cap_5_mont_p_22

OK PUNTOS 0.7

=CAPITULO 5. EJERCICIO 22=

Suponga que: math f\left ( x \right ) = 1.5x^{2} math Determinar la media y la varianza de x para -1<x<1.


 * SOLUCION**

MEDIA

math \mu = E_{X}\left [ x \right ]=\int_{-\infty }^{\infty }xf\left ( x \right )dx math

math \mu =\int_{-1}^{1}x*1.5x^{2}dx math

math \mu =\int_{-1}^{1}1.5x^{3}dx math

math \mu =\left. \frac{1.5x^{4}}{4} \right|_{-1}^{1} math

math \mu =\frac{1.5*1^{4}}{4}-\frac{1.5*\left ( -1^{4} \right )}{4} math

math \mu =0 math

VARIANZA

math \sigma ^{2} =\int_{-\infty }^{\infty }x^{2}f\left ( x \right )dx - \mu ^{2} math

math \sigma ^{2} =\int_{-1}^{1}x^{2}*1.5x^{2}dx - \mu ^{2} math

math \sigma ^{2} =\int_{-1}^{1}1.5x^{4}dx - \mu ^{2} math

math \sigma ^{2} =\left. \frac{1.5x^{5}}{5} \right|_{-1}^1 - \mu ^{2} math

math \sigma ^{2} =\left (\frac{1.5*1^{5}}{5}-\frac{1.5*\left ( -1^{5} \right )}{5} \right ) - 0 math

math \sigma ^{2} =\left (\frac{1.5*1^{5}}{5}+\frac{1.5*\left ( 1^{5} \right )}{5} \right ) - 0 math

math \sigma ^{2}=0.6 math

Solucionado por:
 * Juan Manuel López
 * Julian Cadena Isaza