Question

The p.d.f of a continuous random variable is give

P(x) ={ kx (1-x) e^x , 0 ≤ x ≤ 1

{ 0, otherwise

Find k and hence find mean and standard deviation

Answer 1

$\int^1_0 kx(1-x)e^xdx=1$

$\int^1_0 kx(1-x)e^xdx=[ke^x(x-1)-ke^x(x^2-2x+2)]|^1_0=$

$=-ke+k+2k$

$-ke+k+2k=1$

$k=\frac{1}{1-e}$

$E(X)=\int^1_0 kx^2(1-x)e^xdx=[ke^x(x^2-2x+2)-ke^x(x^3-3x^2+6x-6)]|^1_0=$

$=3ke-2k-6k=\frac{3e-8}{1-e}$

$V(X)=E(X^2)-(E(X))^2$

$E(X^2)=\int^1_0 kx^3(1-x)e^xdx=[ke^x(x^3-3x^2+6x-6)-$

$-ke^x(x^4-4x^3+12x^2-24x+24)]|^1_0=-11ke+30k=\frac{30-11e}{1-e}$

$V(X)=\frac{30-11e}{1-e}-(\frac{3e-8}{1-e})^2=\frac{30-30e-11e+11e^2}{(1-e)^2}=\frac{11e^2-41e+30}{(1-e)^2}$

$\sigma=\sqrt{V(X)}=\frac{\sqrt{11e^2-41e+30}}{1-e}$