Suggested Solution of Supplementary Exercise

Suggested Solution of Supplementary Exercise - THE...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1306 INTRODUCTORY STATISTICS (SEMESTER 1 2008/2009) Suggested Solution of Supplementary Exercise Question 1 (2.63) (a) The corresponding probability is 6 2 5 2 3 2 ± 4 6 5 = 15 10 3 4 6 6 6 6 6 = 25 108 : (b) The corresponding probability is 6 5 3 ± 5 4 6 5 = 6 10 5 4 6 6 6 6 6 = 25 162 : (c) The corresponding probability is 6 5 5 3 2 2 ± 6 5 = 6 5 10 1 6 6 6 6 6 = 25 648 : (d) The corresponding probability is 6 5 4 ± 5 6 5 = 6 5 5 6 6 6 6 6 = 25 1296 : Question 2 (2.64) Let M and S be the events of malpractice insurance and surgeons respectively for chosen doctor. Now, we have P ( M ) = 64 78 ;P ( S ) = 36 78 and P ( M \ S ) = 34 78 . Then, we want P ( M \ S ) = P ( M [ S ) = 1 ± P ( M [ S ) = 1 ± ( P ( M ) + P ( S ) ± P ( M \ S )) = 1 ± 64 78 + 36 78 ± 34 78 ± = 2 13 : 1
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Question 3 (3.92) In this problem, we have F Y ( y ) = 1 288 Z y 6 (36 u 2 ) du = 1 864 (6 + y ) 2 (12 y ) : (a) P ( Y ± 2) = 1 P ( Y < 2) = 1 F Y (2) = 1 1 864 (6 + 2) 2 (12 2) = 7 27 : (b) P ( Y 1) = F Y ( 1) = 1 864 (6 + ( 1)) 2 (12 ( 1)) = 325 864 : (c) P (1 ² Y ² 3) = F Y (3) F Y (1) = 1 864 (6 + 3) 2 (12 3) 1 864 (6 + 1) 2 (12 1) = 95 432 : (d) P ( Y = 5) = 0 : Question 4 (3.104) (a) Clearly, the possible values of Z is 0 or 1 while those of W are 0,1,2. Denote p ( z;w ) as the joint pmf of Z and W . Then, p (0 ; 0) = 48 52 ³ 47 51 = 188 221 ; p (0 ; 1) = 48 52 ³ 4 51 = 16 221 ; p (1 ; 1) = 4 52 ³ 48 51 = 16 221 ; p (1 ; 2) = 4 52 ³ 3 51 = 1 221 : Hence, the joint pmf of Z and W is summarized in the following table 2
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Z n W 0 1 2 0 188 = 221 16 = 221 0 1 0 16 = 221 1 = 221 (b) p z (0) = 188 221 + 16 221 + 0 = 12 13 and p z (1) = 0 + 16 221 + 1 221 = 1 13 . (c) Given Z = 1 , the conditional distribution of W is W 0 1 2 P ( W j Z = 1) 0 16 = 17 1 = 17 Question 5 (5.64) Let Y = number of applicants having college degrees. Then, Y Hyper ( N = 16 ;D = 10 ;n = 3) . (a) P ( Y = 0) = 10 0 16 ± 10 3 ± 16 3 ± = 1 28 : (b) P ( Y = 1) = 10 1 16 ± 10 2 ± 16 3 ± = 15 56 : (c) P ( Y = 2) = 10 2 16 ± 10 1 ± 16 3 ± = 27 56 : (d) P ( Y = 3) = 10 3 16 ± 10 0 ± 16 3 ± = 3 14 : Question 6 (5.75) Let Y = number of drivers involved in at least one car accident in any given year. Then, Y Bin ( n = 150 ;p = 0 : 04) which can be approximated by Y Po ( ) 3
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with = (150) (0 : 04) = 6 : (a) P ( Y = 5) 6 5 5! e
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This note was uploaded on 09/06/2010 for the course STAT STAT1306 taught by Professor Prof during the Fall '08 term at HKU.

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Suggested Solution of Supplementary Exercise - THE...

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