estatistica 6 bussab - bussab&morettin...

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Unformatted text preview: bussab&morettin estatstica bsica Captulo 6 Problema 01. 56 3! ! 5 ! 8 3 8 ) ( = = = n combinaes possveis 1 3 3 5 = = X 15 2 3 1 5 1 = = X 30 1 3 2 5 2 = = X 10 3 3 5 3 = = X Ento a distribuio de X dada por: X 1 2 3 P(X=x) 56 1 56 15 56 30 56 10 Problema 02. 512 8 ) ( 3 = = n combinaes possveis 27 3 5 3 = = X 135 3 5 1 3 1 2 1 = = X 225 3 5 2 3 2 1 2 = = X 125 3 5 3 3 3 3 = = X X 1 2 3 P(X=x) 512 27 512 135 512 225 512 125 Problema 03. 2 1 1 = C X 4 1 2 1 2 2 = = RC X 8 1 2 1 3 3 = = RRC X X 1 2 3 4 ..... P(X=x) 2 1 4 1 8 1 16 1 .....- cap.6 pg. 1 -- bussab&morettin estatstica bsica De modo geral, x x x X P = = =- 2 1 2 1 2 1 ) ( 1 , x=1,2,3.... Problema 04. Seguindo o mesmo raciocnio idntico ao Problema 02, tem-se: X 1 2 3 4 P(X=x) 16 1 16 4 16 6 16 4 16 1 Problema 05. No contexto apresentado, a distribuio do nmero de caras dada por: ( 29 4. 3, 2, 1, 0, y , 1 4 ) ( 4 =- = =- y y p p y y Y P Problema 06. Por similaridade, tem-se: ( 29 n. 3,..., 2, 1, 0, y , 1 ) ( =- = =- y n y p p y n y Y P Problema 07. Para o Problema 01, tem-se: 875 , 1 56 105 56 30 56 60 56 15 E(X) = = + + = 018 , 4 56 225 56 90 56 120 56 15 ) E(X 2 = = + + = [ ] [ ] 502 , 875 , 1 018 , 4 E(X)- ) E(X Var(X) 2 2 2 =- = = Para o Problema 02, tem-se: 875 , 1 512 960 512 375 512 450 512 135 E(X) = = + + = 219 , 4 512 2160 512 1175 512 900 512 135 ) E(X 2 = = + + = [ ] [ ] 703 , 875 , 1 219 , 4 E(X)- ) E(X Var(X) 2 2 2 =- = = Problema 08. , 2 16 4 16 12 16 12 16 4 E(Y) = + + + = , 5 16 16 16 36 16 24 16 4 ) E(Y 2 = + + + = [ ] [ ] , 1 , 2 , 5 E(X)- ) E(X Var(X) 2 2 2 =- = = Problema 09. Y=3X 3 6 9 P(Y=y) 56 1 56 15 56 30 56 10- cap.6 pg. 2 -- bussab&morettin estatstica bsica Z=X 2 1 4 9 P(Z=z) 56 1 56 15 56 30 56 10 Problema 10. RRR RRC RCR CRR RCC CRC CCR CCC X 1 1 1 2 2 2 3 Y 1 2 3 2 2 3 2 1 p 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 Do quadro acima obtm-se: X 1 2 3 P(X=x) 8 1 8 3 8 3 8 1 5 , 1 8 1 3 8 3 2 8 3 1 8 1 E(X) = + + + = ( 29 ( 29 ( 29 ( 29 75 , 8 1 5 , 1 8 3 5 , 8 3 5 , 8 1 5 , 1 Var(X) 2 2 2 2 = + + - + - = Y 1 2 3 P(Y=y) 8 2 8 4 8 2 2 8 2 3 8 4 2 8 2 1 E(X) = + + = ( 29 ( 29 ( 29 50 , 8 2 1 8 4 8 2 1 Var(X) 2 2 2 = + + - = Problema 11....
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This note was uploaded on 12/05/2011 for the course MATH 123 taught by Professor Gg during the Spring '11 term at Universidade Federal de Minas Gerais.

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estatistica 6 bussab - bussab&morettin...

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