A continuous random variable X has p.d.f defined by f(x) =A +B x 0≤x≤1. If the mean of the distribution is 1/3 . Find A and B.

Since the total probability is 1

$∫_{-∞}^∞\ f(x) \ dx$ = 1

$∫_0^1(A+Bx) \ dx = [Ax+B .\frac{x^2}{2}]_0^1 = A+\frac{B}{2} $ = 1…….(1)

since, mean is $ \frac{1}{3}; ∫_0^1x(A+Bx) \ dx = \frac{1}{3}$

$∫_0^1(Ax+Bx^2) \ dx = \frac{1}{3}$

$[A\frac{x^2}{2}+B \frac{ x^3}{3}]_0^1 = \frac{1}{3}$ or $ \frac{A}{2}+\frac{B}{3} = \frac{1}{3} $ or $3A +2B = 2$………(2)

Solving (1) & (2) $A = 2, B = -2 $

∴ The p.d.f is $f(x) = 2 -2x 0≤x≤14$