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Unformatted text preview: ISyE 2027 Probability with Applications Fall 2011 R. D. Foley Solution to Homework 7 November 11, 2011 1. Solution: (a) 1% × 60% + 2% × 20% + 3% × 20% = 1.6%. (b) 3% × 20% 1.6% = 0.006 0.016 = 0.375. 2. Solution: (a)No. (b) 99% × 1 1000000 99% × 1 1000000 + 5% × ( 1 1 1000000 ) = 1.98 × 10 5 . (c) 99% × 1 10 99% × 1 10 + 5% × 9 10 = 0.6875. (d)No. The prior probability that a person is HIV positive is very low, just like person A, according to part (b), the probability that a person is HIV positive given the test is positive is very low. So it doesn’t make sense to use this test to do mass screenings of every resident in the state of Georgia. 3. Solution: (a) Y has the p.d.f f ( y ) = fl 1 + y , y < 1 y , y > So, h = 1. (b)Pr { 1 2 < Y 6 1 2 } = R 1 2 1 2 f ( y ) dy = 3 4 . (c)E [ Y ] = R 1 1 yf ( y ) dy = R 1 y ( 1 + y ) dy + R 1 y ( 1 y ) dy = 0. (e) Var ( Y ) = E [ Y 2 ]( E [ Y ]) 2 = E [ Y 2 ] = R 1 1 y 2 f ( y ) dy = R 1 y 2 ( 1 + y ) dy + R 1 y 2 ( 1 y ) dy = 1 6 . (f)E [ Y E [ Y ]] = E [ Y ]  E [ Y ] = 0....
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