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sample_quiz4_solution

# sample_quiz4_solution - h d f k 61 44 33 18 Thus n = 61 44...

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Math 187 Prof. Garsia typed by Alex Brik 05-04-2010 SAMPLE QUIZ 4 solution. 1. (3 5 4 1 6 2) 2. a) ° 2 = (171 ° 1000 = 6) 2 1000 = 6 + (186 ° 1000 = 6) 2 1000 = 6 + (174 ° 1000 = 6) 2 1000 = 6 + (170 ° 1000 = 6) 2 1000 = 6 + + (192 ° 1000 = 6) 2 1000 = 6 + (107 ° 1000 = 6) 2 1000 = 6 ± 27 : 95 b) There are 5 degrees of freedom. Thus the probability that the fair die would produce a chi-square value larger than 27.95 is 0.001%. 3. Recall p ± N ² 0 : 027 ( N ° 1) I c + 1 ° N ² 0 : 038 I c = Z X ° = A N ° ( N ° ° 1) N ( N ° 1) : Then I c = 1 N ( N ° 1) Z X ° = A N 2 ° ° Z X ° = A N ° ! = Z X ° = A N 2 ° N ( N ° 1) ° 1 N ° 1 : Now N = 1000 ; Z X ° = A N ° = 40000 : Thus I c = 40000 1000 ² 999 ° 1 999 = 39 999 p ± 1000 ² 0 : 027 999 ² 39 = 999 + 1 ° 1000 ² 0 : 038 = 27 2 = 13 : 5 : Thus the period is approximately 13 or 14 . 4. We will °rst compute chi-square statistic assuming that "hdfkf" corresponds to "where". Using the table given we obtain the following probabilities for the english letters 1

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letters w h e r frequency 16 35 130 77 probability 16/258 35/258 130/258 77/258 The ciphertext frequencies of the letters in "hdfkf" are
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Unformatted text preview: h d f k 61 44 33 18 . Thus n = 61 + 44 + 33 + 18 = 156 : Then & 2 = & 61 & 16 258 ± 156 ± 2 16 258 ± 156 + & 44 & 35 258 ± 156 ± 2 35 258 ± 156 + & 33 & 130 258 ± 156 ± 2 130 258 ± 156 + & 18 & 77 258 ± 156 ± 2 77 258 ± 156 ² 341 Assuming that "pdlhl" corresponds to "where": the ciphertext frequencies of the letters in "pdlhl" are p d l h 18 44 80 61 n = 18 + 44 + 80 + 61 = 203 : & 2 = & 18 & 16 258 ± 203 ± 2 16 258 ± 203 + & 44 & 35 258 ± 203 ± 2 35 258 ± 203 + & 80 & 130 258 ± 203 ± 2 130 258 ± 203 + & 61 & 77 258 ± 203 ± 2 77 258 ± 203 ² 17 Thus "pdlhl" is the image of "where" . 2...
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sample_quiz4_solution - h d f k 61 44 33 18 Thus n = 61 44...

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