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Sections 5.4 %26 5.5 Recitation Problems (1)

# Sections 5.4 %26 5.5 Recitation Problems (1) - Solutions...

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Solutions for Recitation Problems from Chapter 5 Section 5.4 for the binomial distribution: Problems 28, 29, 31, 33 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 stick 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 = 6 Want Prob ( x =2 | n =6, p =0.23). Chap 5 recitation problems 1 © Harvey Singer 2008

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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 Chap 5 recitation problems 2 © Harvey Singer 2008 ( 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
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 ) Chap 5 recitation problems 3 © Harvey Singer 2008 ( 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 , n | x prob ( 29 ( 29 0.07326680 77 0 1 1 77 0 23 0 10 1 10 77 0 23 0 0 10 0 10 23 0 10 0 10 10 0 0 10 0 = × × = × × × = × × - = = = = - . . . ! ! . . ! ! ! . p , n | x prob

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E ( x ) = np = 10×0.23 = 2.3 So in a sample of 10 travelers, expect 2.3 to stick with the tour group. 2. What is the chance of fewer than half as many as expected? Want Prob (fewer than half as many as expected) By “half as many as expected” mean half of E ( x ) = 2.3 half as many as expected = 0.5×2.3 = 1.15. So Prob (fewer than half as many as expected) = Prob ( x <1.15) But the only integer values of x that are less than 1.15 are less than 1, specifically the two values 0 and 1.
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Sections 5.4 %26 5.5 Recitation Problems (1) - Solutions...

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