ࡱ> oqnlmnopqrstu6$%&'()*+, 5@ Z1bjbj22 XX(k::::EEEEzFj<2GPOOOOMUMUMUiiiiiii$VkRmZiMU;TMUMUMUi::OOikYkYkYMU :lO:OikYMUikYkYYdX,fO&G PR/[EYVlAf(-gli0jifnWnPf::::nf0MUMUkYMUMUMUMUMUii#&OY&U5 - Problem 7 We know that ambient temperature and wind speed are important in determining the current carrying capability of a transmission line. A transmission line owner wants to maximize its sale of capacity for a particular line in order to maximize its profits. However, it must decide how much capacity to sell on a day-ahead basis. Therefore it must predict temperature and wind. Let's denote the event "warm day" as event A and the event "windy day" as event B. The weather forecast indicates that the probability of a windy day is 0.80, the probability of a warm day is 0.30, and the probability of a warm and windy day is 0.28 Are the events A and B independent? What is the probability of the event "windy day will be warm"? What is the probability of the event "warm day will be windy"? What is the probability of the event "windy and not warm"? What is the probability of the event "windy day will be not warm"? What is the probability of the event "not warm day will be windy"? Solution to problem 7 No, since  EMBED Equation.3 =0.28 but  EMBED Equation.3  EMBED Equation.3 =0.8*0.3=0.24  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.28/0.80=0.35  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.28/0.30=0.93 Denote the event "not warm" as a. Then  EMBED Equation.3 since A and a are complements. Therefore,  EMBED Equation.3 = EMBED Equation.3 - EMBED Equation.3 =0.80-0.28=0.52  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.52/0.8=0.65  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.52/0.7=0.74  U5 - Problem 8 Of great interest to the electric power industry today is the detection of the "hot spot" temperature in power transformers. This is the temperature of the hottest part of the winding and is a function of many variables including loading and ambient temperature. A simplified approach assumes that if the hot spot temperature exceeds a certain level, the transformer will fail in one hour; otherwise it will not fail. The hot spot temperature may be approximated if both the oil temperature and the loading current are known. We desire to predict transformer failure for a particular, rather low, loading current I. Let's denote the event "transformer failure" to be A. Let's also denote the event "oil temperature test indicates eminent failure" to be B. It is known that at the loading level I,  EMBED Equation.3 =0.003 and that the oil temperature test has 2% false positives and 1% false negatives. What is  EMBED Equation.3 ? (a is the complement of A) What is  EMBED Equation.3 ? What is  EMBED Equation.3 ? What is  EMBED Equation.3 ? What is  EMBED Equation.3 ? What is  EMBED Equation.3 ? If, at this loading current I, the oil temperature test indicates the transformer will fail, (i) What is the probability that the transformer will fail ? (ii) What is the probability that the transformer will not fail ? Consider that the event A is "having cancer" and the event B is "positive test for cancer". What does the above result suggest about screening tests for diseases with low prevalence? Solution to problem 8 I provide two answers below, depending on whether you though a "positive" test indicated "good" transformer or failed transformer. I will accept either answer. Solution 1 : If you assumed that a "positive" test indicates that the transformer is "good" and a "negative" test indicates that the transformer is failed, then  EMBED Equation.3 =1-0.003=0.997  EMBED Equation.3 =1-0.01=0.99  EMBED Equation.3 =0.02  EMBED Equation.3 = EMBED Equation.3  EMBED Equation.3 =0.003*0.98=0.00294  EMBED Equation.3 = EMBED Equation.3  EMBED Equation.3 =0.997*0.02=0.01994  EMBED Equation.3 = EMBED Equation.3 + EMBED Equation.3 =0.02291 Given B, we use  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.00297/0.02291=0.13  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.01994/0.02291=0.87=1- EMBED Equation.3  Note that each of the above two probabilities could have been computed using Bayes Theorem, e.g.,  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3  where in this case,  EMBED Equation.3 =A and  EMBED Equation.3 =a it suggests that if I am in a population where cancer occurs infrequently (e.g., person under 40), and if a reliable, but imperfect test indicates I have cancer, the probability that I actually do have cancer (13%) is still in my favor (i.e., there is 87% chance I do not have cancer ). Solution 2 : If you assumed that a "positive" test indicates that the transformer is "good" and a "negative" test indicates that the transformer is failed, then  EMBED Equation.3 =1-0.003=0.997  EMBED Equation.3 =1-0.02=0.98  EMBED Equation.3 =0.01  EMBED Equation.3 = EMBED Equation.3  EMBED Equation.3 =0.003*0.98=0.00294  EMBED Equation.3 = EMBED Equation.3  EMBED Equation.3 =0.997*0.01=0.00997  EMBED Equation.3 = EMBED Equation.3 + EMBED Equation.3 =0.01291 Given B, we use  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.00294/0.01291=0.2277  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3 =0.0997/0.01291=0.7723=1- EMBED Equation.3  Note that each of the above two probabilities could have been computed using Bayes Theorem, e.g.,  EMBED Equation.3 = EMBED Equation.3 / EMBED Equation.3  where in this case,  EMBED Equation.3 =A and  EMBED Equation.3 =a (h) it suggests that if I am in a population where cancer occurs infrequently (e.g., person under 40), and if a reliable, but imperfect test indicates I have cancer, the probability that I actually do have cancer (22.77%) is still in my favor (i.e., there is 77.23% chance I do not have cancer ).  U6 - Problem 1 An engineer at a large industrial is looking for an energy seller to supply its electric energy needs. He has the names of five energy sellers, A, B, C, D, and E, and email addresses. He wants to know which ones have load following capability, so he sends them a message. A and B are the only ones that have load following capability. If we assume that the order of reply is random, then we can define the number of replies necessary before a load following supplier is identified a the random variable Y, e.g., y=1 indicates that the first supplier to reply is a load following generator. Find and plot the probability mass function for Y. Hint: Evaluate  EMBED Equation.3  one at a time. The multiplication rule may help. Find and plot the cumulative distribution function for Y. Solution to problem 1 Note that Y is the RV associated with the first positive request. (a)  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  (b)   U7 - Problem 1 Suppose that the temperature T of a certain conductor has a uniform pdf in the range 40 to 60 degrees C. Then compute:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   Solution to problem 1  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   U7 - Problem 2 Let x denote the vibratory stress (psi) on a wind turbine blade for a particular wind speed y. Assume that the pdf for X is given by  EMBED Equation.3  Note that y is just a constant. Verify that  EMBED Equation.3  is a legitimate pdf Suppose y=100. What is the probability that x is less than 200 ? less than or equal to 200 ? Greater than 200 ? What is the probability that x is between 100 and 200 ? 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` ` ` ` ` ` ` ` ` `  ` ` ` ` ` ` ` ` ` ` ?` ` ``S&'Khf`S&'KhfS&'Khf`S&'Khf`S&'Khf`S&'Khf`S&'Khf`S&'hf`S&'Khf`S&' ` ` ` ` ` ` ` ` ` ` ?` ` ` ` ` ` ` ` ` ` `? ` ` ` ` ` ` ` ` ` `  ` ` ` ` ` ` ` ` ` ` ?` ` ` ` ` ` ` ` ` ` `? ` ` ` ` ` ` ` ` ` ` | ` ` ` ` ` LJ\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2?\|e\|e\|e\|e\|e\|e\|e\|e\|e\|eǷH 4H 4H 4H 4H 4@(2@(2@(2s(2@(2? 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