Quantitative Methods for Business chapter 13 pdf

CHAPTER Smooth running – continuous probability distributions and basic queuing theory 13 Chapter objectives This chapter will help you to: ■ make use of the normal distribution and appreciate its import- ance ■ employ the Standard Normal Distribution to investigate nor- mal distribution problems ■ apply the exponential distribution and be aware of its useful- ness in analysing queues ■ analyse a simple queuing system ■ use the technology: continuous probability distribution ■ become acquainted with business uses of the normal distribution In the previous chapter we looked at how two different types of the- oretical probability distributions, the binomial and Poisson distribu- tions, can be used to model or simulate the behaviour of discrete 406 Quantitative methods for business Chapter 13 random variables. These types of variable can only have a limited or finite number of values, typically only whole numbers like the number of defective products in a pack or the number of telephone calls received over a period of time. Discrete random variables are not the only type of random variable. You will also come across continuous random variables, variables whose possible values are not confined to a limited range. In the same way as we used discrete probability distributions to help us investigate the behaviour of discrete random variables we use continuous prob- ability distributions to help us investigate the behaviour of continuous random variables. Continuous probability distributions can also be used to investigate the behaviour of discrete variables that can have many different values. The most important continuous probability distribution in Statistics is the normal distribution. As the name suggests, this distribution rep- resents the pattern of many ‘typical’ or ‘normal’ variables. You may find the distribution referred to as the Gaussian distribution after the German mathematician Carl Friedrich Gauss (1777–1855) who developed it to model observation errors that arose in surveying and astronomy measurements. The normal distribution has a very special place in Statistics because as well as helping us to model variables that behave in the way that the normal distribution portrays, it is used to model the way in which results from samples vary. This is of great importance when we want to use sample results to make predictions about entire populations. 13.1 The normal distribution Just as we saw that there are different versions of the binomial distribu- tion that describe the patterns of values of binomial variables, and dif- ferent versions of the Poisson distribution that describe the behaviour of Poisson variables, there are many different versions of the normal dis- tributions that display the patterns of values of normal variables. Each version of the binomial distribution is defined by n, the num- ber of trials, and p, the probability of success in any one trial. Each ver- sion of the Poisson distribution is defined by its mean. In the same way, each version of the normal distribution is identified by two defining characteristics or parameters: its mean and its standard deviation. The normal distribution has three distinguishing features: ■ It is unimodal, in other words there is a single peak. ■ It is symmetrical, one side is the mirror image of the other. ■ It is asymptotic, that is, it tails off very gradually on each side but the line representing the distribution never quite meets the horizontal axis. Because the normal distribution is a symmetrical distribution with a single peak, the mean, median and mode all coincide at the middle of the distribution. For this reason we only need to use the mean as the measure of location for the normal distribution. Since the average we use is the mean, the measure of spread that we use for the normal distribution is the standard deviation. The normal distribution is sometimes described as bell-shaped. Figure 13.1 illustrates the shape of the normal distribution. It takes the form of a smooth curve. This is because it represents the probabilities that a con- tinuous variable takes values right across the range of the distribution. If you look back at the diagrams we used to represent discrete prob- ability distributions in Figures 12.1 and 12.2 in Chapter 12 you will see that they are bar charts that consist of separate blocks. Each distinct block represents the probability that the discrete random variable in question takes one of its distinct values. Because the variable can only take discrete, or distinct, values we can represent its behaviour with a diagram consisting of discrete, or distinct, sections. If we want to use a diagram like Figure 12.1 or 12.2 to find the prob- ability that the discrete variable it describes takes a specific value, we can simply measure the height of the block against the vertical axis. In contrast, using the smooth or continuous curve in Figure 13.1 to find the probability that the continuous variable it describes takes a particu- lar value is not so easy. To start with we need to specify a range rather than a single value because we are dealing with continuous values. For instance, referring to the probability of a variable, X being 4 is inadequate as in a continuous Chapter 13 Continuous probability distributions and basic queuing theory 407 0.0 0.1 0.2 0.3 0.4 x P(X ϭ x) Figure 13.1 The normal distribution distribution it implies the probability that X is precisely 4.000. Instead we would have to specify the probability that X is between 3.500 and 4.499. This probability would be represented in a diagram by the area below the curve between the points 3.500 and 4.499 on the horizontal axis as a proportion of the total area below the curve. The probability that a con- tinuous variable takes a precise value is, in effect zero. This means that in practice there is no difference between, say, the probability that X is less than 4, P(X Ͻ 4.000) and the probability that X is less than or equal to 4, P(X р 4.000). Similarly the probability that X is more than 4, P(X Ͼ 4.000) is essentially indistinguishable from the probability that X is more than or equal to 4, P(X у 4.000). For convenience the equal- ities are left out of the probability statements in what follows. When we looked at the binomial and Poisson distributions in Chapter 12 we saw how it was possible to calculate probabilities in these distributions using the appropriate formulae. In fact, in the days before the sort of software we now have became available, if you needed to use a binomial or a Poisson distribution you had to start by consulting pub- lished tables. However, because of the sheer number of distributions, the one that you wanted may not have appeared in the tables. In such a situation you had to calculate the probabilities yourself. Calculating the probabilities that make up discrete distributions is tedious but not impossible, especially if the number of outcomes involved is quite small. The nature of the variables concerned, the fact that they can only take a limited number of values, restricts the number of cal- culations involved. In contrast, calculating the probabilities in continuous distributions can be daunting. The variables, being continuous, can have an infinite number of different values and the distribution consists of a smooth curve rather than a collection of detached blocks. This makes the mathematics involved very much more difficult and puts the task beyond many people. Because it was so difficult to calculate normal distribution probabil- ities, tables were the only viable means of using the normal distribution. However, the number of versions of the normal distribution is literally infinite, so it was impossible to publish tables of all the versions of the normal distribution. The solution to this problem was the production of tables describing a benchmark normal distribution known as the Standard Normal Dis- tribution. The advantage of this was that you could analyse any version of the normal distribution by comparing points in it with equivalent points in the Standard Normal Distribution. Once you had these equiva- lent points you could use published Standard Normal Distribution tables to assist you with your analysis. 408 Quantitative methods for business Chapter 13 Chapter 13 Continuous probability distributions and basic queuing theory 409 Although modern software means that the Standard Normal Distribution is not as indispensable as it once was, it is important that you know something about it. Not only is it useful in case you do not have access to appropriate software, but more importantly, there are many aspects of further statistical work you will meet that are easier to understand if you are aware of the Standard Normal Distribution. 13.2 The Standard Normal Distribution The Standard Normal Distribution describes the behaviour of the vari- able Z, which is normally distributed with a mean of zero and a stand- ard deviation of one. Z is sometimes known as the Standard Normal Variable and the Standard Normal Distribution is known as the Z Distribution. The distribution is shown in Figure 13.2. If you look carefully at Figure 13.2 you will see that the bulk of the distribution is quite close to the mean, 0. The tails on either side get closer to the horizontal axis as we get further away from the mean, but they never meet the horizontal axis. They are what are called asymptotic. As you can see from Figure 13.2, half of the Standard Normal Distribution is to the left of zero, and half to the right. This means that half of the z values that make up the distribution are negative and half are positive. Table 5 on pages 621–622, Appendix 1 provides a detailed breakdown of the Standard Normal Distribution. You can use it to find the probabil- ity that Z, the Standard Normal Variable, is more than a certain value, z, Ϫ3 Ϫ2 Ϫ1012 3 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.2 The Standard Normal Distribution or less than z. In order to show you how this can be done, a section of Table 5 is printed below: Suppose you need to find the probability that the Standard Normal Variable, Z, is greater than 0.62, P(Z Ͼ 0.62). Begin by looking for the value of z, 0.62, to just one decimal place, i.e. 0.6, amongst the values listed in the column headed z on the left hand side. Once you have found 0.6 under z, look along the row to the right until you reach the figure in the column headed 0.02. The figure in the 0.6 row and the 0.02 column is the proportion of the distribution that lies to the right of 0.62, 0.2676. This area represents the probability that Z is greater than 0.62, so P(Z Ͼ 0.62) is 0.2676 or 26.76%. If you want the probability that Z is less than 1.04, P(Z Ͻ 1.04), look first for 1.0 in the z column and then proceed to the right until you reach the figure in the column headed 0.04, 0.1492. This is the area to the right of 1.04 and represents the probability that Z is more than 1.04. To get the probability that Z is less than 1.04, subtract 0.1492 from 1: P(Z Ͻ 1.04) ϭ 1 Ϫ P(Z Ͼ 1.04) ϭ 1 Ϫ 0.1492 ϭ 0.8508 or 85.08% In Example 13.1 you will find a further demonstration of the use of Table 5. 410 Quantitative methods for business Chapter 13 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.2266 0.2236 0.2206 0.2177 0.2148 0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 Example 13.1 Use Table 5 to find the following: (a) The probability that Z is greater than 0.6, P(Z Ͼ 0.6). (b) The probability that Z is less than 0.6, P(Z Ͻ 0.6). (c) The probability that Z is greater than 2.24, P(Z Ͼ 2.24). Chapter 13 Continuous probability distributions and basic queuing theory 411 (d) The probability that Z is greater than Ϫ1.37, P(Z ϾϪ1.37). (e) The probability that Z is less than Ϫ1.37, P(Z ϽϪ1.37). (f) The probability that Z is greater than 0.38 and less than 2.71, P(0.38 Ͻ Z Ͻ 2.71). (g) The probability that Z is less than Ϫ0.88 and more than Ϫ1.93, P(Ϫ1.93 Ͻ Z ϽϪ0.88). (h) The probability that Z is less than 1.59 and more than Ϫ0.76, P(Ϫ0.76 Ͻ Z Ͻ 1.59). Until you are used to dealing with the Standard Normal Distribution you may find it helpful to make a small sketch of the distribution and identify on the sketch the z value(s) of interest and the area that represents the probability you want. (a) The probability that Z is greater than 0.6, P(Z Ͼ 0.6). The value of Z in this case is not specified to two places of decimals, so we take the fig- ure to the immediate right of 0.6 in Table 5, in the column headed 0.00, which is 0.2743. This is the probability that Z is greater than 0.6. We could also say that 27.43% of z values are greater than 0.6. This is represented by the shaded area in Figure 13.3. (b) The probability that Z is less than 0.6, P(Z Ͻ 0.6). In part (a) we found that 27.43% of z values are bigger than 0.6. This implies that 72.57% of z values are less than 0.6, so the answer is 1 Ϫ 0.2743 which is 0.7257. This is represented by the unshaded area in Figure 13.3. Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.3 Example 13.1 (a): P(Z Ͼ 0.6) 412 Quantitative methods for business Chapter 13 (c) The probability that Z is greater than 2.24, P(Z Ͼ 2.24). The figure in the row to the right of 2.2 and in the column headed 0.04 in Table 5 is 0.0125. This means that 1.25% of z values are bigger than 2.24. The probability that Z is bigger than 2.24 is therefore 0.0125. This is represented by the shaded area in Figure 13.4. (d) The probability that Z is greater than Ϫ1.37, P(Z ϾϪ1.37). The figure in the row to the right of Ϫ1.3 and the column headed 0.07 in Table 5 is 0.9147. This is the area of the distribution to the right of Ϫ1.37 and represents the probability that Z is greater than Ϫ1.37. This is shown as the shaded area in Figure 13.5. (e) The probability that Z is less than Ϫ1.37, P(Z ϽϪ1.37). From part (d) we know that the probability that Z is greater than Ϫ1.37 is 0.9147, so the probability that Z is less than Ϫ1.37 (by which we mean Ϫ1.4, Ϫ1.5 and so on) is 1 Ϫ 0.9147, which is 0.0853. This is represented by the unshaded area in Figure 13.5. (f) The probability that Z is greater than 0.38 and less than 2.71, P(0.38 Ͻ Z Ͻ 2.71). The probability that Z is greater than 0.38, P(Z Ͼ 0.38), is shown in Table 5 in the row for 0.3 and the column headed 0.08, 0.3520. You will find the probability that Z is Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.4 Example 13.1 (c): P(Z Ͼ 2.24) Chapter 13 Continuous probability distributions and basic queuing theory 413 greater than 2.71 in the row for 2.7 and the column headed 0.01, 0.0034. We can obtain the probability that Z is more than 0.38 and less than 2.71 by taking the probability that Z is more than 2.71 away from the probability that Z is more than 0.38: P(0.38 Ͻ Z Ͻ 2.71) ϭ P(Z Ͼ 0.38) Ϫ P(Z Ͼ 2.71) ϭ 0.3520 Ϫ 0.0034 ϭ 0.3486 This is represented by the shaded area in Figure 13.6. Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.5 Example 13.1 (d): P(Z ϾϪ1.37) Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.6 Example 13.1 (f ): P(0.38 Ͻ Z Ͻ 2.71) 414 Quantitative methods for business Chapter 13 Another way of approaching this is to say that if 35.20% of the area is to the right of 0.38 and 0.34% of the area is to the right of 2.71, then the difference between these two per- centages, 34.86%, is the area between 0.38 and 2.71. (g) The probability that Z is greater than Ϫ1.93 and less than Ϫ0.88, P(Ϫ1.93 Ͻ Z ϽϪ0.88). In Table 5 the figure in the Ϫ1.9 row and the 0.03 column, 0.9732, is the probability that Z is more than Ϫ1.93, P(Z ϾϪ1.93). The probability that Z is more than Ϫ0.88, P(Z ϾϪ0.88) is the figure in the Ϫ0.8 row and the 0.08 column, 0.8106. The probability that Z is between Ϫ1.93 and Ϫ0.88 is the probability that Z is more than Ϫ1.93 minus the probability that Z is more than 0.88: P(Ϫ1.93 Ͻ Z ϽϪ0.88) ϭ P(Z ϾϪ1.93) ϪP(Z ϾϪ0.88) ϭ 0.9732 Ϫ 0.8106 ϭ 0.1626 This is represented by the shaded area in Figure 13.7. (h) The probability that Z is greater than Ϫ0.76 and less than 1.59, P(Ϫ0.76 Ͻ Z Ͻ 1.59). The probability that Z is greater than Ϫ0.76, P (Z ϾϪ0.76), is in the Ϫ0.7 row and the 0.06 column of Table 5, 0.7764. P(Z Ͼ 1.59) is in the row for 1.5 and the column Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13.7 Example 13.1 (g): P(Ϫ1.93 Ͻ Z ϽϪ0.88) [...]... miles (a) What proportion of bearings will fail before the warranty mileage is reached? 432 Quantitative methods for business 13. 12* 13. 13 13. 14 13. 15 Chapter 13 (b) An alternative supplier offers the company bearings with a mean lifetime of 69,230 miles and a standard deviation of 4620 miles What proportion of these bearings can be expected to fail before the warranty mileage is reached? (c) Should... Example 13. 3 (a) type ϭ NORMDIST(10,9.8,0.5,TRUE) ■ 426 Quantitative methods for business Chapter 13 ■ Click on the √ button to the left of the formula bar or press Enter The probability that X is greater than x should appear in the cell you originally clicked The probability shown for Example 13. 3 (a) should be 0.655422, so the probability that X is greater than 10 is 0.344578 13. 5.2 MINITAB For a normal... 0.2, P(Z Ͼ 0.2), is 0.4207 Quantitative methods for business Chapter 13 This is represented by the shaded area in Figure 13. 9 0.4 P(Z ϭ z) 0.3 0.2 0.1 0.0 Ϫ3 Ϫ2 Ϫ1 0 z 1 2 3 Figure 13. 9 Example 13. 2 (a): 0.4207 ϭ P(Z Ͼ 0.2) (b) To find the value of Z that has an area of 0.0505 to the right of it you will have to look further down Table 5 The figure 0.0505 appears in the row for 1.6 and the column headed... mpg? (e) What is the minimum mpg of the 15% most fuel efficient cars? (f) What is the maximum mpg of the 10% least fuel efficient cars? 430 Quantitative methods for business 13. 5 13. 6 13. 7 Chapter 13 A large company insists that all job applicants who are invited for interview take a psychometric test The results of these tests follow a normal distribution with a mean of 61% and the standard deviation... then press ϩ/Ϫ then SHIFT (or possibly 2nd for second function) then ex The exponential distribution for the service times in Example 13. 5 is shown in Figure 13. 13 The shaded area represents the probability that the service time exceeds 5 minutes P(X ϭ x) 0.3 0.2 0.1 Figure 13. 13 Distribution of service times in Example 13. 5 0.0 0 5 10 Service time (x) 15 Chapter 13 Continuous probability distributions... Review Questions 13. 2 to 13. 11 at the end of the chapter 13. 3 The exponential distribution The importance of the normal distribution and the attention rightly devoted to it in quantitative methods programmes often obscures the fact that it is not the only continuous probability distribution The normal distribution is a symmetrical distribution and is therefore an entirely appropriate model for continuous... value, which we can find using: P(X Ͼ x) ϭ eϪx/␮ 422 Quantitative methods for business Chapter 13 Or the probability that the variable is less than a particular value, which is: P(X Ͻ x) ϭ 1 Ϫ eϪx/␮ To use these expressions for any specific distribution you need to know only the mean of the distribution and the specific value of interest Example 13. 5 The times it takes to serve customers visiting a... mean of the distribution This would lead us to ask what the mean and standard deviation of the distribution of waiting times would 428 Quantitative methods for business Chapter 13 have to be in order to fulfil this objective, how do these parameters compare with current performance and by how much would we have to improve one or both in order to achieve the objective In this illustration 1% sounds an acceptably... Here we will look at one of the simpler models The point of any queuing theory model is to provide us with information about the operation of the queuing system it represents, specifically the average waiting time and the average length of the queue 424 Quantitative methods for business Chapter 13 If we make certain assumptions about the behaviour of individuals in the queue and know the patterns of... rest of the distribution 418 Quantitative methods for business Chapter 13 In later work you will find that particular z values are often referred to in this style because it is the area of the tail that leads us to use a particular z value and we may want to emphasize the fact Values of Z that cut off tails of 5%, 2.5%, 1% and 1⁄2% crop up in the topics we will look at in Chapter 16 The z values that . 5 minutes. 422 Quantitative methods for business Chapter 13 151050 0.3 0.2 0.1 0.0 Service time (x) P(X ϭ x) Figure 13. 13 Distribution of service times in Example 13. 5 Example 13. 5 The times. the unshaded area in Figure 13. 3. Ϫ3 Ϫ2 Ϫ10123 0.0 0.1 0.2 0.3 0.4 z P(Z ϭ z) Figure 13. 3 Example 13. 1 (a): P(Z Ͼ 0.6) 412 Quantitative methods for business Chapter 13 (c) The probability that. associated with it. z x ϭ Ϫ ␮ ␴ 418 Quantitative methods for business Chapter 13 Example 13. 3 The Plotoyani Restaurant offers a 10 oz Steak Special. The steaks they use for these meals have uncooked