MATH 2332 Simulating Probability Distributions

Transcription

1 MATH 2332 Simulating Probability Distributions Demonstration 1: Binomial distribution example of estimating values and probabilities A company has developed a very inexpensive explosive-detection machine for use in airports. However, if an explosive is actually in a suitcase, the probability of it being detected by this machine is only 0.60 (so the probability the machine fails to detect an explosive is 0.40). A medium traffic airport is interested in purchasing five of the machines. To assess the likelihood that multiple machines will fail to detect an explosive, the airport decides to simulate a 35-trial sequence to determine if the low cost is worth the risk. The scenario to simulate consists of the five machines being used independently per trial to simultaneously screen luggage containing an explosive. [For the exercise, also assume a trial is done at a randomly selected time and there is no wear-n-tear or depreciation of the machines from trial to trial.] A. Over the 35-trials, how many times during screening did machines fail to detect an explosive? How does your value compare to the expected value (where the expected value is four tenths of the total screenings performed.) B. What is the probability of observing this number of failures during the trial period? How does this probability compare to the probability of observing the expected value? C. Based on the simulated sample, what is the probability more than one machine failed to detect an explosive. How comfortable are you with the accuracy of this value as an estimate of the actual probability? Explain. D. Write three to four brief statements to the company summarizing the findings of your simulation. Write up for the simulation and exercises Description of the simulation The cumulative probability function (CDF X in Appendix 1 Table 1.1) was used to assign the results of each trial. For example, any random number less than indicated 0 machines failed to detect an explosive. A random number between and indicated 1 failed to detect an explosive. The remaining value assignments are as follows: ( , ] 2, ( , ] 3, ( , ] 4, and ( , 1] 5. The first ten trial results and the summary statistics of the simulated sample are in the Appendix 1 (Table 1.2 and Figure 1.1 respectively). Dr. Gardner S2017 1

2 Exercise solutions A. The sum of screenings (SumC) failing to detect an explosive in the simulated sample was 72. The expected total count of failures (E(SumC) = (5)(35)(4/10) = 70) was slightly lower than the estimated sum. That is to say, SumC slightly over-estimated the expected total count. B. P(SumC = 72) = 72/ [(5)(35)] = and the probability of the expected count is The two probabilities are very close, the probability for the simulated sample slightly over-estimated the probability of the expected total count. C. In reference to the simulated sample, there were 24 screenings in which more than one machine did not detect an explosive. Therefore, P(S > 1) = 24/35 = Based on the comparisons made between the simulated values and expected value in parts A and B, we assume the estimated probability is slightly over-estimates the actual probability. [Note: The actual probability (0.6630) from the PMF in Appendix 1 Table 1 supports this observation. The simulated data represent a transformation of the binomial distribution X~Bin (5, 0.4) This is FYI, and optional for your paper.] D. There is approximately a 69% probability that at least two machines will not detect an explosive device in luggage during screening. The risk of failure is too high to justify purchasing the machines, even if the machines are inexpensive. The simulated data slightly over-estimated the actual values, but increasing the number of trials should improve the estimates. Demonstration 2: Normal distribution example of checking normalcy and estimating values A cable is made up of four wires. The breaking strength of each wire is a normally distributed random variable with mean 10 kn and standard deviation 1 kn. The strength of the cable, using the brittle wire method, is estimated to be the strength of the weakest wire multiplied by the number of wires. [Hint: In this case, S = (4) min(w1, W2, W3, W4).] A. Use a simulated sample of size That is, simulate 1000 cable strength estimate where a cable is made up of four random wires. Use the simulated data to complete the exercise. B. Construct the normal probability plot of cable strengths. Is there evidence to suggest the cable strengths are non-normal? Explain. C. Estimate the mean strength of this type of cable. D. Estimate the median cable strength. E. Estimate the standard deviation of the cable strength. F. To be acceptable for a certain application, the probability that the cable breaks under a load of 28 kn must be less than Does the cable appear to be acceptable? Explain. Dr. Gardner S2017 2

3 Write up for the simulation and exercise solutions Description of the simulation A. Four columns (W1 W4) were created each with 1000 rows. Random values from the normal distribution N(10, 1) were generated using the Random Normal formula in JMP. The strength (S) column contained the formula 4*Minimum (W1,W2,W3,W4) and the Accept column contained a conditional formula returning 1 for a strength value less than 28 kn. The sum of the Accept column was also determined. Simulated samples 983 to 1000 are provided in Appendix 2, Table 2.1. Solutions to the remaining exercises B. The normal probability plot (Appendix 2, Figure 2.1) is relatively linear, so there is no strong evidence that the distribution of S is non-normal although a few outliers are present. Using Figure 2.1, C. the estimated mean strength of this type of cable is 35.9 kn, D. the standard deviation is 2.8 kn, E. and the median is 36 kn. F. There were five cables that broke under a load of 28 kn (Figure 2.2). The probability is P(S < 28) = 5/1000 = 0.005, which is less than the 0.01 making this type of cable acceptable. Since 1000 samples were generated, the probability estimate can be assumed reliable. Demonstration 3: Exponential distribution example of estimating a probability An engineer has to choose between two types of cooling fans to install in a computer. The lifetimes, in months, for fans of type A are exponentially distributed with mean 50 months, and the lifetime for fans of type B are exponentially distributed with mean 30 months. Since type A fans are more expensive, the engineer decides that she will choose type A fans if the probability that a type A fan will last more than twice as long as a type B fan is greater than 0.5 [Hint: P(A > 2B) > 0.5]. Estimate this probability using simulated samples of size Write up of the simulation and exercise solutions Lambda for the exponential distribution of fan type A was determined to be lambda A = 1/50 =0.02 and for fan type B, lambda B was approximately Columns A and B were created using the Random Exponential formula, for example the column A formula was Round(Exp()/0.02, 2). Each column had 1000 rows. A column to compare A to twice B was created using a conditional formula an 1 was returned if A was greater. The count of lifetimes of A that were larger than twice B was determined using the Analyze > Distribution tool. Dr. Gardner S2017 3

4 In Figure 3.1, the count of A > 2B is 461. Therefore P(A > 2B) = 461/1000 = which is less than 0.5. The engineer should not choose the type A fans because the expense is not feasible for the lifetime. Since 1000 samples were generated, the probability estimate can be assumed reliable. Dr. Gardner S2017 4

5 APPENDIX 1 JMP MACHINES OUTPUT Table 1.1 The probability mass function and cumulative distribution of X~B(5, 0.4) Table 1.2. The first ten trials of the simulation. Figure 1.1 Summary statistics and distribution of the simulated sample. Dr. Gardner S2017 1

6 APPENDIX 2 JMP CABLE OUTPUT Table 2.1. Simulated samples Figure 2.1. Summary statistics and distribution of the simulated sample cables. Dr. Gardner S2017 2

7 Figure 2.2. Count of cables that broke under a load of 28 kn. Dr. Gardner S2017 3

8 APPENDIX 3 JMP FANS OUTPUT Table 3.1. Simulated samples Figure 3.1. The count of lifetimes of A that are twice as long as lifetimes of B. Dr. Gardner S2017 4