μ σ μ σ 081 P Y P Y 10 6 348 2 31 10 6 348 2 31 P Z P Z 1 58 1 1 58 1 0 719 0

Μ σ μ σ 081 p y p y 10 6 348 2 31 10 6 348 2 31 p

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= = μ σ μ σ 081 P Y P Y ( ) , , , , ' ' > = > 10 6 348 2 31 10 6 348 2 31 = > ( ) = ( ) = = = P Z P Z 1 58 1 1 58 1 0 719 0 281 , , , , np n p Y B = > = > 6 348 5 1 33 652 5 40 0 1 , ( ) , ( ; , 587 6 348 2 31 ) ( , ; , ) N e) Y B ' ( ; , ) 40 0 1587 Distribuciones binomial y normal
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637 El sueldo mínimo de los directivos es de 2.810,29 euros. PARA FINALIZAR… Calcula el valor de k para que la siguiente función sea una función de densidad de una variable aleatoria continua. a) Obtén la función de distribución F ( x ). b) Representa gráficamente ambas funciones. c) Calcula las probabilidades. P ( X = 1) = 0 c) P X P X P X 1 2 3 2 3 2 1 < < = < < 2 3 2 1 2 = F F = = 39 48 15 48 1 2 1 f ( x ) 1 X Y 1 F ( x ) 1 X Y b) F x x x x x x ( ) = < < + < < + 0 0 12 2 3 0 2 1 2 2 si si si a) 1 1 6 1 2 0 2 = = + = − + f x dx x k dx x ( ) 2 1 3 2 2 4 3 2 3 0 2 + =− + = = kx k k k P X P X 1 2 3 2 1 < < = ( ) f x x k x x ( ) [ , ] [ , ] = + 1 6 0 2 0 0 2 si si 083 b) P X a P X a ( ) , . , . , = 0 05 1 500 794 12 1 500 794 12 = = P Z a 1 500 794 12 0 0 . , , 5 1 500 794 12 0 95 1 500 P Z a a < = . , , . 794 12 1 65 2 810 29 , , . , = = a a) P X P X ( . ) . , . . , > = > 1 600 1 500 794 12 1 600 1 500 794 12 0 13 1 0 13 1 0 5 = > ( ) = − = = − P Z P Z , ( , ) , 517 0 4483 = , P X P X ( ) , . . > = > 960 0 75 1 500 960 1 500 σ σ = > − = = < P Z P Z 540 540 σ σ = = = 0 75 540 0 68 , , σ σ 794,12 14 SOLUCIONARIO
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638 La probabilidad de que un reloj sea defectuoso es del 4 %. Halla. a) El número de relojes defectuosos que se estima en un lote de 1.000. b) La probabilidad de menos de 10 defectuosos. a) μ = 1.000 0,04 = 40 relojes b) B (1.000; 0,04) N (40; 6,19) En una distribución normal, el 3 % de los valores es inferior a 19 y el 5 % es superior a 28,6. Calcula P ( X < 18). Las bolas para rodamiento se someten a un control de calidad consistente en eliminar las que pasan por un orificio de diámetro d y, también, las que no pasan por otro orificio de diámetro D , con d < D . Calcula la probabilidad de eliminar una bola, sabiendo que la medida de sus diámetros sigue una distribución normal de parámetros: . P X d P X D P X D d D d d D d D d ( ) ( ) ( ) ( < + > = + < + 2 2 0,3 0,3 ) ( ) + + > + P X D d D d D D 2 0,3 d D d P Z d D 2 0 3 2 , ( ) = = < 0,3 ( ) ( ) D d P Z D d D d + >
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