Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: CADENA DE MARKOV ESTABLE DE DIMENSIÓN 3. EJERCICIO DE APLICACIÓN PROBLEMA DE LOS ESTADOS DE UN PACIENTE DE UN HOSPITAL En un hospital, cada paciente es clasificado de acuerdo a tres estados: crítico (C), serio (S) o estable (E). Estas clasificaciones son actualizadas cada mañana por un médico internista, teniendo en cuenta la evaluación experimentada por cada paciente. Las probabilidades con las que cada paciente se mueve de un estado a otro se resume mediante la siguiente matriz de transición: M = [[0.84,0.054,0.11],[0.82,0.099,0.08],[0.97,0.026,0.0037]] Esta situación se puede modelar con el siguiente grafo: Las componentes de la matriz M se corresponden con el grafo de la siguiente manera: a11 = P(C/C)=0.84 a12 = P(S/C)=0.054 a13 = P(E/C)=0.11 a21 = P(C/S)=0.82 a22= P(S/S)= 0.099 a23 = P(E/S)=0.08 a31 = P(C/E)=0.97 a31=P(S/E)= 0.026 a33=P(E/E)=0.0037 Calcula: a) La potencia 5 de la matriz M es [M to the power of # n end exponent]={[[0.85,0.053,0.098],[0.85,0.053,0.098],[0.85,0.053,0.098]]; [[0.85,0.053,0.095],[0.85,0.053,0.095],[0.85,0.053,0.095]]; [[0.86,0.0041,0.14],[0.86,0.0041,0.14],[0.86,0.0041,0.14]]} b) El vector de estabilidad es W = {{x=0.85,y=0.053,z=0.098}; {x=0.85,y=0.053,z=0.095}; {x=0.86,y=0.0041,z=0.14}}
Right answer summary: part 1: [[0.85,0.053,0.098],[0.85,0.053,0.098],[0.85,0.053,0.098]]; part 2: {x=0.85,y=0.053,z=0.098}
Question state: todo