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Question 6: Let X and Y be two random variables such that: Y = 75 - 3X – X2, E(x²) = 12 and E(Y) = 54. Then V(X) =

Question 6: Let X and Y be two random variables such that: Y = 75 - 3X – X2, E(x²) = 12 and E(Y) = 54. Then V(X) =

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.7: Probability
Problem 39E: Assume that the probability that an airplane engine will fail during a torture test is 12and that...
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41% D 11:47 + Math 310_Key Prac... Question 6: Let X and Y be two random variables such that: Y = 75 – 3X – X², E(x²) = 12 and E(Y) = 54. Then V(X) = a. 6. b. 12 C. 3 d. 9 Question 7: Let X be a random variable such that E(X – 2)2 = 10 and E(X + 1)2 = 4, then a. E(X) = - b. E(X) = c. E(X) =- d. E(X) = - Question 8: Let X and Y be a random variables such that E(X) = 2, E(Y) = 5, V(X) = 9, V(Y) = 8 and Cov(X, Y) = 3. Then E(3X? – 2Y² + XY) = a. 15 b. 0 c. -14 d. None of the above Question 9: 80% of people those who purchase pet insurance are women. If the owners of 10 pet insurance are randomly selected, then the probability that more than 3 are women: (a) 0.9991 (b) 0.9999 (c) 0.0009 (d) None of the above
Transcribed Image Text:41% D 11:47 + Math 310_Key Prac... Question 6: Let X and Y be two random variables such that: Y = 75 – 3X – X², E(x²) = 12 and E(Y) = 54. Then V(X) = a. 6. b. 12 C. 3 d. 9 Question 7: Let X be a random variable such that E(X – 2)2 = 10 and E(X + 1)2 = 4, then a. E(X) = - b. E(X) = c. E(X) =- d. E(X) = - Question 8: Let X and Y be a random variables such that E(X) = 2, E(Y) = 5, V(X) = 9, V(Y) = 8 and Cov(X, Y) = 3. Then E(3X? – 2Y² + XY) = a. 15 b. 0 c. -14 d. None of the above Question 9: 80% of people those who purchase pet insurance are women. If the owners of 10 pet insurance are randomly selected, then the probability that more than 3 are women: (a) 0.9991 (b) 0.9999 (c) 0.0009 (d) None of the above
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