 Expected Value and Variance The expected value, or mean, of a random variable is a measure of its central location. • Expected value of a discrete random variable: E(x) = m = Sxf(x) The variance summarizes the variability in the values of a random variable. • Variance of a discrete random variable: Var(x) = s2 = S(x - m)2f(x) The standard deviation, s, is defined as the positive square root of the variance. Slide 8 Example: JSL Appliances Expected Value of a Discrete Random Variable x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40 1.20 = E(x) The expected number of TV sets sold in a day is 1.2 Slide 9 Example: JSL Appliances Variance and Standard Deviation of a Discrete Random Variable x x-m (x - m)2 f(x) (x - m)2f(x) _____ _________ ___________ _______ _______________ 0 -1.2 1.44 .40 .576 1 -0.2 0.04 .25 .010 2 0.8 0.64 .20 .128 3 1.8 3.24 .05 .162 4 2.8 7.84 .10 .784 1.660 = s 2 The variance of daily sales is 1.66 TV sets squared. The standard deviation of sales is 1.29 TV sets. Slide 10 Example: JSL Appliances Formula Spreadsheet for Computing Expected Value and Variance A B C D 1 x f(x) xf(x) (x- m ) 2 f(x) 2 0 0.40 =A2*B2 =(A2-C$7)^2*B2 3 1 0.25 =A3*B3 =(A3-C$7)^2*B3 4 2 0.20 =A4*B4 =(A4-C$7)^2*B4 5 3 0.05 =A5*B5 =(A5-C$7)^2*B5 6 4 0.10 =A6*B6 =(A6-C$7)^2*B6 7 =SUM(C2:C6) =SUM(D2:D6) 8 Expected Value Variance Slide 11 Binomial Probability Distribution Properties of a Binomial Experiment • The experiment consists of a sequence of n identical trials. • Two outcomes, success and failure, are possible on each trial. • The probability of a success, denoted by p, does not change from trial to trial. • The trials are independent. Slide 12 Example: Evans Electronics Binomial Probability Distribution Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year? Let: p = .10, n = 3, x = 1 Slide 13 Binomial Probability Distribution Binomial Probability Function n! f ( x) p x (1 p ) (n x ) x !( n x )! where f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial Slide 14 Example: Evans Electronics Using the Binomial Probability Function n! f ( x) p x (1 p ) (n x ) x !( n x )! 3! f (1) ( 0.1)1 ( 0. 9) 2 1!( 3 1)! = (3)(0.1)(0.81) = .243 Slide 15 Example: Evans Electronics Using the Tables of Binomial Probabilities p n x .10 .15 .20 .25 .30 .35 .40 .45 .50 3 0 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250 1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750 2 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750 3 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250 Slide 16 Example: Evans Electronics Using a Tree Diagram First Second Third Value Worker Worker Worker of x Probab. L (.1) 3 .0010 Leaves (.1) S (.9) 2 .0090 Leaves (.1) L (.1) 2 .0090 Stays (.9) 1 .0810 S (.9) L (.1) 2 .0090 Leaves (.1) S (.9) 1 .0810 Stays (.9) L (.1) 1 .0810 Stays (.9) 0 .7290 S (.9) Slide 17 Example: Evans Electronics Using an Excel Spreadsheet • Step 1: Select a cell in the worksheet where you want the binomial probabilities to appear. • Step 2: Select the Insert pull-down menu. • Step 3: Choose the Function option. • Step 4: When the Paste Function dialog box appears: Choose Statistical from the Function Category box. Choose BINOMDIST from the Function Name box. Select OK. continued Slide 18 Example: Evans Electronics Using an Excel Spreadsheet (continued) • Step 5: When the BINOMDIST dialog box appears: Enter 1 in the Number_s box (value of x). Enter 3 in the Trials box (value of n). Enter .1 in the Probability_s box (value of p). Enter false in the Cumulative box. [Note: At this point the desired binomial probability of .243 is automatically computed and appears in the right center of the dialog box.] Select OK (and .243 will appear in the worksheet cell requested in Step 1). Slide 19 The Binomial Probability Distribution Expected Value E(x) = m = np Variance Var(x) = s 2 = np(1 - p) Standard Deviation SD( x ) s np(1 p) Example: Evans Electronics E(x) = m = 3(.1) = .3 employees out of 3 Var(x) = s 2 = 3(.1)(.9) = .27 SD( x) s 3(.1)(.9) .52 employees Slide 20 MINITAB APPLICATION Minitab Program Slide 21 Poisson Probability Distribution Properties of a Poisson Experiment • The probability of an occurrence is the same for any two intervals of equal length. • The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval. Slide 22 Poisson Probability Distribution Poisson Probability Function m x e m f ( x) x! where f(x) = probability of x occurrences in an interval m = mean number of occurrences in an interval e = 2.71828 Slide 23 Example: Mercy Hospital Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening? Using the Poisson Probability Function m = 6/hour = 3/half-hour, x = 4 34 ( 2. 71828) 3 f ( 4) .1680 4! Slide 24 Example: Mercy Hospital Using the Tables of Poisson Probabilities m x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .0498 1 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .1494 2 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .2240 3 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .2240 4 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .1680 5 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .1008 6 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .0504 7 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .0216 8 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081 9 .0003 .0004 .0005 .0007 .0009 .0011 .0014 .0018 .0022 .0027 10 .0001 .0001 .0001 .0002 .0002 .0003 .0004 .0005 .0006 .0008 11 .0000 .0000 .0000 .0000 .0000 .0001 .0001 .0001 .0002 .0002 12 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0001 Slide 25 Poisson Probability Distribution The Poisson probability distribution can be used as an approximation of the binomial probability distribution when p, the probability of success, is small and n, the number of trials, is large. • Approximation is good when p < .05 and n > 20 • Set m = np and use the Poisson tables. Slide 26 MINIAB APPLICATION Minitab Program Slide 27 The End of Discrete Probability Distribution Slide 28 Chapter 3, Part II Continuous Random Variables Continuous Random Variables Normal Probability Distribution Exponential Probability Distribution Slide 29 Continuous Probability Distributions A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval. The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2. Slide 30 Uniform Probability Distribution A random variable is uniformly distributed whenever the probability is proportional to the length of the interval. Uniform Probability Density Function f(x) = 1/(b - a) for a < x < b = 0 elsewhere Expected Value of x E(x) = (a + b)/2 Variance of x Var(x) = (b - a)2/12 where: a = smallest value the variable can assume b = largest value the variable can assume Slide 31 Example: Slater's Buffet Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Probability Density Function f(x) = 1/10 for 5 < x < 15 = 0 elsewhere where x = salad plate filling weight Slide 32 Example: Slater's Buffet What is the probability that a customer will take between 12 and 15 ounces of salad? f(x) P(12 < x < 15) = 1/10(3) = .3 1/10 x 5 10 12 15 Salad Weight (oz.) Slide 33 MINITAB APPLICATION Minitab Program Slide 34 The Normal Probability Distribution Graph of the Normal Probability Density Function f(x) x m Slide 35 Normal Probability Distribution The Normal Curve • The shape of the normal curve is often illustrated as a bell-shaped curve. • The highest point on the normal curve is at the mean, which is also the median and mode of the distribution. • The normal curve is symmetric. • The standard deviation determines the width of the curve. • The total area under the curve is 1. • Probabilities for the normal random variable are given by areas under the curve. Slide 36 Normal Probability Distribution Normal Probability Density Function 1 ( x m )2 / 2s 2 f (x) e s 2p where m = mean s = standard deviation p = 3.14159 e = 2.71828 Slide 37 Standard Normal Probability Distribution A random variable that has a normal distribution with a mean of zero and a standard deviation of one is said to have a standard normal probability distribution. The letter z is commonly used to designate this normal random variable. Converting to the Standard Normal Distribution xm z s We can think of z as a measure of the number of standard deviations x is from m. Slide 38 Example: Pep Zone Pep Zone sells auto parts and supplies including a popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that leadtime demand is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stockout, P(x > 20). Slide 39 Example: Pep Zone Standard Normal Distribution Area = .2967 z = (x - m)/s Area = .2033 = (20 - 15)/6 = .83 Area = .5 z 0 .83 The Standard Normal table shows an area of .2967 for the region between the z = 0 line and the z = .83 line above. The shaded tail area is .5 - .2967 = .2033. The probability of a stockout is .2033. Slide 40 Example: Pep Zone Using the Standard Normal Probability Table z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 .1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 .2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 .3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 .4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 .5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 .6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549 .7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 .8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133 .9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389 Slide 41 Example: Pep Zone Using an Excel Spreadsheet • Step 1: Select a cell in the worksheet where you want the normal probability to appear. • Step 2: Select the Insert pull-down menu. • Step 3: Choose the Function option. • Step 4: When the Paste Function dialog box appears: Choose Statistical from the Function Category box. Choose NORMDIST from the Function Name box. Select OK. continue Slide 42 Example: Pep Zone Using an Excel Spreadsheet (continued) • Step 5: When the NORMDIST dialog box appears: Enter 20 in the x box. Enter 15 in the mean box. Enter 6 in the standard deviation box. Enter true in the cumulative box. Select OK. At this point, .7967 will appear in the cell selected in Step 1, indicating that the probability of demand during lead time being less than or equal to 20 gallons is .7967. The probability that demand will exceed 20 gallons is 1 - .7967 = .2033. Slide 43 Example: Pep Zone If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Area = .05 Area = .5 Area = .45 cutting the z.05 area of .05. Let z.05 represent the z value 0 tail Slide 44 Example: Pep Zone Using the Standard Normal Probability Table We now look-up the .4500 area in the Standard Normal Probability table to find the corresponding z.05 value. z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 . 1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 z.05 = 1.645 is a reasonable estimate. . Slide 45 Example: Pep Zone The corresponding value of x is given by x = m + z.05s = 15 + 1.645(6) = 24.87 A reorder point of 24.87 gallons will place the probability of a stockout during leadtime at .05. Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05. Slide 46 Exponential Probability Distribution Exponential Probability Density Function 1 f ( x) e x/m for x > 0, m > 0 m where m = mean e = 2.71828 Cumulative Exponential Distribution Function P ( x x0 ) 1 e xo / m where x0 = some specific value of x Slide 47 Example: Al’s Carwash The time between arrivals of cars at Al’s Carwash follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less. P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866 Slide 48 Example: Al’s Carwash Graph of the Probability Density Function f(x) .4 .3 P(x < 2) = area = .4866 .2 .1 x 1 2 3 4 5 6 7 8 9 10 Slide 49 Relationship Between the Poisson and Exponential Distributions The continuous exponential probability distribution is related to the discrete Poisson distribution. The Poisson distribution provides an appropriate description of the number of occurrences per interval. The exponential distribution provides a description of the length of the interval between occurrences. Slide 50 The End of Chapter 3 Slide 51