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fin08sol - Final Exam 45-730 Probability Spring 2008 Name...

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Final Exam 45-730 Probability Spring 2008 Name SOLUTION This test is open notes, open book. This is an individual work. Communication with anybody except the proctor or instructor is forbidden. Laptops are allowed but Internet communication is strictly forbidden. There are twenty one questions total: sixteen multiple-choice questions and five problems. For each of the multiple-choice questions, select the one best answer in your judgment: marking more than one answer will result in no credit. For choices between values, pick the one closest to your answer if your answer is not listed. For each of the remaining problems, please write your answer in fractional form (for example 3/4, 23/41, etc) or with at least four digits of accuracy (for example 0.7500, 0.5609, etc). Show your work. The test is two hours and fifty minutes long. Maximum score is 75 points. 1
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PART I: Multiple choice questions (2 points each) 1. The probability that an employee at a company eats lunch at the company cafeteria is 0.23. The probability that an employee is female is 0.52. The probability that an employee eats lunch at the employee cafeteria and is female is 0.11. What is the probability that a randomly chosen employee either eats at the cafeteria or is female? 0.12 0.34 0.44 0.64 0.84 Solution: P (female or lunch) = P (female) + P (lunch) - P (female and lunch) = 0 . 52 + 0 . 23 - 0 . 11 = 0 . 64. 2. A recent survey found that 14% of secretaries have experienced some form of wrist pain from typing. It also found that 6% of all secretaries have both experienced some form of wrist pain from typing and taken aspirin on a daily basis. What is the probability that a secretary who has wrist pain takes aspirin on a daily basis? 0.14 0.39 0.43 0.45 0.57 Solution: It is given that P (wrist pain) = 0 . 14 and P (wrist pain and aspirin) = 0 . 06. Then, by the formula for conditional probability, P (aspirin | wrist pain) = P (wrist pain and aspirin) /P (wrist pain) = 0 . 06 / 0 . 14 = 0 . 43. 3. On average, you receive 3.2 pieces of junk mail a day. Assuming that the number of pieces of junk mail you receive each day follows a Poisson distribution, what is the standard deviation of the number of pieces of junk mail you receive daily? 0.31 1.79 3.2 10.24 There is not sufficient information to determine this Solution: The variance of a Poisson random variable with mean λ is λ . Therefore, the standard deviation we look for is 3 . 2 = 1 . 79. 2
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4. Five fair and balanced coins are tossed. What is the probability that exactly one head appears? 1/32 5/32 1/5 10/32 16/32 Solution: C 5 1 2 5 = 5 32 5. A single fair and balanced coin is tossed twice. Which of the following is not an independent set of events? Two heads in total, first coin toss is a head One head in total, first coin toss is a head One head in total, second coin toss is a head First coin toss is a head, second coin toss is a head First coin toss is a head, second coin toss is a tail Solution: If there are two heads in total and we only tossed the coin twice, then clearly the first coin must be a head. Therefore, these two events are dependent. All the others can be verified to be independent.
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