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quiz8-key

# quiz8-key - E X =(1(0.1(1.5(0.3(1.5(0.2(2.5(0.15(3(0.25 =...

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Quiz 8. March 30, 2011 Seat # _________ Name: ___ < KEY > ___ Closed book and notes. No calculator. From Problem 5--1, Montgomery and Runger, fourth edition. Consider the probability mass function in the following table. x y f X , Y ( x , y ) 1 1 0.1 1.5 2 0.3 1.5 3 0.2 2.5 4 0.15 3 5 0.25 1. (2 points) Determine the value of F X , Y (2.2, 3.5) = P( X 2.2, Y 3.5). ____________________________________________________________ F X , Y (2.2, 3.5) = P( X 2.2, Y 3.5) = P( X = 1, Y = 1) + P( X = 1.5, Y = 2) + P( X = 1.5, Y = 3) = 0.1 + 0.3 + 0.2 = 0.6 ____________________________________________________________ 2. (2 points) Determine the value of F X (2.5) = P( X 2.5). ____________________________________________________________ F X (2.5) = P( X 2.5) = P( X = 1, Y = 1) + P( X = 1.5, Y = 2) + P( X = 1.5, Y = 3) + P( X = 2.5, Y = 4) = 0.1 + 0.3 + 0.2 + 0.15 = 0.75 ____________________________________________________________ 3. (2 points) Determine the value of E( X ). ____________________________________________________________ E( X ) = (1)(0.1)
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Unformatted text preview: E( X ) = (1)(0.1) + (1.5)(0.3) + (1.5)(0.2) + (2.5)(0.15) + (3)(0.25) = 1.975 ← ____________________________________________________________ 4. (2 points) Determine the value of f Y | X = 1.5 (1) = P( Y = 1 | X = 1.5). ____________________________________________________________ f Y | X = 1.5 (1) = P( Y = 1 | X = 1.5) = P( X = 1.5, Y = 1) / P( X = 1.5) = P( X = 1.5, Y = 1) / [(P( X = 1.5, Y = 2) + P( X = 1.5, Y = 3)] = 0.0 / (0.3 + 0.2) = 0.0 ← ____________________________________________________________ 5. (2 points) T F ← X and Y are independent. ______________________________________________________________________ Write on the back any concerns about the weekly quizzes or the course in general. IE 230 – Page 1 of 1 – Schmeiser...
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