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Unformatted text preview: 2 ,b = 4 . The pdf is shown below: P ( X < 0) = Z2 1 / 6 d x = x 6 2 = 1 / 3 . (b) We use LOTUS E h  X  i = Z  X  f X ( x ) d x = 1 / 6 h Z2 (x ) d x + Z 4 x d x i = 1 / 6 hx 2 2 2 + x 2 2 4 i = 5 / 3 . (c) Y =  X  takes values in [0 , 4], hence F Y ( ) = 0 for < 0 and F Y ( ) = 1 for > 4. For 0 2, F Y ( ) = P { Y } = P { X } = / 3. For 2 4, F Y ( ) = P { Y } = P { X } = ( + 2) / 6. Hence, f Y ( ) = d F Y ( ) d = 1 / 3 , 2 , 1 / 6 , 2 < 4 , , else . Problem 4. (a) P ( X k ) = + X i = k i e i ! (b) Let P ( X k ) = P r , then P (all bikes are in use in on exactly 3 days) = 6 3 ! P 3 r 1P r 3...
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This note was uploaded on 09/08/2009 for the course GE 331 taught by Professor Negarkayavash during the Spring '09 term at University of Illinois at Urbana–Champaign.
 Spring '09
 NegarKayavash

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