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Unformatted text preview: Midterm Examination # 1 Sta 113: Probability and Statistics in Engineering Thursday, 2007 Sep. 20, 1:15 2:30 pm This is a closedbook exam so do not refer to your notes, the text, or any other books (please put them on the floor). You may use a single sheet of notes or formulas and a calculator, but materials may not be shared. A formula sheet and two blank worksheets are attached to the exam. You must show your work to get partial credit. Even correct answers will not receive full credit without justification. Please give all numerical answers to at least two correct digits or as exact fractions reduced to lowest terms . Write your solutions as clearly as possible and make sure its easy to find your answers (circle them if necessary), since you will not receive credit for work that I cannot understand or find. Good Luck! If you find a question confusing please ask me to clarify it. Cheating on exams is a breach of trust with classmates and faculty, and will not be tolerated. After completing the exam please acknowledge the Duke Honor Code with your signature below: I have neither given nor received unauthorized aid on this exam . Signature: Print Name: 1. /25 2. /25 3. /30 4. /25 Total: /100 Name: Sta 113 Problem 1: One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 90% detection rate for carriers and a 5% detection rate for noncarriers. Suppose the test is applied independently to two different blood samples from the same randomly selected individual. a. What is the probability that both tests yield the same results? b. If both tests are positive, what is the probability that the selected individual is a carrier? Answer 1 Let events B , C and BP respectively be B oth tests yielding the same result , tested person being a C arrier , and B oth tests yielding a P ositive result . a. P ( B ) = P ( B  C ) P ( C )+ P ( B  C ) P ( C ) = 0 . 01(0 . 1 2 +0 . 9 2 )+0 . 99(0 . 05 2 + . 95 2 ) = 0 . 9042 b. P ( BP ) = P ( BP  C ) P ( C ) + P ( BP  C ) P ( C ) = 0 . 9 2 . 01 + 0 . 05 2 . 99 = . 0106, P ( C  BP ) = P ( BP  C ) P ( C ) /P ( BP ) = 0 . 9 2 . 01 / . 0106 = 0 . 7660. Spring 2009 Page 1 of 10 Feb. 17, 2009 Name: Sta 113 Problem 2: The mode of a discrete random variable X with pmf p ( x ) is that value x * for which p ( x ) is largest (the most probable x value). Let X Bin ( n,p ). By considering the ratio b ( x + 1; n,p ) /b ( x ; n,p ), show that b ( x ; n,p ) increases with x as long as x < np (1 p ). Conclude that the mode x * is the integer satisfying ( n + 1) p 1 x * ( n + 1) p ....
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 Statistics, Probability

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