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Unformatted text preview: STATISTICS 116 MIDTERM EXAM SOLUTIONS Thursday May 10, 2007 Name: Student ID: Instructions: 1. Print your name and student ID number. 2. There are six problems, each worth 10 marks each. 3. You must show all your work to get full credit. 4. If you get stuck on a problem, move onto another one. 5. Good luck! 1 1. Suppose ( X,Y ) has the following joint probability density function: f X,Y ( x,y ) = k ( x y ) , y x 1 (a) Find k . (b) Find the marginal probability density functions f X ( x ) and f Y ( y ). (c) Find the conditional probability density functions f X  Y ( x  y ) and f Y  X ( y  x ). (d) Find E ( X ) and E ( Y ). Solution (a) We know the joint density has to integrate to one. Z 1 Z 1 y k ( x y ) dxdy = k Z 1 ( x 2 / 2 yx )  1 x = y dy = k Z 1 y 2 / 2 y + 1 / 2 dy = k ( y 3 / 6 y 2 / 2 + y/ 2)  1 y =0 = k/ 6 Hence k = 6 in order for the integral to be 1. (b) To get the marginal density, we integrate the joint density with respect to the other variable. f X ( x ) = Z x 6( x y ) dy = 6( xy y 2 / 2)  x y =0 = 6( x 2 x 2 / 2) = 3 x 2 for 0 x 1, and 0 otherwise f Y ( y ) = Z 1 y 6( x y ) dx = 6( x 2 / 2 yx )  1 x = y = 3 6 y + 3 y 2 for 0 y 1, and 0 otherwise (c) f X  Y ( x  y ) = f X,Y ( x,y ) f Y ( y ) = 6( x y ) 3 6 y + 3 y 2 = 2( x y ) ( y 1) 2 f Y  X ( y  x ) = f X,Y ( x,y ) f X ( x ) = 6( x y ) 3 x 2 = 2( x y ) x 2 (d) E ( X ) = Z 1 xf X ( x ) dx = Z 1 x 3 x 2 dx = 3 x 4 / 4  1 x =0 = 3 / 4 E ( Y ) = Z 1 yf Y ( y ) dy = Z 1 y (3 6 y + 3 y 2 ) dy = (3 y 2 / 2 6 y 3 / 3 + 3 y 4 / 4)  1 y =0 = 1 / 4 2 2. Your nextdoor neighbor has a rather old and temperamental burglar alarm. If someone breaks into his house, the probability of the alarm sounding is 0 . 95. In the last 2 years, though, it has gone off on 5 different nights, each time for no apparent reason. Police records show that the chances of a particular home being burglarized in your community on any given night are 2 in 10...
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 Spring '07
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 Statistics, Probability

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