Section 5.4 Recitation Problems

Section 5.4 Recitation Problems - ASW ESBE (5e) Chapter 5...

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ASW ESBE (5e) Chapter 5 Solutions to Selected Problems Section 5.4 for the binomial distribution: Problems 28, 29, 31, 33, 34, 35. 28. What kind of problem is this? Your choices are either binomial or Poisson. International travelers either stick with their tour group or they don’t. One or the other, can’t be both, and no third choice. So this is success vs. failure problem. For all parts of this problem, the “success” is selecting an international traveler who sticks with the tour group. (Looking at what is asked in the questions.) So this is a binomial problem. Let X be the r.v. that counts the number of international travelers who stick with their tour group. Given: p = 0.23, because 23% of all of all international travelers stick with their tour group (from last sentence of paragraph). Binomial with p = 0.23; n and x are given in the questions. (Note if p = 0.23, then 1 – p = 1 – 0.23 = 0.77) Then Prob ( X = x | n , p ) is calculated by ( 29 ( 29 x n x . . ! x n ! x ! n . p , n | x prob - × × - = = 77 0 23 0 23 0 a. “In a sample of six” means n = 6. “… that two will stick …” means x = 2 Want Prob ( x =2 | n =6, p =0.23). Some Section 5.4 Problem Solutions 1
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So Prob ( x =2 | n =6, p =0.23) = 0.2789. b. n = 6. Want Prob (at least two of the n = 6). “at least two” means two or more. In terms of x , “at least two” means x ≥ 2, so x = 2, 3, 4, 5, 6. So want Prob ( x ≥2 | n =6, p =0.23) The only values not in this list are x < 2, specifically, x = 0 and 1. This is a shorter list of values, meaning fewer binomial calculations. So using the concept of complements: Prob ( x ≥2 | n =6, p =0.23) = 1 – Prob ( x < 2 | n =6, p =0.23) = 1 – [ Prob ( x = 0 | n =6, p =0.23) + Prob ( x = 1 | n = 6, p =0.23)] Calculating for x = 0 Some Section 5.4 Problem Solutions 2 ( 29 ( 29 0.278939 0.35153041 0.0529 15 77 0 23 0 15 77 0 23 0 1 4 1 2 1 4 5 6 77 0 23 0 4 2 6 23 0 23 0 2 6 2 6 23 0 6 2 4 2 4 2 4 2 2 6 2 = × × = × × = × × × × × × × × × × = × × = × × - = = = = - . . . . . . ! ! ! . . ! ! ! . p , n | x prob ( 29 ( 29 0.208422 77 0 23 0 1 77 0 1 1 77 0 23 0 6 1 6 23 0 23 0 0 6 0 6 23 0 6 0 4 2 6 6 0 0 6 0 = × × = × × = × × × = × × - = = = = - . . . . . ! ! . . ! ! ! . p , n | x prob
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and calculating for x = 1 So Prob ( x = 0) + Prob ( x = 1) = 0.2084 + 0.3735 = 0.5819 and So Prob ( x ≥2 | n =6, p =0.23) = 1 – 0.5819 = 0.4181 c. n = 10. Want Prob (none) “none” means x = 0. Want Prob ( x =0 | n =10, p =0.23) So Prob ( x =0 | n =10, p =0.23) = 0.0733. Additional questions. 1. In a sample of 10, how many are expected? Want E ( x ) Some Section 5.4 Problem Solutions 3 ( 29 ( 29 0.373536 0.270678 23 0 6 77 0 23 0 6 77 0 23 0 5 1 6 23 0 23 0 1 6 1 6 23 0 6 1 5 5 1 1 6 1 = × × = × × = × × × = × × - = = = = - . . . . . ! ! . . ! ! ! . p
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This note was uploaded on 02/19/2011 for the course OM 210 taught by Professor Singer during the Spring '08 term at George Mason.

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Section 5.4 Recitation Problems - ASW ESBE (5e) Chapter 5...

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