HW5-Solutions

# HW5-Solutions - 69 According to the problem description X...

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Unformatted text preview: 69. According to the problem description, X is hypergeometric with n = 6, N: 12, and 1M: 7. P(X:4) : W_ 350 {162] 924 (213121; 1151 [1?) 1M 7 1276 7 7 b. E :n- :6-—:3.5;V : — 6 — 1—— :0.795;0:0.892.So. (X) . (X) [12—1) [121 12] ixr>[email protected]:Pa¢45+08%):HX>4sm):HX:5méyauiammpmuo fl. : .379 . P(X: 4) : 1 —P(X>> 4): 1 —[P(X: 5) + P(X: 6)] : i — 1 [.114 + .007] = l 7.121 = .879. c. W e can approximate the hypergeometric distribution with the binomial if the population size and the number of successes are large. Here. n : 15 and ELI/N: 40.1400 : .1, so h(x;15, 40, 400) 1: b(x;15, .10). Using this approximation. P(XS 5) 2 3(5; 15. .10) I .998 from the binomial tables. (This agrees with the exact answer to 3 decimal places.) 73. a. The successes here are the top M = 10 pairs. and a sample of n = 10 pairs is drawn from among the N [’10‘*‘[’ 10 . . . x _. 10 —.r, I 20. The probability is therefore Mr: 10. 10. 20) I . 1 101 1). Let X: the number among the top 5 who play east-west. (Now. M = 5.) Then P( all of top 5 play the same direction) = P(X: 5) 7 P(X= 0) = '5‘. '15‘ “'5‘. '15) 51 5 l . 01 101 M5: 10. 5. 20)—l?(5; 10. 5. 20) = ‘ l ‘ ’+' l ‘ ._ 2.033. 0‘ [ 20 \ 10 , \ 10 75. Let X : the number of boxes that do M contain a prize until you ﬁnd 2 prizes. ThenX ‘- NB(2, .2). a. x + 2 *1 P(X: x) : nb(x; 2, .2) : [ ](.2)2(1— .2)‘ : (x + 1)(.2)2(.8)~‘. 271 b. P(4 boxes purchased) =P(2 boxes Without prizes) I P(X= 2) = nb(2_; 2, .2) = (2 + '1)(.2)2(.8)2 = .0768. c. P(at most 4 boxes purchased) = P(X: 2) = an(x;2,.8) = .04 + .064 + .0768 = .1808. x:0 1'(l*p) i 2(172) d. E : (X) p .2 : 8. The total number of boxes you expect to buy is 8 + 2 : 10. 89. a. Jet = 8 when 1‘: 1. so P(X= 6) = —86 6' = .122: FLY: 6)=1—F{5: 8):.809La11d P(X210)=1—F(9;8)=.283. b. t = 90 111111 = 1.5 hours. so IN = 12; thus the expected number of arrivals is 12 and the standard deviation is a =J1_2=3.464. 111 this example. a = rate of occmrence = 11"(111ea11 time between occmrences) = 1.=".5 = 2. a. For a two-year period. 111‘ = or = (2)0) = 4 loads. b. Apply a Poisson model with ,1: = 4: P(X> 5)=1—P(X:2 5) = 1 —F(5; 4) = 1 — \$85 = .215. c. For a. = 2 and the value of 1‘ unknown. ano loads occur during the period of length I) = eel: DO — 0‘ 9—31. Solve for 1: 6—215 .1 :> —2t£ 111(1) :> 133 1.1513 years. P(X= 0) = ...
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