Answer to Question #183630 in Statistics and Probability for Robert

Question

For a continuous function f(x) =kx²e^{-x where x>0.Find i) k, ii) mean, iii) variance}

Answer 1

$f(x) =kx^2e^{-x}, \; x>0$

i) $\int f(x) dx =1$

$\int_0^\infty kx^2 e^{-x}dx = k \int_0^\infty x^2 e^{-x} dx =/ \text{integrating by parts}/ = 2k =1$

$k=1/2$

ii) mean

$M[x] = \int_0^\infty xf(x) dx= k\int_0^\infty x^3 e^{-x} dx = / \text{integrating by parts}/ = 6k =3$

$M[x] = 3$

iii) variance

$D[x] = M[x^2] - (M[x])^2$

$M[x^2] = \int_0^\infty x^2f(x) dx= k\int_0^\infty x^4 e^{-x} dx = / \text{integrating by parts}/ = 24k =12$

$D[x] = 12 - (3)^2 = 12 - 9 = 3$