A continuous random variable X has p.d.f defined by $f(x) =kx^2e^{-x} ,x ≥ 0$ .Find mean and variance.

Since the total probability is 1

$ ∫_{-∞}^∞f(x) =1$

$∴ ∫_0^∞kx^2 e^{-x}\ dx =1 $

$k∫_0^∞x^2 e^{-x}\ dx = k[x^2 . \frac{e^{-x}}{(-1)}-2x. \frac{e^{-x}}{1}+2.\frac{e^{-x}}{(-1)}]_0^∞ = 2k = 1$

k = $\frac{1}{2}$

Now, Mean $ \bar{X} = ∫_0^∞xf(x)\ dx = \frac{1}{2} ∫_0^∞x^3 e^{-x}\ dx$

$μ_1'= \frac{1}{2} [x^3. \frac{e^{-x}}{(-1)} - 3x^2. \frac{e^{-x}}{1} + 6x.\frac{e^{-x}}{(-1)} - 6 .\frac{e^{-x}}{1}]_0^∞ = 3$

$μ_2' = ∫_0^∞x^2 f(x)\ dx = \frac{1}{2} ∫_0^∞ x^2.x^2 e^{-x}\ dx = \frac{1}{2} ∫_0^∞x^4 e^{-x} \ dx $

$= \frac{1}{2} [x^4 .\frac{e^{-x}}{(-1)} - 4x^3.\frac{e^{-x}}{1}+12x^2.\frac{e^{-x}}{(-1)} - 24x. \frac{e^{-x}}{1} + 24.\frac{e^{-x}}{(-1)}]_0^∞ $

= 12

Variance = $μ_2' - [μ_1' ]^2$ = 12 -9 = 3