a El espacio muestral es E 1 C 2 C 3 C 4 C 5 C 6 C 1 X 2 X 3 X 4 X 5 X 6 X La

A el espacio muestral es e 1 c 2 c 3 c 4 c 5 c 6 c 1

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a) El espacio muestral es: E = {(1, C ), (2, C ), (3, C ), (4, C ), (5, C ), (6, C ), (1, X ), (2, X ), (3, X ), (4, X ), (5, X ), (6, X )} La probabilidad de cada suceso elemental es . 1 12 002 0,18 0,16 0,14 0,12 0,1 0,08 0,06 0,04 0,02 2 3 4 5 6 7 8 9 10 11 12 X P ( X = x i ) P ( X x i ) 2 1 36 1 36 3 1 18 1 12 4 1 12 1 6 5 1 9 5 18 6 5 36 5 12 7 1 6 7 12 8 5 36 13 18 9 1 9 5 6 10 1 12 11 12 11 1 18 1 12 12 1 36 1 X (1, 1) = 2 X (1, 2) = 3 X (1, 3) = 4 X (1, 4) = 5 X (1, 5) = 6 X (1, 6) = 7 X (2, 1) = 3 X (2, 2) = 4 X (2, 3) = 5 X (2, 4) = 6 X (2, 5) = 7 X (2, 6) = 8 X (3, 1) = 4 X (3, 2) = 5 X (3, 3) = 6 X (3, 4) = 7 X (3, 5) = 8 X (3, 6) = 9 X (4, 1) = 5 X (4, 2) = 6 X (4, 3) = 7 X (4, 4) = 8 X (4, 5) = 9 X (4, 6) = 10 X (5, 1) = 6 X (5, 2) = 7 X (5, 3) = 8 X (5, 4) = 9 X (5, 5) = 10 X (5, 6) = 11 X (6, 1) = 7 X (6, 2) = 8 X (6, 3) = 9 X (6, 4) = 10 X (6, 5) = 11 X (6, 6) = 12 001 Distribuciones binomial y normal
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603 b) Respuesta abierta. La función X asigna a cada suceso el número obtenido en el dado. La función Y asigna a cada suceso el número elemental 1 si sale cara en la moneda y 2 si sale cruz. Consideramos la variable aleatoria que cuenta la suma de las puntuaciones al lanzar dos dados de 6 caras. Calcula los parámetros de esta variable aleatoria. Media: μ = 7 Desviación típica: ¿Puedes encontrar una variable aleatoria discreta que proceda de una variable estadística continua? ¿Y lo contrario? Consideramos la variable estadística cuantitativa continua «altura de las personas de un país, medida en metros». Definimos sobre esta variable estadística la variable aleatoria: Para cada altura Esta variable está definida para cualquier suceso elemental de la variable estadística, es decir, cada una de las alturas; además, es discreta, pues solo toma dos valores. Por tanto, de una variable estadística continua se puede obtener una variable aleatoria discreta, pero no a la inversa, pues un número finito de valores no puede tener un número infinito de imágenes. h X h h h ( ) = > 0 1 1 1 si si 004 σ = = 5,852 2,419 003 0,5 0,1 2 1 Y P ( Y = y i ) P ( Y y i ) 1 1 2 1 2 2 1 2 1 Y (1, C ) = 1 Y (2, C ) = 1 Y (3, C ) = 1 Y (4, C ) = 1 Y (5, C ) = 1 Y (6, C ) = 1 Y (1, X ) = 2 Y (2, X ) = 2 Y (3, X ) = 2 Y (4, X ) = 2 Y (5, X ) = 2 Y (6, X ) = 2 0,16 0,14 0,12 0,1 0,08 0,06 0,04 0,02 2 3 4 5 6 1 0,18 X P ( X = x i ) P ( X x i ) 1 1 6 1 6 2 1 6 1 3 3 1 6 1 2 4 1 6 2 3 5 1 6 5 6 6 1 6 1 X (1, C ) = 1 X (2, C ) = 2 X (3, C ) = 3 X (4, C ) = 4 X (5, C ) = 5 X (6, C ) = 6 X (1, X ) = 1 X (2, X ) = 2 X (3, X ) = 3 X (4, X ) = 4 X (5, X ) = 5 X (6, X ) = 6 14 SOLUCIONARIO
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604 En el experimento aleatorio que consiste en lanzar dos dados de 6 caras, consideramos la variable aleatoria X , que asocia a cada suceso elemental el producto de las puntuaciones que se ven. Halla y representa las funciones de probabilidad y de distribución. La función de probabilidad es: f x x x ( ) , , , , , , , , , = = = 1 36 1 9 16 25 36 1 18 2 3 5 8 10 1 si si 5 18 20 24 30 1 12 4 1 9 6 12 0 , , , , , si si en el resto x x = = X P ( X = x i ) P ( X x i ) 10 1 18 19 36 12 1 9 23 36 15 1 18 25 36 16 1 36 13 18 18 1 18 7 9 20 1 18 5 6 24 1 18 8 9 25 1 36 11 12 30 1 18 35 36 36 1 36 1 X P ( X = x i ) P ( X x i ) 1 1 36 1 36 2 1 18 1 12 3 1 18 5 36 4 1 12 2 9 5 1 18 5 18 6 1 9 7 18 8 1 18 4 9 9 1 36 17 36 X (1, 1) = 1 X (1, 2) = 2 X (1, 3) = 3 X (1, 4) = 4 X (1, 5) = 5 X (1, 6) = 6 X (2, 1) = 2 X (2, 2) = 4 X (2, 3) = 6 X (2, 4) = 8 X (2, 5) = 10 X (2, 6) = 12 X (3, 1) = 3 X (3, 2) = 6 X (3, 3) = 9 X (3, 4) = 12 X (3, 5) = 15 X (3, 6) = 18 X (4, 1) = 4 X (4, 2) = 8 X (4, 3) = 12 X
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