The normal distribution revisited UNIFORM DISTRIBUTION (OVER A SPHERE) The Normal Distribution The Normal Distribution is perhaps the most important distribution in probability and statistics. It often arises in practice due to a result in probability known as “The Central Limit Theorem”, which roughly states that sums and averages of independent identically distributed random variables tend to be normally distributed. © 2002 Thomson / South-Western Slide 6-2 The Normal Distribution For example, if “n” is large and “p” is “not small”, then binomial (n, p)-distribution can be approximated well by the normal distribution with mean np and variance np(1 – p). © 2002 Thomson / South-Western Slide 6-3 The Normal Distribution A normal distribution is completely characterized by its mean µ and standard deviation σ and has the probability density function given by: 1 − 2  x−       2 1  2 e Where:  = mean of X  = standard deviation of X  = 3.14159 . . . e = 2.71828 . . . f ( x) = © 2002 Thomson / South-Western Slide 6-4 The Normal Distribution Normal random variable is often denoted as N(µ, σ). The probability density function of normal distribution is bell shaped and symmetric around its mean.  © 2002 Thomson / South-Western Slide 6-5 The Normal Distribution In applications, one is often interested in computing the probability that a normal random variable is smaller than a given value and vice versa. We indicate two computational methods: one uses Microsoft Excel functions, while the other is based on the traditional approach of standardizing a normal random variable. Given a r.v. X that is N(µ, σ), we often face two problems. © 2002 Thomson / South-Western Slide 6-6 The Normal Distribution Problem 1: Given any number x, find a probability p = P{X ≤ x}. Method 1. Use the Microsoft Excel function NORMDIST: p = NORMDIST(x, µ, σ, True). © 2002 Thomson / South-Western Slide 6-7 The Normal Distribution Problem 1: Given any number x, find a probability p = P{X ≤ x}. Method 2. Transform the random variable X and the number x into a new random variable Z and number z as: 𝑥−𝜇 𝑍 = 𝑋−𝜇 and 𝑧 = 𝜎 𝜎 Then Z is also normally distributed with mean 0 and a standard deviation 1. © 2002 Thomson / South-Western Slide 6-8 The Normal Distribution Problem 1: Method 2. (continued) That is, Z = N(0, 1), which is called “standard normal” random variable. Clearly 𝑝=𝑃 𝑋≤𝑥 =𝑃 𝑍≤𝑧 which can be read from the table of cumulative distribution of the standard normal r.v. For negative values of z, notice by symmetry that 𝑃 𝑍 ≤ 𝑧 = 𝑃 𝑍 ≥ −𝑧 = 1 − 𝑃 𝑍 ≤ −𝑧 © 2002 Thomson / South-Western Slide 6-9 The Normal Distribution Problem 2: Conversely, given any probability p, find a number x, such that P{X ≤ x} = p. Method 1. Use the Microsoft Excel function NORM.INV: x = NORM.INV(p, µ, σ) © 2002 Thomson / South-Western Slide 6-10 The Normal Distribution Problem 2: Method 2. Transform the random variable X and the number x into their “standard” counterparts Z and z as described above. Given p, read z backwards from the table such that P{Z ≤ z} = p and find the quantity x by transforming back: x = µ + zσ. © 2002 Thomson / South-Western Slide 6-11 Discussion Problem W03-02 • Assume that weekly demand for phones at B&M Office Supplies is normally distributed and independent across weeks, with a mean of 2,500 and a standard deviation of 500. The manufacturer takes two weeks (lead time) to fill an order placed by the B&M manager. Copyright © 2016 Pearson Education, Inc. 12 – 12 Discussion Problem W03-02 a) The store manager currently orders 10,000 phones when the inventory on hand drops to the Reorder Point (ROP) = 6,000. What is the probability that demand during lead time (ddlt) does not exceed ROP = 6,000? That is, what is the probability of no stock-out? b) The store manager orders 10,000 phones when the inventory on hand drops to ROP. The probability that demand during lead time (ddlt) does not exceed ROP is required to be at least 90%. What is the ROP? Copyright © 2016 Pearson Education, Inc. 12 – 13 Normal distribution Generate Normal Distribution: x = NORM.INV(RAND(), µ, σ) 0.918842 0.395963 -0.79772 0.11585 0.987444 -0.24365 -0.71719 -0.74743 2.297748 1.169751 -1.21575 0.663374 0.470682 -1.16578 -0.77616 2.033189 -0.44258 0.727747 -0.09513 -1.293 0.88108 -0.91086 -0.16578 -0.31745 0.514011 1.505128 -0.44205 1.003868 -0.26488 0.309634 -1.55528 -0.27688 1.318288 -1.48123 0.428877 1.039995 -1.50424 -0.24944 1.112608 -0.90563 0.22384 1.134955 -1.90455 0.051121 -0.63878 -0.03582 0.147168 1.032384 0.763318 -1.58178 0.190792 0.116342 0.299409 -0.35769 0.020187 -1.70079 1.54783 0.523125 -0.67707 -0.40655 -3.19294 0.24598 -0.62806 -1.56228 -2.02277 -0.4383 1.002191 0.753205 -0.47117 1.479776 -0.05458 -2.40283 0.587668 -0.06385 1.013893 -0.14877 0.821971 0.447103 0.662095 -0.18641 -1.03276 -0.79103 0.245152 1.432146 -0.2499 -0.17436 -0.14949 0.79584 -0.67783 -1.53133 0.272352 -0.99651 1.039476 -0.07641 0.020781 0.223408 1.827398 -0.72093 0.476805 -3.11503 Normal distribution Generate Normal Distribution: x = NORM.INV(RAND(), µ, σ) Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) (The data should all be in one column.) Load the Analysis Toolpak. Under the Tools menu, choose Data Analysis, and then Regression. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) Follow the directions given in the dialog box: 1. Enter values for the Input Y Range. The Input Y Range contains the data for which you want the probability plot. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) Follow the directions given in the dialog box: 2. Enter values for the Input X Range. These are irrelevant in this case. We are only interested in the Normal Probability Plot option. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) Follow the directions given in the dialog box: 3. Check the Normal Probability Plots option. Click OK. Excel creates a normal probability plot. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) Follow the directions given in the dialog box: 4. Click on the newly created chart. Under Chart in the menu, choose Add Trendline. Under the Type tab, choose the linear option. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) Follow the directions given in the dialog box: 5. Under the Options tab, check Display Rsquared value on chart. The square root of the R-squared value is the correlation value. Normal distribution How can we check if the distribution is normal? ( http://www.engr.mun.ca/~ggeorge/4421/demos/NormPlot.html ) 0.918842 -1.21575 0.88108 -1.55528 0.22384 0.190792 -3.19294 -0.05458 -1.03276 0.272352 0.395963 0.663374 -0.91086 -0.27688 1.134955 0.116342 0.24598 -2.40283 -0.79103 -0.99651 -0.79772 0.470682 -0.16578 1.318288 -1.90455 0.299409 -0.62806 0.587668 0.245152 1.039476 0.11585 -1.16578 -0.31745 -1.48123 0.051121 -0.35769 -1.56228 -0.06385 1.432146 -0.07641 0.987444 -0.77616 0.514011 0.428877 -0.63878 0.020187 -2.02277 1.013893 -0.2499 0.020781 -0.24365 2.033189 1.505128 1.039995 -0.03582 -1.70079 -0.4383 -0.14877 -0.17436 0.223408 -0.71719 -0.44258 -0.44205 -1.50424 0.147168 1.54783 1.002191 0.821971 -0.14949 1.827398 -0.74743 0.727747 1.003868 -0.24944 1.032384 0.523125 0.753205 0.447103 0.79584 -0.72093 2.297748 -0.09513 -0.26488 1.112608 0.763318 -0.67707 -0.47117 0.662095 -0.67783 0.476805 1.169751 -1.293 0.309634 -0.90563 -1.58178 -0.40655 1.479776 -0.18641 -1.53133 -3.11503 Normal distribution Normal Probability Plot 3 R² = 0.9302 2 1 0 Y 0 20 40 60 -1 -2 -3 -4 Sample Percentile 80 100 120 Discussion Problems W05-01 Check if the 100 numbers in Excel Worksheet data of normalPlot.xlsx are normally distributed. 0.918842 -1.21575 0.88108 -1.55528 0.22384 0.190792 -3.19294 -0.05458 -1.03276 0.272352 0.395963 0.663374 -0.91086 -0.27688 1.134955 0.116342 0.24598 -2.40283 -0.79103 -0.99651 -0.79772 0.470682 -0.16578 1.318288 -1.90455 0.299409 -0.62806 0.587668 0.245152 1.039476 0.11585 -1.16578 -0.31745 -1.48123 0.051121 -0.35769 -1.56228 -0.06385 1.432146 -0.07641 0.987444 -0.77616 0.514011 0.428877 -0.63878 0.020187 -2.02277 1.013893 -0.2499 0.020781 -0.24365 2.033189 1.505128 1.039995 -0.03582 -1.70079 -0.4383 -0.14877 -0.17436 0.223408 -0.71719 -0.44258 -0.44205 -1.50424 0.147168 1.54783 1.002191 0.821971 -0.14949 1.827398 -0.74743 0.727747 1.003868 -0.24944 1.032384 0.523125 0.753205 0.447103 0.79584 -0.72093 2.297748 -0.09513 -0.26488 1.112608 0.763318 -0.67707 -0.47117 0.662095 -0.67783 0.476805 1.169751 -1.293 0.309634 -0.90563 -1.58178 -0.40655 1.479776 -0.18641 -1.53133 -3.11503 More about normal distribution (Spherically Uniform) How can we pick a set of random points uniformly distributed on the unit circle x12 + x22=1? More about normal distribution (Spherically Uniform) How can we generate uniformly distributed points on the surface of the 3-d unit sphere x2 + y2 + z2 = 1? More about normal distribution (Spherically Uniform) • A standard method is to generate three standard normals and construct a unit vector from them. That is, when Xi ∼ N(0,1) and λ2=X12 + X22+X32, then (X1/λ,X2/λ,X3/λ) is uniformly distributed on the sphere. This method works well for d-dimensional spheres, too. More about normal distribution (Spherically Uniform) Here is a picture of 100 independent draws from a uniform spherical distribution obtained with the method: More about normal distribution (Spherically Uniform) Spherical Uniform Distribution: • How can we pick a set of random points uniformly distributed on the 4-dimensional unit sphere x12 + x22 + x32 + x42 =1? Discussion Problem W05-02 Generate 100 random points (x1, x2, x3, x4) uniformly distributed on the 4-dimensional unit sphere x12 + x22 + x32 + x42 =1. Chi-Square Distribution • In probability theory and statistics, the chisquared distribution with k degrees of freedom is the distribution of a sum of the squares of k independent standard normal random variables. Discussion Problem W05-03 • A student claims that the 20 points on the right are uniformly distributed on the unit sphere. Check if his claim is true. Use an Excel function CHISQ.INV(RAND(),k) to scale the points on the unit sphere back to normal. ID 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x1 0.430004 -0.02946 0.229287 -0.17224 -0.49227 -1.04505 1.189759 0.2092 -0.15866 -0.4369 0.443128 -0.44789 -0.28041 0.611842 -0.64634 -0.23787 -0.92584 0.078051 -0.71522 -0.48723 x2 -0.90149 0.091992 0.086981 -0.02726 0.340813 -0.33236 -0.36662 0.046161 0.494648 -0.55992 0.126155 -0.33003 0.097483 -0.70488 0.227253 0.217872 0.067467 0.087426 -0.41675 -0.4449 x3 -0.04914 0.662165 0.324414 0.086176 0.492164 -0.84766 -0.74931 -0.11354 -0.2732 0.266089 -0.358 -0.19669 0.00271 0.289633 -0.3257 -0.58871 0.430977 -0.12261 -0.27674 0.662477 Independent Normal Random Variables If X and Y are independent random variables that are normally distributed, then their sum is also normally distributed. That is, if 𝑋~𝑁 𝜇𝑋 , 𝜎𝑋2 𝑌~𝑁 𝜇𝑌 , 𝜎𝑌2 𝑍 = 𝑋 + 𝑌, then 𝑍~𝑁 𝜇𝑋 + 𝜇𝑌 , 𝜎𝑋2 + 𝜎𝑌2 . © 2002 Thomson / South-Western Slide 6-33 Independent Normal Random Variables This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances (i.e., the square of the standard deviation is the sum of the squares of the standard deviations). © 2002 Thomson / South-Western Slide 6-34 PERT Application of normal distribution and independence Program Evaluation and Review Technique (www.se.cuhk.edu.hk/~seem3530/files/ProjMgt-IA-PERT.ppt) SHOPPING MALL RENOVATION Discussion Problem W05-04 The durations of activities A, B, C, and D are independent and normally distributed with [mean, standard deviation] in the process flow chart below. What is the probability to complete the project within 17 weeks? PERT SEEM 3530 36 Example: Modified Calculations ▪ PERT Then the durations of the two paths are normally distributed as follows: length (A-B) = X1 ~ N(17, 3.61) length (C-D) = X2 ~ N(16, 3.35) SEEM 3530 37 The Probability Density Functions PERT SEEM 3530 38 Project Completion Probabilities ▪ The project can be completed in 17 weeks only if both (A-B) and (C-D) are completed within that time. The probabilities for the two paths to be completed in that time are given below: 17-17 P(X1  17) = P(Z  ----------- ) = P(Z  0)=0.5 3.61 17-16 P(X2  17) = P(Z  ----------- ) = P(Z  0.299)=0.62 3.35 ▪ Thus, the probability of completing the project within 17 weeks is P(X  17) = P(X1  17) P(X2  17) = (0.5)(0.62) = 0.31 = 31 %. PERT SEEM 3530 39 Project Completion Probabilities ▪ ▪ ▪ Assume there are n paths, with completion times X1, X2, …, Xn. Then, the probability of completing the project is P(X  T) = P(X1  T) P(X2  T) … P(Xn  T) Assume that the paths are statistically independent (i.e. the time to traverse each path in the network is independent of what happens on the other paths). Although this additional assumption is rarely true in practice, empirical evidence suggests that good results can be obtained. PERT SEEM 3530 40 PERT Application of normal distribution and independence Program Evaluation and Review Technique (MBPF) WONDER SHED INC. Process Flow Time Wonder Shed Inc. is a manufacturer of storage sheds. The manufacturing process involves the procurement of sheets of steel that will be used to form both the roof and the base of each shed. 42 Process Flow Time The first step involves separating the material need for the roof from that needed for the base. Then the roof and the base can be fabricated in parallel, or simultaneously. Roof fabrication involves first punching and then forming the roof to shape. Base fabrication entails the punching-and-forming process plus a subassembly operation. 43 Process Flow Time Fabricated roofs and bases are then assembled into finished sheds that are subsequently inspected for quality assurance. A list of activities needed to fabricate a roof, fabricate a base, and assemble a shed is given in Table 4.2. a flowchart of the process is shown in Figure 4.1. 44 Flow time and critical paths 45 Process Flow Time Table 4.2 adds the mean duration, called flow time, of each activity. See that the flowchart (Figure 4.1) shows that two paths connecting the beginning and the end of the process – Path 1 (roof): Start→1 → 3 → 5 → 7 → 8 → End • 120 minutes = 20 + 25 + 20 + 15 + 40 – Path 2 (base): Start →1 →2 →4 →6 →7 →8 →End • 150 minutes = 20 + 35 + 10 + 30 + 15 + 40 46 Process Flow Time Assume that the duration of each activity is independent and normally distributed with the mean and the variance in the following table: Activity Mean Time Variance 1 20 9 2 35 36 3 25 16 4 10 4 5 20 10 6 30 28 7 15 7 8 40 50 47 Process Flow Time We also assume that Path 1 and Path 2 are independent. What is the probability that both of them finish within 140 minutes? Activity Mean Time Variance 1 20 9 2 35 36 3 25 16 4 10 4 5 20 10 6 30 28 7 15 7 8 40 50 48 Process Flow Time • The mean time and the variance of Path 1 are – Path 1 (roof): Start→1 → 3 → 5 → 7 → 8 → End • 120 minutes = 20 + 25 + 20 + 15 + 40 • 92 minutes2 = 9 + 16 + 10 + 7 + 50 • The duration of Path 1 is N(µ = 120, σ2 = 92). 49 Process Flow Time • The mean time and the variance of Path 2 are – Path 2 (base): Start →1 →2 →4 →6 →7 →8 →End • 150 minutes = 20 + 35 + 10 + 30 + 15 + 40 • 134 minutes2 = 9 + 36 + 4 + 28 + 7 + 50 • The duration of Path 2 is N(µ = 150, σ2 = 134). 50 Process Flow Time • The duration of Path 1 is N(µ = 120, σ2 = 92). • The duration of Path 2 is N(µ = 150, σ2 = 134). P[X1 ≤ 140] = NORMDIST(140,120,SQRT(92),1) P[X2 ≤ 140] = NORMDIST(140,150,SQRT(134),1) What is the probability that both of them finish within 140 minutes? P[X1 ≤ 140, X2 ≤ 140] = P[X1 ≤ 140] * P[X2 ≤ 140] 51 Process Flow Time ▪ ▪ ▪ We assumed that the paths are statistically independent (i.e. the time to traverse each path in the network is independent of what happens on the other paths). Strictly speaking, this additional assumption is hardly true because activities 1, 7, and 8 belong to both of Paths 1 and 2, Although this additional assumption is rarely true in practice, empirical evidence suggests that good results can be obtained. PERT SEEM 3530 52 Discussion Problem W05-05 Wonder Shed Inc. is a manufacturer of storage sheds. The manufacturing process involves the procurement of sheets of steel that will be used to form both the roof and the base of each shed. 53 Discussion Problem W05-05 The first step involves separating the material need for the roof from that needed for the base. Then the roof and the base can be fabricated in parallel, or simultaneously. Roof fabrication involves first punching and then forming the roof to shape. Base fabrication entails the punching-and-forming process plus a subassembly operation. 54 Discussion Problem W05-05 Fabricated roofs and bases are then assembled into finished sheds that are subsequently inspected for quality assurance. A list of activities needed to fabricate a roof, fabricate a base, and assemble a shed is given in Table 4.2. A flowchart of the process is shown in Figure 4.1. 55 Flow time and critical paths 56 Discussion Problem W05-05 Assume that the duration of each activity is independent and normally distributed with the mean and the variance in the following table: Activity Mean Time Variance 1 20 9 2 35 36 3 25 16 4 10 4 5 20 10 6 30 28 7 15 7 8 40 50 57 Discussion Problem W05-05 We also assume that Path 1 and Path 2 are independent. What is the probability that both of them finish within 140 minutes? Activity Mean Time Variance 1 20 9 2 35 36 3 25 16 4 10 4 5 20 10 6 30 28 7 15 7 8 40 50 58 Application of Independent Normal Distributions IMPACT OF AGGREGATION ON INVENTORY Impact of Aggregation on Inventory A BMW dealership has k = 4 retail outlets serving the entire Chicago area (disaggregate option). Weekly demand at each outlet is normally distributed, with a mean of D = 25 cars and a standard deviation of σD = 5. The lead time for replenishment from the manufacturer is L = 2 weeks. The outlet manager orders when the inventory on hand drops to the Reorder Point (ROP). Copyright © 2016 Pearson Education, Inc. 12 – 60 Impact of Aggregation on Inventory Each outlet covers a separate geographic area, and the demand of each outlet is independent of the other. The dealership is considering the possibility of replacing the four outlets with single large outlet (aggregate option). Copyright © 2016 Pearson Education, Inc. 12 – 61 Impact of Aggregation on Inventory Assume that the demand in the central outlet is the sum of the demand across all four areas. The probability of demand during lead time does not exceed the reorder point is required to be at least 90%. a) What is the reorder point of each outlet in disaggregate option? b)What is the reorder point of the central outlet in aggregate option? Copyright © 2016 Pearson Education, Inc. 12 – 62 Impact of Aggregation on Inventory a) What is the reorder point of each outlet in disaggregate option? Let X be the random variable of demand during lead time for one outlet and let X1 and X2 be the random variables of demand for the 1st week and the 2nd week. Then X = X1 + X2. The average demand during lead time (2 weeks) is E[X] = D + D = 25 + 25 = 50 and the variance is V[X] = σ2 + σ2 = 25 + 25 = 50. Copyright © 2016 Pearson Education, Inc. 12 – 63 Impact of Aggregation on Inventory a) What is the reorder point of each outlet in disaggregate option? 𝑃 𝑋 < 𝑅𝑂𝑃 = 𝑃 𝑍 < 𝑅𝑂𝑃−𝐸[𝑋] 𝑉[𝑋] = 𝑃 𝑍 < 𝑅𝑂𝑃−50 50 =𝐹 Therefore, 𝑅𝑂𝑃−50 50 𝑅𝑂𝑃−50 50 = 0.90 = z0.90 Or, ROP = NORM.INV(0.90,50,SQRT(50)) = 59.06193802 Copyright © 2016 Pearson Education, Inc. 12 – 64 Impact of Aggregation on Inventory b) What is the reorder point of the central outlet in aggregate option? Let Y be the random variable of demand during lead time for the central outlet. The average demand during lead time (2 weeks) is E[Y] = 4 * 50 = 200 and the variance is V[Y] = 4 * 50 = 200. Let cROP denote the reorder point of the central outlet. Copyright © 2016 Pearson Education, Inc. 12 – 65 Impact of Aggregation on Inventory b) What is the reorder point of the central outlet in aggregate option? 𝑃 𝑌 < 𝑐𝑅𝑂𝑃 = 𝑃 𝑍 < 𝑐𝑅𝑂𝑃−𝐸[𝑌] 𝑉[𝑌] = 𝑃 𝑍 < 𝑐𝑅𝑂𝑃−200 =𝐹 Therefore, 𝑐𝑅𝑂𝑃−200 200 200 𝑐𝑅𝑂𝑃−200 = 200 0.90 = z0.90 Or, cROP = NORM.INV(0.90,200,SQRT(200)) = 218.1239 < 236.2478 = 4 * 59.06193802 = 4 * ROP Copyright © 2016 Pearson Education, Inc. 12 – 66 Discussion Problem W05-06 A BMW dealership has k = 4 retail outlets serving the entire Chicago area (disaggregate option). Weekly demand at each outlet is normally distributed, with a mean of D = 25 cars and a standard deviation of σD = 5. The lead time for replenishment from the manufacturer is L = 2 weeks. The outlet manager orders when the inventory on hand drops to the Reorder Point (ROP). Copyright © 2016 Pearson Education, Inc. 12 – 67 Discussion Problem W05-06 Each outlet covers a separate geographic area, and the demand of each outlet is independent of the other. The dealership is considering the possibility of replacing the four outlets with single large outlet (aggregate option). The dealership is considering the possibility of replacing the four outlets with a single large central outlet (aggregate option). Copyright © 2016 Pearson Education, Inc. 12 – 68 Discussion Problem W05-06 Assume that the demand in the central outlet is the sum of the demand across all four areas. The probability of demand during lead time does not exceed the reorder point is required to be at least 90%. a) What is the reorder point of each outlet in disaggregate option? b)What is the reorder point of the central outlet in aggregate option? Copyright © 2016 Pearson Education, Inc. 12 – 69 Normal Approximation of the binomial distribution For large n (say n > 20) and p not too near 0 or 1 (say 0.05 < p < 0.95) the distribution approximately follows the Normal distribution. This can be used to find binomial probabilities. If X ~ binomial (n, p) where n > 20 and 0.05 < p < 0.95 then approximately X has the Normal distribution with mean E(X) = np and variance V(X) = np(1 – p). © 2002 Thomson / South-Western Slide 6-70 Normal Approximation of Binomial: Parameter Conversion • Conversion equations  = n p  = n pq • Conversion example: Given that X has a binomial distribution , find P( X  25| n = 60 and p =. 30 ).  = n  p = (60 )(. 30 ) = 18  = n  p  q = (60 )(. 30 )(. 70 ) = 3. 55 © 2002 Thomson / South-Western Slide 6-71 Normal Approximation of Binomial: Interval Check   3 = 18  3(355 . ) = 18  10.65  − 3 = 7.35  + 3 = 28.65 0 10 20 © 2002 Thomson / South-Western 30 40 50 60 n 70 Slide 6-72 Normal Approximation of Binomial: Correcting for Continuity Values Being Determined Correction X X X X X X +.50 -.50 -.50 +.05 -.50 and +.50 +.50 and -.50 © 2002 Thomson / South-Western The binomial probability , P( X  25| n = 60 and p =. 30) is approximated by the normal probability P(X  24.5|  = 18 and  = 3. 55). Slide 6-73 Normal Approximation of Binomial: Graphs 0.12 0.10 0.08 0.06 0.04 0.02 0 6 8 10 12 14 16 18 20 22 24 26 28 30 © 2002 Thomson / South-Western Slide 6-74 Normal Approximation of Binomial: Computations X P(X) 25 26 27 28 29 30 31 32 33 Total 0.0167 0.0096 0.0052 0.0026 0.0012 0.0005 0.0002 0.0001 0.0000 0.0361 © 2002 Thomson / South-Western The normal approximation, P(X  24.5|  = 18 and  = 355 . ) 24.5 − 18   = P Z     355 . = P( Z  183 . ) =.5 − P( 0  Z  183 . ) =.5−.4664 =.0336 Slide 6-75 3-10 Normal Approximation to the Binomial and Poisson Distributions Normal Approximation to the Binomial Discussion Problem W05-07-(a, b) Fill in the blank rectangles: a) b) 3-10 Normal Approximation to the Binomial and Poisson Distributions Normal Approximation to the Poisson 3-10 Normal Approximation to the Binomial and Poisson Distributions Normal Approximation to the Poisson Example 3-36: Water Contaminants 3-10 Normal Approximation to the Binomial and Poisson Distributions Normal Approximation to the Poisson Example 3-36: Water Contaminants (continued) Discussion Problem W05-08 Normal Approximation to the Poisson Example 3-36: Water Contaminants Estimate the probability based on Normal Approximation. Discussion Problem W05-01 The durations of activities A, B, C, and D are independent and normally distributed with [mean, standard deviation) in the process flow chart below. What is the probability to complete the project within 17 weeks? (mean, standard deviation) 60 A [9,3) (8,21 с D [10,3] [6,1.5] 3 PERT SEEM 3530 1 Discussion Problem W05-02 Wonder Shed Inc. is a manufacturer of storage sheds. The manufacturing process involves the procurement of sheets of steel that will be used to form both the roof and the base of each shed. 2 Discussion Problem W05-03 A BMW dealership has k = 4 retail outlets serving the entire Chicago area (disaggregate option). Weekly demand at each outlet is normally distributed, with a mean of D = 25 cars and a standard deviation of op = 5. The lead time for replenishment from the manufacturer is L = 2 weeks. The outlet manager orders when the inventory on hand drops to the Reorder Point (ROP). Copyright © 2016 Pearson Education, Inc. 12-8 Discussion Problem W05-03 Each outlet covers a separate geographic area, and the demand of each outlet is independent of the other. The dealership is considering the possibility of replacing the four outlets with single large outlet (aggregate option). The dealership is considering the possibility of replacing the four outlets with a single large central outlet (aggregate option). Copyright © 2016 Pearson Education, Inc. 12-9 Discussion Problem W05-03 Assume that the demand in the central outlet is the sum of the demand across all four areas. The probability of demand during lead time does not exceed the reorder point is required to be at least 90%. a) What is the reorder point of each outlet in disaggregate option? b)What is the reorder point of the central outlet in aggregate option? Copyright © 2016 Pearson Education, Inc. 12-10
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