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Midterm20Exam20-201995

Midterm20Exam20-201995 - CEE 597 — Risk Analysis and...

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Unformatted text preview: CEE 597 — Risk Analysis and Management Midterm Exam 1995 Test is open book and open notes. Show work to justify answers. You have 50 minutes to complete this 50 point exam. 1. (8 points) (a) What is the leading cause of deaths in the United States? (b) The number of cancer deaths, and the fraction of all deaths due to cancer, has been increasing over time in the United States. 15 this cause for alarm? (c) Name six of the major errors, decisions, or initiating events that lead to the tragedy and loss of life at Bhopal, India, related to release of MIC. 2. (8 points) We have discussed the many dimensions and criteria useful to desuibe the significant dimensions of risk issues. Bungy jumping has become a new form of recreation. I was wondering how its risk might be compared with that of sky diving (with a parachute) and hung gliding. Define 3 different quantitative criteria to describe risk of injury (or even death) related to these three activities. In 2-3 sentences indicate for each criteria why it is important, what dimension of risk it emphasizes, and what it neglects. 3. (12 points) Consider the network below. a) What are the minimal cut sets ? b) If all components have a failure probability of 1%, except component A which has a failure probability of 0.1%, provide a good estimate of the system’s reliability. (It doesn’t have to be exact.) c) If you had the opportunity to improve the reliability of just one component, which should you upgrade ? d) Draw an equivalent fault tree. (All inputs should be independent.) 4. (12 points) You participate in FEMA special teams to investigate destructive natural events that occur with little warning: major floods, tornadoes, hurricanes, and earthquakes. Below is the frequency with which FEMA receives calls to investigate events of different types, and the probability that events of different types turn out to be a disaster (killing 10 or more people). When a call comes, there is a 30% chance you are included on the team that responds to the event, a 25% chance you are a standby that doesn’t go, and a 45% chance you are not involved at all. s CEE 597 — Risk Analysis and Management Midterm Exam 1995 Event type Frequency PIDisuster] major floods 0.6/ yr 0.15 tornadoes " 3.0/ yr 0.05 hurricanes 02/yr 0.7 earthquakes 0.2/yr 0.9 (a) Draw an event tree. (b) What is the probability in a two—year period that you are on a team that responds to an event that is a disaster ? (c) What is mean and variance of time until you investigate five disasters ? (d) Would it be reasonable to consider the arrival of such natural disasters to be a Poisson process? 5. (10 points) Pickup trucks whose gas tanks explode in side collusions have been a concern. Assume that model C pickups have been alledged to be vulnerable to such explosions, and New York State data , shows that 20,000 such vehicles are registered in the state and there have been 3 such accidents. On the "6 other hand, there are 60,000 other pickups registered of the same model year and they have also experienced 3 such explosions. For an appropriate randomized test, what would be (a) the appropriate hypotheses for this situation, (b) a reasonable test statistic, (c) the distribution of the test statistic under the hypotheses, (d) the p—value for this data, and (e) the type 11 error probability for an a = 5% test if the model G pickups are really 3 times more likely to be in a collusion that results in an explosion? P[ X = x l n: 6, p ] for binomial distribution with various values of p 0-1 Q-Z 9225 m M 25 D-_6 Q-fifl M Qé 9.2 0.5314 0.2621 0.1780 0.0881 0.0467 0.0156 0.0041 0.0014 0.0002 0.0001 0.0000 0.3543 0.3932 0.3560 0.2638 0.1866 0.0938 0.0369 0.0164 0.0044 0.0015 0.0001 0.0984 0.2458 0.2966 0.3292 0.3110 0.2344 0.1382 0.0821 0.0330 0.0154 0.0012 0.0146 0.0819 0.1318 0.2191 0.2765 0.3125 0.2765 0.2191 0.1318 0.0819 0.0146 0.0012 0.0154 0.0330 0.0821 0.1382 0.2344 0.3110 0.3292 0.2966 0.2458 0.0984 0.0001 0.0015 0.0044 0.0164 0.0369 0.0938 0.1866 0.2638 0.3560 0.3932 0.3543 0.0000 0.0001 0.0002 0.0014 0.0041 0.0156 0.0467 0.0881 0.1780 0.2621 0.5314 U\U‘IpfiWN'-‘DX Critical Points for a Standard Normal Distribution Q 0.5 0.4 0.25 0.2 0.1 0.05 0.01 0.002 0.001 2m 0000 0.253 0.675 0.842 1.282 1.645 2.326 2.878 3.090 CEE 597 — Risk Analysis and Management SOLUTIONS - Midterm Exam 1995 1) (a) Heart disease. (b) Population increasing so numbers increase. In addition, population not dying young from contageous diseases, so cancer among older citizens represents a larger fraction of all deaths. Age-adjusted cancer rates flat or down, except lung cancer. (c) Runaway chemical reaction initiated intentially by employee or accidential water leak during maintenance, refrigeration system that kept tanks cool turned off for months, vent-gas scrubber underdesigned, vent-gas scrubber not refilled with castic soda (neutralizing agent), flame tower not working, high personnel turned over at plant but received few months instead of a year of training, operators didn’t understand manuals, gauges in control room may not have worked, storing large amounts of dangerous chemical, overconfident management that did no conceive of such a disaster. PICK SIX! 2) (9 points) Students need to be precise 8: complete in definition of criteria. Criteria 1: # of serious injuries or deaths per year in the US. Describes total magnitude of the problem in the US and thus its national importance; ignores number of participants in each activity and potency. Criteria 2: # of injuries or deaths per jump or flight. Describes potency of jump or flight and thus direct riskiness to participants for that unit of activity. Does not indicate total social impact or total US significance. Criteria 3: Avg Cost/ injury or Avg days lost work/ injury, or some other measure of severity to participant of injuries; ignores total US cost or total lost hours of work. Problem: Injuries per hour doesn’t seem right. Sky diving jumps take a lot longer than a single bungy jump. Does one count the ride in the plane? Different individuals may fall at different rates. Likewise hang gliding flights could be very long. Most of risk is in the landings which should be fairly independent of the rate of fall or length of flight. What is time for bungy jumping: include the wait in line or just time from jump until one is again on the ground? Other ideas: Injuries per participant per year. Decreased life expectancy/ jump. How might we account for experience of participants? 3. a) Minimal cut sets: A, EB, EDC (1 pt each) b) Failure probability z 10‘3 + 10*1 + 10"6 = 1.1x10‘3 --> Reliability = 0.9989 (3 pts) Will accept R = 0.999 as answer. Note: intersection A and EB has probability 107. c) Improve A. A Clearly dominates failure probability. (2 pts) d) System __ QZG—E GIGS Failure t? CEE 597 — Risk Analysis and Management SOLUTIONS -- Midterm Exam 1995 0.30 — Yes you go. I ter SK . No you dont go 0.2 Not a disaster Arrival of call (2 pts) 0.15 Start 3) (21) Whether or not you go can be included in analysis before or after one resolves what type of natural event it is, or whether or not the event is a disaster. 0 Need to calculate probability that you go on a team that investigates a disaster. Frequency of teams investigating events is A. = 0.6 (.15) + 3.0 (0.05) + 0.2 (0.7) + 0.2 (0.9) = 056/ yr (3 pts) Chances that you are on the team is 30%, so frequency you are on team investigating a disaster is N = 0.3 (0.8/ yr) = 0.168/ yr (1 pt) (b) Probability in 2-year period you are on a team that responds to a disaster is given by P[ on a team over 2 years] = 1 — exp[ — 2 yr ’* (0.168/yr)] = 0.29 or about 29%. (2 pts) Requested probability of 29%, not frequency of 0.34 per 2-year period. Some clever people computed P(at least one call) * 0.3 - which gives wrong answer. (c) Mean until 5 disasters is 5 / (0.168/ yr) = 29.8 about 30 yrs. (1 pt) Variance = 5/ (027/ yr)2 = 177 (yr)2 (1 pt) (d) These disasters occur (1) fairly independently in general without memory [earthquakes in different locations, tornadoes, and hurricanes don’t pay much attention to each other]. (2) one—at-a-time, (3) at a constant rate. This first assumption is not perfect because one earthquake can trigger another, or release tension. Likewise the weather that causes a hurricanes may inhibit tornadoes in locations were they occur. But it is probably not too had an assumption. We also need a constant rate within the year which is probably not true for floods, tornadoes and hurricanes, which are all seasonal. On average over a year assumptions reasonable bad. Should not occur in pairs, but occur one at a time. (2 pts) 5) (a) Ho: probability each truck suffers an accident is the same. Ha: probability a G model truck suffers an accident is larger. (2 pts) (b) X = number of accidents suffered by model G trucks (1 pt) (c) H0: X ~ Bin[ n=6, p = 20/ (20+60) = 0.25 ] Ha: X ~ Bin[ n=6, p > 0.25] (2 pts) (d) p-value for X = 3 is Pr[ X 2 3 | Ho] = 0.0002 + 0.0044 + 0.330 + ,0. 1318 = 0.17 (2 pts) (e) For 5% test reject if X 2 4 (1 pt) Ha1 becomes X ~ Bin[ n=6, p 3*20/(3*20+60) = 0.5 ] so [5 = Pr[ X S 3 I Ha] = 0.016 + 0.094 + 0.234 + 0.313 = 0.66 (2 pts) ...
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