Math 3C - Hw#9

Math 3C - Hw#9 - = 1 2 p + 0 2 (1-p )-p 2 = p (1-p ) = 0 ....

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Math 3C Lec 2 DIS 2E/2F (Spring 2007) Ilhwan Jo 19. (a) Suppose S n is binomially distributed with parameters n = 200 and p = 0 . 3. Use the central limit theorem to find an approximation for P (99 S n 101) without the histogram correction. Solution Since S n is binomially distributed, S n = n i =1 X i where X k = 1 if the k -th trial is successful 0 otherwise . We have μ = EX i = 1 · p + 0 · (1 - p ) = p = 0 . 3 , σ 2 = var( X i ) = EX 2 i - ( EX i ) 2
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Unformatted text preview: = 1 2 p + 0 2 (1-p )-p 2 = p (1-p ) = 0 . 3 . 7 = 0 . 21 . By the central limit theorem, P (99 S n 101) = P 99-200 . 3 200 . 21 S n-n n 2 101-200 . 3 200 . 21 = P 6 . 02 S n-n n 2 6 . 33 F (6 . 33)-F (6 . 02) = 0 . 1...
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This document was uploaded on 02/02/2009.

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