HW 5 Solutions

# HW 5 Solutions - =.25 25.25 4 2(5(2(10(4.25 100(1 16 =...

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STATW1211.005 Prof. Zhang TR 02:40P-03:55P Homework 5 – Solutions Textbook: Probability and statistics for engineering and the sciences (7 th edition) by Jay L. Devore Chapter 5: Question 60 Y is normally distributed with , and . Thus, and Question 66 a. With M = 5X 1 + 10X 2 , E(M) = 5(2) + 10(4) = 50, Var(M) = 5 2 (.5) 2 + 10 2 (1) 2 = 106.25, σ M = 10.308. b. P( 75 < M ) = c. M = A 1 X 1 + A 2 X 2 with the A I ’s and X I ’s all independent, so E(M) = E(A 1 X 1 ) + E(A 2 X 2 ) = E(A 1 )E(X 1 ) + E(A 2 )E(X 2 ) = 50 d. Var(M) = E(M 2 ) – [E(M)] 2 . Recall that for any r.v. Y, E(Y 2 ) = Var(Y) + [E(Y)] 2 . Thus, E(M 2 ) = (by independence)
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Unformatted text preview: = (.25 + 25)(.25 + 4) + 2(5)(2)(10)(4) + (.25 + 100)(1 + 16) = 2611.5625, so Var(M) = 2611.5625 – (50) 2 = 111.5625 e. E(M) = 50 still, but now Cov(X 1 ,X 2 ) = (.5)(.5)(1.0) = .25, so = 6.25 + 2(5)(10)(.25) + 100 = 131.25 Question 87 a. Substituting yields , so b. Same argument as in a c. Suppose . Then , which implies that (a constant), so , which is of the form . Chapter 6: Question 12 . Question 13 Question 16 a. b. . Setting the derivative with respect to equal to 0 yields , from which ....
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