Solutions Chapter 7 E7.1 Some of the problems that might be encountered in collecting data on check-in times are: Need to collect separate data for each airline (time and cost). Need to collect data for different staff experience levels (time and cost). If staff know they are being timed they may work faster or slower (accuracy of data). How does the length of queue affect the speed with which staff work? Should environmental issues be taken into account, e.g. staff may work slower on hotter days? Changes to security requirements may alter the speed with which staff can work (the past is not always a good predictor of the future). E7.2 Process 1: The Weight of a Bottle Leaving a Filling Process A normal distribution. Process 2: The Time Between Failure of a Machine A negative exponential distribution or a Weibull distribution. Process 3: The Check-in Time at an Airport A skewed distribution, i.e. gamma, Erlang or lognormal. Process 4: The Number of Orders for a Product Received at a Warehouse from a Retail Outlet A Poisson distribution or possibly an integer uniform if very little data is available. 1
E7.3 Process 1: The Weight of a Bottle Leaving a Filling Process Histogram of the data looks roughly normal as would be expected for this type of process. 4 3 3 2 2 1 1 8. - 8. 8. - 9. 9. - 9. 9. - 1. 1. - 1. 1. - 11. 11. - 11. 11. - 12. Comparison of empirical data and normal (mean = 1.11, st. dev. =.2): 4 3 3 2 2 1 1 8. - 8. 8. - 9. 9. - 9. 9. - 1. 1. - 1. 1. - 11. 11. - 11. 11. - 12. Empirical Normal 2
P P plot of normal (mean = 1.11, st. dev. =.2) versus empirical data: Cumulative probability: normal 1 8 6 4 2 2 4 6 8 1 Cumulative probability: empirical Chi-square test (9% confidence) for fit of normal (mean = 1.11, st. dev. =.2) to empirical data: range Observed (empirical data) Expected: Normal (mean = 1.11, st. dev. =.2) (O i -E i ) 2 /E i <9. 14 12.19.27 9. 1. 29 29.62.1 1. 1. 36 3.9. >1. 22 22.9.2 Total (chi-square value).3 Critical value (1 degree of freedom) 3.841 Accept Conclusion The normal (mean = 1.11, st. dev. =.2) appears to be a reasonable fit. 3
Process 2: The Time Between Failure of a Machine Histogram of the data looks roughly negative exponential as might be expected for this type of process. 6 4 3 2 1-1 1-2 2-3 3-4 4- -6 6-7 Comparison of empirical data and negative exponential (mean = 13.41): 6 4 3 2 1-1 1-2 2-3 3-4 4- -6 6-7 Empirical Exponential 4
P P plot of negative exponential (mean = 13.41) versus empirical data: 1 Cumulative probability: exponential 8 6 4 2 2 4 6 8 1 Cumulative probability: empirical Chi-square test (9% confidence) for fit of negative exponential (mean = 13.41) to empirical data: range Observed (empirical data) Expected: Neg. exp. (mean = 13.41) (O i -E i ) 2 /E i <1 1 47.89.2 1--2 2 24.96. 2--3 11 13..31 >3 12 13.11.9 Total (chi-square value).6 Critical value (2 degree of freedom).991 Accept Conclusion The negative exponential (mean = 13.41) appears to be a reasonable fit.
Process 3: The Check-in Time at an Airport A histogram of the data shows that the data are skewed (as expected for this type of process) and so the data are likely to come from the Erlang, gamma or lognormal distributions. Here, a gamma distribution is fitted to the data. 4 3 3 2 2 1 1-1 1-2 2-3 3-4 4- -6 6-7 In order to fit a gamma distribution the shape and scale parameters must be estimated. These can be calculated from the mean and standard deviation: mean = shape * scale = 2.2 st. dev. = shape * scale = 1. Solving this simultaneous equation gives: shape =.6 scale =.44 Comparison of empirical data and gamma (shape =.6, scale =.44): 4 4 3 3 2 2 1 1-1 1-2 2-3 3-4 4- -6 6-7 Empirical Gamma 6
P P plot of gamma (shape =.6, scale =.44) versus empirical data: Cumulative probability: gamma 1 8 6 4 2 2 4 6 8 1 Cumulative probability: empirical Chi-square test (9% confidence) for fit of gamma (shape =.6, scale =.44) to empirical data: range Observed (empirical data) Expected: Gamma (shape =.6, scale =.44) (O i -E i ) 2 /E i <1 9 7.6.26 1 2 33 38.93.9 2 3 38 33.72.4 3 4 14 14.27.1 >4 6.42.6 Total (chi-square value) 1.77 Critical value (2 degree of freedom).991 Accept Conclusion The gamma (shape =.6, scale =.44) appears to be a reasonable fit. 7
Process 4: The Number of Orders for a Product Received at a Warehouse from a Retail Outlet Histogram of the data looks roughly Poisson as would be expected for this type of process. 3 2 2 1 1 1 2 3 4 6 7 8 9 1 11 12 Comparison of empirical data and Poisson (mean =.16): 3 2 2 1 1 1 2 3 4 6 7 8 9 1 11 12 Empirical Poisson 8
P P plot of Poisson (mean =.16) versus empirical data: Cumulative probability: Poisson 1 8 6 4 2 2 4 6 8 1 Cumulative probability: empirical Chi-square test (9% confidence) for fit of Poisson (mean =.16) to empirical data: range Observed (empirical data) Expected: Poisson (mean =.16) (O i -E i ) 2 /E i <3 1 11.18.12 3 12 13.1.1 4 16 16.96. 24 17. 2.41 6 12 1..62 7 11 11.1. 8 9 7.16.47 >8 6 7.81.42 Total (chi-square value) 4.21 Critical value (6 degree of freedom) 12.916 Accept Conclusion The Poisson (mean =.16) appears to be a reasonable fit. 9