Answer
E(x) = 2. As it is in the middle of 1 and 3, hence the distribution is expected to be uniform.
Work Step by Step
f(x) = 1/d-c = 1/3-1 = 1/2
$P(1 \leq x \leq 3) = ∫_{1}^{3} x.f(x).dx$
= $∫_{1}^{3} \frac{1}{d-c}.x.dx$
= $∫_{1}^{3} \frac{1}{2}x..dx$
= $1/2 [x^{2}/2]_{1}^{3} $
= 1/2. [9/2 - 1/2 = 1/2. [8/2] = 2
