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Math 3C - Hw#9

# Math 3C - Hw#9 - = 1 2 Â p 0 2 Â(1-p-p 2 = p(1-p = 0 3 Â 7...

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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 - (
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Unformatted text preview: = 1 2 Â· p + 0 2 Â· (1-p )-p 2 = p (1-p ) = 0 . 3 Â· . 7 = 0 . 21 . By the central limit theorem, P (99 â‰¤ S n â‰¤ 101) = P Â± 99-200 Â· . 3 âˆš 200 Â· . 21 â‰¤ S n-nÎ¼ âˆš nÏƒ 2 â‰¤ 101-200 Â· . 3 âˆš 200 Â· . 21 Â¶ = P Â± 6 . 02 â‰¤ S n-nÎ¼ âˆš nÏƒ 2 â‰¤ 6 . 33 Â¶ â‰ˆ F (6 . 33)-F (6 . 02) = 0 . Â¡ 1...
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