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Unformatted text preview: Solution to homework 2 1. a) The exact value of the probability is P ( X ≥ 2) = 1 P ( X ≤ 1) = 1 P ( X = 0) P ( X = 1) = 1 (1 p ) 20 20 p (1 p ) 19 = 0 . 264 . Using normal approximation, we have to set μ = np = 1 and σ = q np (1 p ) = 0 . 975. Then we obtain P ( X ≥ 2) ≈ F (20 . 5) F (1 . 5) = 0 . 304 . We see that the approximation is not very good. The reason is that the value of np is low, and the normal approximation is not really justified. b) The exact value is P ( X ≥ 20) = 1 P ( X ≤ 19) = 1 19 X i =0 P ( X = i ) = 0 . 00266 . Using normal approximation, we have to set μ = np = 10 and σ = q np (1 p ) = 3 . 082. So P ( X ≥ 20) ≈ F (200 . 5) F (19 . 5) = 0 . 00103 . We see that, in spite of np being equal to 10 (and n (1 p ) nearly 200) the precision of the normal approximation is pretty bad. The reason for this is that we are trying to approximate a very small probability....
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This note was uploaded on 03/17/2008 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .
 Spring '08
 Perevalov

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