251y0321 3/20/03Show your work in problems 7-9.TABLE 5-4TABLE 4-2TABLE 4-4TABLE 4-7251y0321 3/20/03 ECO251 QBA1 SECOND HOUR EXAMMarch 21, 2003 Name: ____KEY_______________ Social Security Number: _____________________Part I: (48 points) Do all the following: All questions are 2 points each except as marked. Exam is normedon 50 points including take-home. (There are 57 possible points.)1. If two events are mutually exclusive, what is the probability that both occur at the same time?a) *0.b) 0.50.c) 1.00.d) Cannot be determined from the information given.2. If two events are collectively exhaustive, what is the probability that one or the other occurs?a) 0.b) 0.50.c) *1.00.d) Cannot be determined from the information given.3. If two events are independent, what is the probability that they both occur?a) 0.b) 0.50.c) 1.00. Explanation: BPAPBAP , but we don’t know either.d) *Cannot be determined from the information given.4. A business venture can result in the following outcomes (with their corresponding chance of occurring in parentheses): Highly Successful (10%), Successful (25%), Break Even (25%), Disappointing (20%), and Highly Disappointing (?). If these are the only outcomes possible for the business venture, what is the chance that the business venture will be considered Highly Disappointing?a) 10%b) 15%c) *20%d) 25%1251y0321 3/20/03Show your work in problems 7-9.TABLE 5-4The following table contains the probability distribution for X = the number of traffic accidents reported in a day in Corvallis, Oregon.X 0 1 2 3 4 5P(X) 0.10 0.20 0.45 0.15 0.05 0.055. Referring to Table 5-4, the probability of at least 1 accident is ___.90_____. Explanation: 111 PxP 6. Referring to Table 5-4, the mean or expected value of the number of accidents is __2.00__.7. Referring to Table 5-4, the standard deviation of the number of accidents is _1.1832_. (4) 00.2xxPxE, 40.522xPxxE, 40.1240.52222xE,1832.140.1 x xP xxP xPx20 .10 0.00 0.001 .20 0.20 0.202 .45 0.90 1.803 .15 0.45 1.354 .05 0.20 0.805 .05 0.25 1.25Sum 1.00 2.00 5.408. We sell a product with a markup of $7 per unit and have fixed costs of $500 monthly. Thus, if we sell 100 units, our profits will be $7(100) – 500 =$200. If our expected sales are 250 units and the standard deviation is 20, what is the mean and standard deviation of our profits? (4) Solution: Use the solution of problem J2. The formula says that if ,baxy baxyand 222xya , so, if ,5007 xy7a and 500b. Then 125050025075007 xy, and 1960020722222xya. Thus .1402072071960022y.2251y0321 3/20/03TABLE 4-2An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following resultwas obtained.Did Well on Midterm Did Poorly on MidtermStudying for Exam 80 20Went Bar Hopping 30 70If we finish the table we get:Did Well on MidtermDid Poorlyon MidtermTotalStudying for Exam 80 20 100Went Bar Hopping 30 70 100Total 110 90 2009. Referring to Table 4-2, what is the probability that a randomly selected student who went bar hopping will do well on the midterm?a. *30/100 Explanation: This is a conditional probability. Out of the 100 who went bar hopping, only 30 did well on the exam. b. 30/110 c. 30/200d. (100/200)*(110/200)10. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" area) *statistically dependent. Explanation: To be independent the probability that someone did well on the midterm, which is 110 out of 200, has to be the same as the conditional probability that someone did well on the midterm, given that he/she studied for the exam, which is 80 out of 100. These fractions are not equal.b) mutually exclusive.c) collective exhaustive.d) None of the above.11. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A B) = 0.1. Then P(B|A) = .25 . Explanation: Note that 1. ABPBAP and, by the multiplication rule, 25.4.1.APABPABP. Note that you can’t use Bayes’ Rule here because you don’tknow .BAP12. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A B) = 0.1. Then P(AB)= .8 . Explanation: Note that by the addition rule, 8.1.5.4. BAPBPAPBAP.13. Evaluate 524C 270725 . Explanation: !!!rrnnCnr. So 270725123449505152!4!48!52524C3251y0321 3/20/0314. If I am playing a card game with hands of 4 cards. The deck contains 52 cards of which 13 are hearts. What is the chance that I get 2 hearts? Do not finish this problem – write out the answer using factorial notation. (4) 270725!2!37!11!39!13!2!37!2!11!52!4!48!39!13!4!48!52!2!37!39!2!11!132 Explanation: !!!rrnnCnr. There are 13 hearts and 39 non-hearts in the deck. We need 2of each. There are !2!11!13132C to pick the hearts and !2!37!39392C ways to pick the non-hearts out of a total of 270725123449505152!4!48!52524Cways to pick a hand of 4. The answer is 2135.495051521212123438391213!2!37!2!11!52!4!48!39!13!4!48!52!2!37!39!2!11!132524392132CCCPTABLE 4-4Suppose that patrons of a restaurant were asked whether they preferred beer or whether they preferred wine. 70% said that they preferred beer. 60% of the patrons were male. 80% of the males preferred beer.15. Referring to Table 4-4, the probability arandomly selected patron is a female is.40. 16. Referring to Table 4-4, the probability arandomly selected patron is a female who prefers beer is .22. 17. Referring to Table 4-4, suppose a randomly selected patron prefers wine. Then the probability the patron is a male is .40. (4)Explanation: Assume that 100 people walk into the restaurant. Let M be ‘male,’ F be
View Full Document
Unlocking...