Answer to Question #233400 in Statistics and Probability for Reyad

Question

Calculate the expected value of a random variable X if f(x) = x – 5⁄2 for 0<x<1 and 2x for 1<x<2 and 0 otherwise.

Answer 1

$E(x) = \int\limits_{ - \infty }^{ + \infty } {xf(x)dx = } \int\limits_0^1 {x\left( {x - \frac{5}{2}} \right)} dx + \int\limits_1^2 {x \cdot 2xdx} = \int\limits_0^1 {\left( {{x^2} - \frac{{5x}}{2}} \right)} dx + \int\limits_1^2 {2{x^2}} dx = \left. {\frac{{{x^3}}}{3}} \right|_0^1 - \left. {\frac{{5{x^2}}}{4}} \right|_0^1 + \left. {\frac{{2{x^3}}}{3}} \right|_1^2 = \frac{1}{3} - \frac{5}{4} + \frac{{2(8 - 1)}}{3} = \frac{{15}}{4}$

Answer: $E(x) = \frac{{15}}{4}$