cap_5_mont_p_87

OK PUNTOS 0.5

=CAPITULO 5, PROBLEMA 87=

Si la variable aleatoria X tiene una distribución exponencial con media teta, determine lo siguiente:


 * Solución:**

a) math P\left ( X> \theta \right ) math

La solucion es la siguiente:

math E\left [ X \right ]=\theta math

math E\left [ X \right ]=\frac{1}{\lambda }=\theta math

math \frac{1}{\theta }=\lambda math

math P\left ( X> \theta \right )=\int_{\theta }^{\infty }\frac{1}{\theta }*e^{\frac{-1}{\theta }*X}*dX math

math P\left ( X> \theta \right )=e^{\frac{-1}{\theta }*\theta } math

math P\left ( X> \theta \right )=e^{-1} math

math P\left ( X> \theta \right )=0.368 math

b) math P\left ( X>2 \theta \right ) math

La solucion es la siguiente:

math E\left [ X \right ]=\theta math

math E\left [ X \right ]=\frac{1}{\lambda }=\theta math

math \frac{1}{\theta }=\lambda math

math P\left ( X> 2\theta \right )=\int_{2\theta }^{\infty }\frac{1}{\theta }*e^{\frac{-1}{\theta }*X}*dX math

math P\left ( X> 2\theta \right )=e^{\frac{-1}{\theta }*2\theta } math

math P\left ( X> 2\theta \right )=e^{-2} math

math P\left ( X> 2\theta \right )=0.135 math

c) math P\left ( X> 3\theta \right ) math

La solucion es la siguiente:

math E\left [ X \right ]=\theta math

math E\left [ X \right ]=\frac{1}{\lambda }=\theta math

math \frac{1}{\theta }=\lambda math

math P\left ( X>3 \theta \right )=\int_{3\theta }^{\infty }\frac{1}{\theta }*e^{\frac{-1}{\theta }*X}*dX math

math P\left ( X>3 \theta \right )=e^{\frac{-1}{\theta }*3\theta } math

math P\left ( X> 3\theta \right )=e^{-3} math

math P\left ( X> 3\theta \right )=0.049 math

Solucionado por:
 * Juan Manuel López
 * Julián Cadena Isaza