Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2014

By definition of gamma function

$$\int\limits_0^{\infty}e^{-t}.t^n.dt=\overline n=(n-1)!......... (1)$$

For any probability density function

$$\int\limits_{-\infty}^{\infty} f(x).dx=1$$

$\therefore \int\limits_0^{\infty}kx^2e^{-x}.dx=1\\ \therefore k\int\limits_0^{\infty}e^{-x}.x^2dx=1\\ \therefore k.2!=1\hspace {1cm} from (1)\\ \therefore k=\dfrac 12\\ \therefore f(x)=\dfrac 12x^2.e^{-x}\\ Mean= E(x)=\int\limits_{-\infty}^{\infty}x.f(x).dx \\ =\int\limits_0^{\infty}x\dfrac 12x^2e^{-x}.dx\\ =\dfrac 12\int\limits_0^{\infty}x^3.e^{-x}.dx\\ \dfrac 12\times 3! \hspace {1cm} from(1)\\ =3\\ E(x)^2=\int\limits_{-\infty}^{\infty}x^2f(x).dx\\ Now \\ =\int\limits_0^{\infty}x^2.\dfrac 12x^2e^{-x}.dx\\ =\dfrac 12\int\limits_0^{\infty}x^4.e^{-x}.dx\\ =\dfrac 12\times 4! \hspace {1cm} from (1)\\ =12\\ \therefore \text { variance }= E(x^2)-[E(x)]^2\\ 12-3^3\\ =3$

Hence, $k = ½,$ Mean $= 3$ and variance $= 3$