sec_3_4_and_3_5_sol

# Sec_3_4_and_3_5_sol - APPM 4/5570 Solutions to Problems from Sections 3.4 and 3.5 46 b(x n p is notation for P(X = x when X bin(n p P(X = x = n x

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APPM 4/5570 Solutions to Problems from Sections 3.4 and 3.5 46. b ( x ; n, p ) is notation for P ( X = x ) when X bin ( n, p ) P ( X = x ) = n x ! p x (1 - p ) n - x B ( x ; n, p ) is notation for P ( X x ) which involves summing up appropriate values of b ( x ; n, p ). a. P ( X = 3) = 8 3 ! (0 . 35) 3 (1 - 0 . 35) 8 - 3 0 . 2786 b. P ( X = 5) = 8 5 ! (0 . 6) 5 (1 - 0 . 6) 8 - 5 0 . 2787 c. P (3 X 5) = P ( X = 3) + P ( X = 4) + P ( X = 5) = 7 3 ! (0 . 6) 3 (1 - 0 . 6) 7 - 3 + 7 4 ! (0 . 6) 4 (1 - 0 . 6) 7 - 4 + 7 5 ! (0 . 6) 5 (1 - 0 . 6) 7 - 5 0 . 7451 d. P (1 X ) = P ( X 1) = 1 - P ( X = 0) = 1 - 9 0 ! (0 . 1) 0 (1 - 0 . 1) 9 - 0 = 1 - (0 . 9) 9 0 . 6126 49. For a randomly selected goblet, let a “success” be that it is “a second”. The the probability of success when we look at any one goblet is p = 0 . 10. Let X be the number of “seconds” in a random sample of 6 goblets. Then X bin (6 , 0 . 10)

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a. P ( X = 1) = 6 1 ! (0 . 10) 1 (0 . 90) 6 - 1 = 0 . 354294 b. P
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## This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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Sec_3_4_and_3_5_sol - APPM 4/5570 Solutions to Problems from Sections 3.4 and 3.5 46 b(x n p is notation for P(X = x when X bin(n p P(X = x = n x

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