quiz9-key

quiz9-key - 3. (1 point) T F E( X ) = w f X ( w ) dw . 4....

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Quiz 9. April 6, 2011 Seat # _________ Name: ___ < KEY > ___ Closed book and notes. No calculator. From Problem 5--17, Montgomery and Runger, fourth edition. Consider the probability density function (pdf) f X , Y ( x , y ) = c x y for 0 x 3, 0 y 3 and zero elsewhere. 1. (2 points) Explain why c = 4 / 81. ____________________________________________________________ Because the area under the density must integrate to one. ____________________________________________________________ 2. (2 points) Determine the value of P( X 2, Y < 1). ____________________________________________________________ P( X 2, Y < 1) = 0 2 0 1 (4 / 81) x y dx dy ____________________________________________________________
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Unformatted text preview: 3. (1 point) T F E( X ) = w f X ( w ) dw . 4. (1 point) T F E( XY ) = vw f X , Y ( v , w ) dv dw . 5. (1 point) T F f X | Y = 4 (1) is undefined. 6. (1 point) T F f X | Y = 2 ( 1) = 0. 7. (1 point) T F f X | Y = 2 ( 1) = P( X = 1 | Y = 2). 8. (1 point) T F corr( X , Y ) = corr( Y , X ). ______________________________________________________________________ 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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