Econ_221-4_W2007_Assignment_1_Answers

Econ_221-4_W2007_Assignment_1_Answers - 1 Concordia...

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Concordia University Department of Economics ECON 221/4 – Sections AA, B, E Instructors: M. DaPalma, V. Terekhov, G. Tsoublekas Winter 2007 - ASSIGNMENT 1 - ANSWERS Due at noon on Friday, January 26, 2007 Name: I.D. Section: THEORY – 50 points 1. A restaurant is famous for its roast beef recipe. The restaurant recommends to its patrons that its roast beef be accompanied with wine of either Merlot or Cabernet varieties. In a specific year, it was estimated that 80% of its customers ordered the famous dish. Furthermore, 35% of the customers had Merlot with their food and 55% of the customers had Cabernet with their food. a)If having Merlot and Cabernet are mutually exclusive, what is the probability that a randomly chosen customer had at least one of the two wine varieties? P (M) = 0.35 P (C) = 0.55 P (RB) = 0.80 Events M and C are mutually exclusive. Therefore, P (M C) = P (M) + P (C) = 0.35 + 0.55 = 0.90 or 90% b) If drinking Merlot and eating Roast Beef are statistically independent, what is the probability that a randomly chosen customer will order at least one of these two? (3 points) Events M and RB are not mutually exclusive. Therefore, P (M RB) = P (M) + P (RB) – P (M RB) = 0.35 + 0.8 – 0.35*0.8 = 0.87 or 87% c)Of the customers who ordered Cabernet, 70% also ordered Roast Beef. What is the probability that a randomly chosen customer will have ordered at least one of these two? (4 points) P (RB | C) = 0.70 then P (C RB) = P (C) * P (RB | C) = 0.55 * 0.70 = 0.385 And P (C RB) = P (C) + P (RB) – P (C RB) = 0.55 + 0.80 – 0.385 = 0.965 or 96.5% 1
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2. The labor force data on gender and number of jobs held indicate that 70% of workers are men. Furthermore, 5% of women hold two or more jobs, while the same figure for men is 4 times larger. (20 points) a) Construct a table or a tree diagram of joint probabilities. (4 points) Jobs Held Men Women Total One job (1J) 2 or more jobs (2J) 0.560 0.140 0.285 0.015 0.845 0.155 Total 0.700 0.300 1.000 b) What is the probability that a randomly chosen worker is a man and holds one job? (3 points) P (M 1J) = P (M) * P (1J │M) = 0.70 * (0.56 / 0.70) = 0.56 = 56% c) What is the probability that two randomly chosen workers are both men and hold one job? (4 points) 2 nd Worker 1 st Worker Total Man with 1 Job (M1J) Other Man with 1 Job (M1J) Other 0.3136 0.2464 0.2464 0.1936 0.5600 0.4400 Total 0.5600 0.4400 1.0000 The men are independent of each other. Therefore, P (M1J M1J) = P (M1J) * P (M1J) = 0.56 * 0.56 = 0.3136 = 31.36% d) What is the probability that a randomly chosen worker is either a woman or is holding 2 jobs (or both)? (3 points) P (W 2J) = P (W) + P (2J) – P (W 2J) = 0.30 + 0.155 – 0.015 = 0.44 = 44% 2
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e) What is the probability that, a person holding more than one job is a woman? (3 points) P (W│2J) = P (W 2J) / P (2J) = 0.015 / 0.155 = 0.0968 = 9.68% f) Are the two variables “number of jobs held” and “gender” statistically independent? (3 points)
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This note was uploaded on 02/25/2010 for the course ECON 00000112 taught by Professor Ianirvine during the Spring '09 term at Concordia Canada.

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Econ_221-4_W2007_Assignment_1_Answers - 1 Concordia...

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