IE 336 Seat # Name (clearly) < KEY > Open book and notes. No calculators. 60 minutes. Cover page and five pages of exam.

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1 Open book and notes. No calculators. 60 minutes. Cover page and five pages of exam. This test covers through Chapter 2 of Solberg (August 2005). All problems are worth five points. To receive full credit, show enough work to indicate your logic. Do not spend time calculating. You will receive full credit if someone with no understanding of probability could simplify your answer to obtain the correct numerical solution. Both A and A denote the complement of A. Score Exam #1, Fall 2005 Schmeiser

2 Open book and notes. 60 minutes. 1. Suppose that C and D are events, with P(D C )=0.3 and P(C )=0.3. Are C and D independent, dependent, or not enough information to know? Independence would be implied by P(D C ) = P(D ). Because P(D ) is unknown, there is not enough information. < NEITK > 2. Suppose that A, B, C, and D are mutually exclusive events, with P(A )=0.4, P(B )=0.2, P(C )=0.3, and P(D )=0.1. Determine the value of P(A B C D). P[ A B C D ]=P[ A B C D ] = P(S)=P( )=0, where S is the sample space. < 0 > 3. Consider a particular class of IE336 students. Of those who took IE230 with the current instructor, 20% expect an "A" grade in IE336. Of the other students, 30% expect an "A" grade. If seventy percent of the class took IE230 with the current instructor, what fraction of the class expects an "A" grade? Let A denote that the randomly chosen student expects an "A". Let C denote that the randomly chosen student had the current instructor for IE230. We know that P(C )=0.7, P(A C )=0.2, and P(A C)=0.3. The total probability yields P(A )=P(A C ) P(C )+P(A C) P(C)=(0.2)(0.7)+(0.3)(0.3). < 0.23 > 4. Suppose that 30% of all Purdue students celebrate Constitution Day. Twenty percent of all Purdue students are graduate students; 10% of them celebrate Constitution Day. What fraction of undergraduate students celebrate Constitution Day? Let C denote that the randomly chosen student celebrates Constitution Day. Let G denote that the randomly chosen student is a grad student. We know that P(C )=0.3, P(G )=0.2, and P(C G )=0.1. Then total probability yields P(C )=P(C G ) P(G )+P(C G) P(G). Therefore, 0.3=(0.1)(0.2)+P(C G)(0.8), which yields P(C G)=[0.3 (0.1)(0.2)] / 0.8 < 0.35 > Exam #1, Fall 2005 Page 1 of 5 Schmeiser

3 IE 336 Seat # Name (clearly) < KEY > For Problems 5 8, consider the joint distribution given by Y X Determine the marginal distribution of X. Sum the rows to obtain P(X = 1)= =0.3 P(X = 3)= =0.7 P(X = x )=0for other values of x. 6. Determine the value of F X (2.5), the cumulative distribution function of X evaluated at 2.5. F X (2.5)=P(X 2.5)=P(X = 1) < 0.3 > 7. Determine the conditional distribution of X given that Y = 4. P(X = 1 Y = 4)=P(X = 1, Y = 4) / P(Y = 4)=0.10 / 0.35=2/7 P(X = 3 Y = 4)=P(X = 3, Y = 4) / P(Y = 4)=0.25 / 0.35=5/7 P(X = x Y = 4)=P(X = x, Y = 4) / P(Y = 4)=0.0 / 0.35=0elsewhere. 8. Determine the value of E(X 2 Y ). E(X 2 Y )=(1 2 )(2)(0.05)+(1 2 )(4)(0.10)+(1 2 )(6)(0.15). + (3 2 )(2)(0.20)+(3 2 )(4)(0.25)+(3 2 )(6)(0.25) < 27.5 > Exam #1, Fall 2005 Page 2 of 5 Schmeiser

4 IE 336 Seat # Name (clearly) < KEY > For Problems 9 10, consider the joint distribution Y X ? 9. What is the value of P(X = 3, Y = 6)? The joint probabilities must sum to one, so P(X = 3, Y = 6)=1 [ ] < 0 > 10. Determine the value of the covariance of X and Y. Cov(X, Y )=E(XY ) E(X ) E(Y ), where E(X)=(1)(0.6)+(3)(0.4)=1.8, E(Y)=(2)(0.25)+(4)(0.45)+(6)(0.3)=4.1, and E(XY)=(1)(2)(0.1)+(1)(4)(0.2)+(1)(6)(0.3)+(3)(2)(0.15)+(3)(4)(0.25)+(3)(6)(0). < 0.68 > 11. Suppose that customer service for the Federal Income Tax answers 90% of all questions correctly. What is the probability that a set of ten questions, answered by ten different agents, yields no more than one error? Assume independence, because the questions are answered by different agents. Let X denote the number of questions answered correctly. Then X is binomial with n = 10 and p = Therefore, P(X 9)=P(X = 9)+P(X = 10)=C 9 p 9 (1 p) 1 + C p 10 (1 p) 0 = (10)(0.9) 9 (0.1) 1 + (1)(0.9) 10 (0.1) 0 = (0.9) 9 + (0.9) 10 = (0.9) 9 (1.9) < > 12. Your instructor bets (with his son) that he can serve a volleyball 100 times without error. For him to have a 50% chance of winning the bet, what is the probability of each serve being successful? (You may assume that the serves are Bernoulli trials.) Let X denote the number of successful serves. We want P(X = 100)=0.5. If X is binomial with n = 100 and probability of success p, then P(X=100)=p 100 = 0.5, which implies that p = / 100. < > Exam #1, Fall 2005 Page 3 of 5 Schmeiser

5 13. The infinite series (1)(0.6)(0.4 0 )+(2)(0.6)(0.4 1 )+(3)(0.6)(0.4 2 )+(4)(0.6)(0.4 3 )+... is the expected value of a random variable X. Determine the value of P(X 2). We know that, by definition, E(X )=Σall x x P(X = x ). Therefore, P(X 2)=P(X = 1)+P(X = 2)=(0.6)(0.4 0 )+(0.6)(0.4 1 ). < 0.84 > For Problems 14 16, consider this situation. From past experience, an airline knows that forty percent of passengers request a chicken dinner. The airline loads forty chicken dinners on a flight having one-hundred passengers. (You may assume that customer preferences are independent.) 14. What is the expected number of disappointed passengers? Let X denote the number of chicken-dinner requests. Then X is binomial with n = 100 and p = 0.4. Let Y = max{0, X 40}, the number of disappointed passengers. Then E(Y )=Σx=0 100 max{0, x 40} P(X = x ) 100 (x 40) P(X = x ), =Σx=41 where P(X = x )=C x 100 (0.4) x (0.6) 100 x. 15. Let µ D denote the expected number of disappointed passengers (the answer to Problem 14). Suppose that the goodwill cost of not meeting a customer s chicken-dinner preference is $10. What is the expected goodwill cost for the flight? E($ 10Y )=$ 10E(Y )=$ 10µ D. 16. Determine the standard deviation of the number of requests for a chicken dinner. State the units. Because X is binomial with n = 100 and p = 0.4, V(X )=n p (1 p)=(100)(0.4)(0.6)=24. Therefore, std(x )= 24 requests. < 4.9 requests > Exam #1, Fall 2005 Page 4 of 5 Schmeiser

6 For problems 17 18, suppose that world-wide commercial airline accidents occur according to a Poisson process with rate one accident per week. 17. If there have been no accidents during the past two weeks, what is the probability that there will be no accidents during the next two weeks? Let X denote the number of accidents during the next two weeks. Because accidents form a Poisson process, the previous two weeks are irrelevant and X is Poisson, with meanµ=λ t, whereλ=1accident per week and t = 2 weeks. Therefore, P(X = 0)=e µ µ 0 / 0!=e 2. < > 18. What family of distributions models the time until three accidents occur? Time until the next accident is exponential, with rateλ=1accident per week. The time until the third accident is the sum of three independent exponentials. < Erlang > 19. Identify an appropriate family of distributions to model the number of named storms in a single year. The possible number of hurricanes is the set {0, 1,...}. The two families that we have studied with this range are geometric and Poisson. The mode of the geometric is zero, which doesn t seem reasonable. If hurricanes are independent of each other, then the number would be Poisson. < Poisson > 20. Let X denote temperature in degrees Fahrenheit; let Y denote temperature in degrees Celsius. (Recall: Given Fahrenheit degrees, compute Celsius degrees by subtracting 32 and then dividing by 1.8.) Write std(y ) in terms of std(x ). The conversion formula yields Y = (X 32) / 1.8. Therefore, V(Y )=V[(X 32) / 1.8]=V(X / 1.8)=V(X ) / Taking the square root yields < std(y )=std(x ) / 1.8 > Exam #1, Fall 2005 Page 5 of 5 Schmeiser

7 Exam #1, Fall 2005 Page 6 of 5 Schmeiser