Qualitative Methods for Risk Analysis - Assignment Example

QUALITATIVE METHODS FOR RISK ANALYSIS Name: Institution affiliated: Date of submission: Tutor: Total marks: 89 QUESTION 1 [27 marks] (a) State, with reasons, the probability distribution of the number of cork-spoiled bottles in a case of 12bottles. 4%* 12 bottles=0.48 This means that for every 100 bottles bought approximately four to five bottles are spoilt. (b) In a case of 12 bottles, what is the probability that there will be (i) At least 1 cork-spoiled bottle? P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =1/12 =0.0833 (ii) Between 2 (inclusive) and 4 (inclusive) cork-spoiled bottles? This means that at least three have cork-spoiled bottles. P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =3/12 =0.25 (c) For cases of 12 bottles, obtain the (i) Mode of the number of cork-spoiled bottles per case. The mode number of cork spoiled bottles per case is 1. (ii) Median of the number of cork-spoiled bottles per case. The medium of the number of cork-spoilt bottles per case is also 1 (iii) Mean of the number of cork-spoiled bottles per case. Must the mean be quoted as a whole number? Explain. The average cork-spoiled bottles is 0.48 It is not a must that the mean number be quoted as a whole number since the mean may also be used for the calculation of fractions on the same. Therefore one can round of to leave the evaluation as a whole number or it may be left as a fraction. (d)Yesterday, John Smith’s wife Mary also bought a case of 12 bottles of 2007CasaMiaRojo Blanco. From these two purchases, what is the probability that the Smith household will have cork-spoilage? P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =4% of 24 =0.96 (e) Critically discuss TWO of the assumptions which underlie your calculations in Question1. There are two main assumptions that are considered when calculating the sums above. The first assumption is exclusivity and the other is the independence of the events or variables. It is important to consider whether the events are mutually exclusive or not. This means that the events A and B cannot happen both at the same time. It is either the corks are spoiled or they are in perfect condition. The second important assumption to be considered is to determine whether the events are independent of each other. The independence of the events is determined by the likelihood that one event will be influenced to happen by the other. QUESTION 2 [6 marks] In a manufacturing process, it is estimated that 40 percent of industrial accidents have operator error attributed as a cause, 45 percent have machine failure attributed as a cause and 10 percent have both. (a) Calculate the probability that an accident is attributable to some other cause. Show your reasoning. Total percentages of accidents are 100%. Of those 40% of the total accidents are as a result of operator error, 45% are as a result of machine failure and 10% are as a result of both. For this reason the remaining 5% are the problems attributable to other causes. P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =5/100 =0.05 (b) Explain whether the events ‘attributable to operator error’ and ‘attributable to machine failure’ are (i) mutually exclusive Mutually exclusive means that a set of events that cannot occur at the same time. This means when one event occurs, the other cannot occur. However, this is not the case with the accidents in the industry that are attributable to the operator errors and those that are attributable to the machine failure. This is because there is 10% likelihood that the accidents occurring in the organization are attributable to both the operator errors and machine failures. (ii) Independent. In the case of independent events, the events in question do not affect the probability of other events from occurring. This means that the probabilities are independent of each other and the occurrence of one event does not affect the occurrence of the other. In this case, both accidents are independent of each other. There is a likelihood that the accidents may happen both at the same time or one may happen and the other may not happen. For this case, the events ‘attributable to operator error’ and ‘attributable to machine failure’ are independent. QUESTION 3 [19 marks] The red cell content of a person’s blood is reported as an RBC value: red blood cell count, in millions of cells per micro litre of blood. RBC values for females in the population are well modelled by a Normal distribution with mean 4.8 and standard deviation 0.30; for males, by a Normal distribution with mean 5.4 and standard deviation 0.35. (a) Sketch, on the same set of axes, the distribution of RBC for females and the distribution of RBC for males. Label your axes. Your sketch, along with its labels and descriptions, should clearly show the similarities/differences of the two distributions. (b) A diagnostic test sheet gives the “expected range” of RBC for females as 4.2 to 5.4; for males as 4.7 to 6.1. What proportion (correct to 4 decimal places) (i) Of females have an RBC value outside the “expected range”? f= =5.4-4.2 =1.2 =0.3/1.2 0.25 (ii) Of males have an RBC value outside the “expected range”? f= =6.1-4.7 =1.4 =0.35/1.4 =0.25 (c) Assuming that Australia’s population is 22 million people, approximately how many Australians would you expect to have an RBC value outside the “expected range”? =0.25 of 22,000,000 =5,500,000 (d) If males with the lowest 2% of RBC values are described as anaemic, what do we know about the RBC values of anaemic males? Approximately half of the population of the males in the extreme conditions are those with lower RBC count in their blood. Therefore about 0.125 of the males in the extreme range have lower RBC count. 2% of this population consists of people who are anaemic. =2%* 0.125 =0.0025 of the male population in Australia is anaemic. (e) Which is the more extreme result: an RBC of 4.0 for a female or an RBC of 6.2 for a male? Explain your answer. Both results are equally on the extreme since they deviation of the results from both the male and the female in this case will be the same. QUESTION 4 [19 marks] In his book To Engineer is Human: the Role of Failure in Successful Design, Henry Petroski says that up to 95% of all structural failures, including those of bridges, aeroplanes and other commonplace products of technology, are believed to be the result of the growth of cracks. In most cases, the cracks grow slowly. However, when there are too many cracks, and they are un-detected, catastrophe can occur. Suppose that a cement retaining wall is considered safe if hairline cracks occur at a mean rate of 0.42 cracks per metre section of the wall. For all calculations in this question, you must show your working. Calculate probabilities to 4 decimal places. (a) We want to calculate probabilities for the number of hairline cracks in a 10 metre section of safe wall. (i) State the random variable X which is of interest here. Random variable X is the growth of hairline cracks in walls. (ii) State the probability distribution of X, including its parameter(s) – if any. For a strong wall, the hairline crack in the wall should be few and for every metre of wall given, the rate at which the cracks should occur should be 0.42 cracks or less. (iii) Explain the reason(s) for your choice of probability distribution. Cumulative probability distribution is the choice of distribution used. This is does on the assumption that the total probability must be equal to 1 and there are two options that are provided for in this case. The first is that there is a probability that the cracks on the walls will appear and the other is based on the probability that no cracks will appear on the walls. The (iv) Calculate the probability of 2hairline cracks. P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =2/10 =0.2 (v) Calculate the probability of at most 2 hairline cracks. Between 1 (inclusive) and 2 (exclusive) Total of one crack likely to occur P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) =1/10 =0.1 (b) Calculate the probability of at least 2 metres between successive hairline cracks. P (an event) = (no. of outcomes in the event)/ (total no. of possible outcomes) = (1/10)* 2 =0.2 (c) Calculate the probability of 3or more hairline cracks in a section of safe wall of length (i) 1 metre. P (0) = (e^-7 * 7^0)/0! P (0) = (0.0009119 * 1)/1 P (0) = 0.0009 P (1) = (e^-7 * 7^1)/1! P (1) = (0.0009119 * 7)/1 P (1) = 0.0064 P (2) = (e^-7 * 7^2)/2! P (2) = (0.0009119 * 49)/2 P (2) = 0.0223 P (0) + P (1) + P (2) = 0.0296, So the probability of observing 3 or more crack in the wall is 1 - 0.0296 = 0.9704. (ii) 5 metres. P (0) + P (1) + P (2) = 0.0296 =0.0296*5 =0.148 =1-0.148 =0.852 (iii) 10 metres. P (0) + P (1) + P (2) = 0.0296 =0.0296*10 =0.296 =1-0.296 =0.704 (iv) 20metres. P (0) + P (1) + P (2) = 0.0296 =0.0296*20 =0.5392 =1- 0.5392 =0.4608 (d) For your calculations in (c), which event has the highest probability? Does this event represent a greater hazard? Explain your answer. Hairline cracks in a section of safe wall of length with the greatest probability are the 1 meter cracks on the wall. The reason to this is that there is little distribution of the probability in the one meter wall as compared to the others. The little distance makes the probability to remain higher as the other probabilities may be distributed evenly as the distance increases. The event does not represent a greater hazard since there is little distribution of the probabilities as compared to the others. QUESTION 5 [18 marks] The Natural Bank of Australia wishes to profile the defaulters among 995 of their clients who have a particular class of loan. The file NBA.xlsx for these clients includes the following variables. Client_ID Client identification number. Age Age, in years. Sex 0 = male, 1 = female. Married Yes or No. Region There are 8 regions, numbered 1, 2, …, 8. Employed 1 = employed, 0 = not employed. Income Gross annual income ($). Defaulted 1 = has defaulted on the loan, 0 = has not defaulted on the loan. [DO NOT print out the file NBA.xlsx. It is unnecessary and will waste paper.] What information about the Income and Employed status of the defaulted clients is provided by the file NBA.xlsx? (For Income, use a grouping starting at $0 and grouping by $50,000.) Income Number of defaults $0-$50,000 122 $50,001- $100,000 189 $100,001-$150,000 0 Employed Number of defaults Yes 158 No 134 The information provided by the banks on the defaulters of loans shows that the high income earning clients rare take loans from the banks. In any case they borrow the loans; they try to repay their loans since they do not have any defaulters of loans at that level. The major clients who borrow loans from the banks are the middle and lower income earning clients. Of these clients a good percentage is employed and still another is unemployed. The highest defaulters of the loans in the banks are the middles income earning class most probably because they take up large sizes of loans that they do not have the capacity to repay and as a result, they end up defaulting. The same case could be told for the lower income earning clients who also have a sizeable number of loan defaulters in the banks. 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