Georgia Institute Of Technology ISYE 6644 all exams combined-ALL ANSWERS CORRECT-GRADED A+

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ISyE 6644 | Sp/Su | Test #1 Solutions (revised 6/17/18 to correct Question 7) This test is 120 minutes. You’re allowed one cheat sheet (both sides). I also give you permission to use your compute ... r (if you want) for Questions 30 and 34. In any case, this test requires a proctor. All questions are 3 points, except 34, which is 1 point | and I’ll probably give you that 1 point as long as you give me any interesting answer. , Good luck! I want you to make this test wish that it had never been born!!! 1. TRUE or FALSE? Simulation can be used to analyze queueing models that are too complicated to solve analytically. Solution: TRUE (of course!) 2 2. Use bisection (or any other method) to find x such that e−x = x. (a) x = 0 (b) x = e (c) x = e=2 (d) x = 0 : :567 (e) None of the above. Solution: Let’s use bisection to find the zero of g(x) = e−x − x. x g(x) comments 0 1 1 −0:6321 look in [0; 1] 0:5 0:1065 look in [0:5; 1] 0:75 −0:2776 look in [0:5; 0:75] 0:625 −0:0897 look in [0:5; 0:625] 0:5625 0:0073 look in [0:5625; 0:625] 0:59375 −0:0415 look in [0:5625; 0:59375] 0.578125 −0:0172 look in [0:5625; 0:578125] 0.5703125 −0:0050 OK, stop here.2 If we keep going, we get closer and closer to x = 0 : :5671; so the answer is (d). 2 3. Suppose that X is a continuous random variable with p.d.f. f(x) = 2x for 0 < x < 1. Find Pr(X < 1=2 j X > 1=4). (a) 0 (b) 0.2 (c) 0.5 (d) 0.8 (e) 1/16 Solution: We have Pr(X < 1=2 j X > 1=4) = Pr(X < 1=2 \ X > 1=4) Pr(X > 1=4) = Pr(1=4 < X < 1=2) Pr(X > 1=4) = R11==42 2x dx R11=4 2x dx = 0:2; after the smoke clears. So the answer is (b). 2 4. Suppose I conduct a series of 4 independent experiments, each of which has a 20% chance of success. What’s the probability that I’ll see at least 3 successes? (a) 0.027 (b) 0.181 (c) 0.819 (d) 0.973 (e) 13 Solution: The number of successes X ∼ Bin(4; 0:2). Thus, Pr(X ≥ 3) = 4Xx =3 x4(0:2)x(0:8)4−x = 0:0272: So the answer is (a). 2 5. If X ∼ Bern(0:5), find E[‘n(X + 1)]. (a) 1 (b) e=2 (c) 0.347 (d) 1.38 (e) None of the above. Solution: By the Unconscious Statistician and the fact that X ∼ Bern(0:5), we have E[‘n(X + 1)] = 1Xx =0 ‘n(x + 1) Pr(X = x) = 1 2 ‘n(1) + 1 2 ‘n(2) = 0:347; so the answer is (c). 2 6. If X has a mean of −2 and a variance of 3, find Var(−2X + 1). (a) −5 (b) −6 (c) −12 (d) 12 (e) 13 Solution: Var(−2X + 1) = 4Var(X) = 12, so the answer is (d). 24 7. Toss a 6-sided die repeatedly. What is the variance of the number of tosses until you observe a 3? (a) 6. (b) 36. (c) 1/6. (d) 5/6. (e) 30. Solution: Let X ∼ Geom(p = 1=6) denote the number of tosses. Thus, Var(X) = q=p2 = (1 − p)=p2 = 30; so that the answer is (e). 2 8. Suppose that X and Y are identically distributed with a mean of −2, a variance of 3, and Cov(X; Y ) = 1. Find Corr(X; Y ). (a) 0 (b) 1/9 (c) 1/3 (d) 1 (e) 3 Solution: Corr = Cov=pVar(X)Var(Y ) = 1=3, so (c) is the answer. 2 9. Again suppose that X and Y are identically distributed with a mean of −2, a variance of 3, and Cov(X; Y ) = 1. Find Var(X − Y ). (a) 3 (b) 4 (c) 5 (d) 65 (e) 8 Solution: Var(X) + Var(Y ) − 2Cov = 4, so the answer is (b). 2 10. Consider a Poisson process with rate λ = 1. What is the probability that the time between the 4th and 5th arrivals is less than 2? (a) 1=e (b) 1 − (1=e) (c) 1=e2 (d) 1 − (1=e2) (e) 0.95 Solution: All interarrivals are i.i.d. Exp(λ). In particular, let X ∼ Exp(λ = 1) denote the time between the 4th and 5th arrivals. Then Pr(X < 2) = 1 − e−λx = 1 − e−2 = 0:865. Thus, the answer is (d). 2 11. If X is Nor(5,4), find Pr(X > 3). (a) 0.05 (b) 0.159 (c) 0.5 (d) 0.841 (e) 0.95 Solution: Pr(X > 3) = Pr(Z > 3p−45) = Pr(Z > −1) = 0:8413. So the answer is (d). 2 12. If X and Y are i.i.d. standard normal random variables, find Pr(X − Y > 1). (a) 0.159 (b) 0.246 (c) 0.5 (d) 0.76 (e) 0.841 Solution: Note that X − Y ∼ Nor(0; 2). Then Pr(X − Y > 1) = PrZ > 1p−20 ≈ 0:24: So (b) is the answer. 2 13. Suppose X and Y are i.i.d. Exp(λ = 2). Find Pr(X + Y ≤ 1). (a) 1=e2 (b) 1 − (1=e2) (c) 0.5 (d) 0.406 (e) 0.594 Solution: Note that S = X + Y ∼ Erlangk=2(λ = 2). Thus, Pr(S ≤ 1) = 1 − k−1 X i =0 e−λs(λs)i i! = 1 − 1X i =0 e−22i i! = 1 − 3e−2 = 0:594: So the answer is (e). 2 14. Suppose X1; : : : ; Xn are i.i.d. from a Pois(λ = 3) distribution. What is the approximate distribution of the sample mean X¯ for n = 300? (a) Pois(3) (b) Pois(0.01) (c) Nor(3, 3) (d) Nor(3, 0.01)7 (e) Nor(0.01, 0.01) Solution: The Central Limit Theorem implies that X¯ ≈ Norµ; σn2  ∼ Nor3; n3 : So the answer is (d). 2 15. Suppose U and V are i.i.d. Unif(0,1) random variables. What does d6Ue + d6V e do? (Recall that dxe is the \ceiling" function.) (a) This gives a continuous Unif(0,12) random variate. (b) This gives a continuous triangular random variate. (c) This gives a normal random variate. (d) This is a simulated 6-sided die toss. (e) This simulates the sum of two 6-sided dice tosses. Solution: (e). 2 [Show More]

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