Gabarito da Lista de Probabilidade

222,00225,50229,0010,0014,56232,50236,0015,0020,0019,1123,65239,43242,9325,0030,0028,2032,74246,43249,9335,0040,0037,2941,83253,43256,9345,0050,0046,3850,92260,43263,8655,0060,00267,36270,8655,4760,0165,0069,90274,36277,8664,5669,0174,9079,9073,5578,1084,90-4-3,589,9082,6487,19-3-2,5-2-1,5-4,00-3,50-3,0014,56-1-0,519,1123,65-2,50-2,00-00,490,9928,2032,74-1,50-1,00-0,501,491,9837,2941,832,482,980,000,5046,3850,923,483,981,0055,4760,011,501,9964,5669,012,492,9973,5578,103,4982,6487,193,990,50,40,40,30,30,20,20,10,10,0X1X2X3X40,50,40,40,30,30,20,20,10,10,0Lista de Exercícios - Modelos Probabilísticos 17Z3Z2Z1Z4XZP(Z>Z 4 ) = 0,1 P(Z>Z 3 ) = 0,3 P(Z>Z 2 ) = 0,7 P(Z>Z 1 ) = 0,9Procurando na tabela da distribuição normal padrão:Z 4 1,28, x 4 = 50 + 1,28 ×10 = 62,8 Z 3 0,53, x 3 = 50 + 0,53 ×10 = 55,3P(Z>Z 2 ) = 0,7 , P(Z>- Z 2 ) = 1 – 0,7 = 0,3 - Z 2 0,53 Z 2 -0,53, x 2 = 50 -0,53 ×10 = 44,7P(Z>Z 1 ) = 0,9, P(Z>- Z 1 ) = 1 – 0,9 = 0,1 - Z 1 1,28 Z 1 -1,28, x 1 = 50 -1,28 ×10 = 37,2As notas então serão 37,2, 44,7, 55,3 e 62,8.66) a) P(X>260). = 250 e = 7 , P(X>260) = P(Z>Z 1 ): Z 1 = (260 – 250)/7 = 1,43.0,50,40,40,30,30,20,20,10,10,00,50,40,40,30,30,20,20,10,10,0P(X1,43)= 0,0764XZb) P(Z50) = P(Z>Z1): Z1 = (50-50,92)/9,09 = -0,10.P(X>50) = P(Z>-0,10) = 1-P(Z>0,1) = 1-0,4602 = 0,5398 > 0,370 => candidato reprovado.P(X>60) = P(Z>Z2): Z2 = (60-50,92)/9,09 = 1,0. P(X>60) = P(Z>1) = 0,1587 < 0,370 => candidato aprovado.0,50,40,40,30,30,20,20,10,10,00,50,40,40,30,30,20,20,10,10,0P(X>50)XXP(X>60)