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Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 4.4 60. The pdf for X is f ( x ) = 0 . 01386 e . 01386 x , x > and zero otherwise. It will be convenient to find the cdf once and use it to answer all questions. F ( x ) = Z x . 01386 e . 01386 u du = 1 e . 01386 x . (This holds for x > 0, for x < 0, F ( x ) = 0.) a. P ( X 100) = F (100) = 1 e . 01386(100) . 7499 P ( X 200) = F (200) = 1 e . 01386(200) . 9375 P (100 < X < 200) = F (200) F (100) . 9375 . 7499 = 0 . 1875 (I used unrounded values to compute 0 . 1875. It is okay to give the answer as 0 . 1876.) b. The mean for an exponential random variable with rate is 1 / . The variance is 1 / 2 . So, in this case, = 1 = 1 . 01386 72 . 15 1 = 1 2 = 1 . 01386 2 5205 . 64 We want P ( X > +2 ) = P ( X > 216 . 45) = 1 F (216 . 45) = 1 h 1 e . 01386(216 . 45) i . 05 66. a. We know that for X ( , ), E [ X ] = and V ar [ X ] = 2 . So, we equate = 20...
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
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