Answer:
[tex]\mathbf{E(X) = \dfrac{3}{4}}[/tex]
Step-by-step explanation:
From the given data, the cumulative distribution function of a random variable can be represented as:
[tex]F_X(t) =\left\{ \begin{array}{c}0........... t <-1 \\ \dfrac{1}{2} ... -1 \leq t < 1\\ \dfrac{1}{2} ....... 1 \leq t < 2 \\ 1 .............. t \geq 2\\\end{array}\right.[/tex]
The objective is to estimate E(X), to do that, let's first evaluate the probability density function by differentiating the cumulative distribution function from above.
[tex]f_X(x) =\left \{ {{\dfrac{1}{2} .......1 \leq x \leq 2 } \atop {0..... otherwise }} \right.[/tex]
∴
[tex]f_X(t) =\left\{ \begin{array}{c} \dfrac{d}{dx}(0)=0........... <-1 \\ \dfrac{d}{dx}(\dfrac{1}{2} ) =0... -1 \leq t < 1\\ \dfrac{d}{dx}(\dfrac{1}{2}x) = \dfrac{1}{2}....... 1 \leq x < 2 \\ \dfrac{d}{dx}(1) = 0 .............. x \geq 2\\\end{array}\right.[/tex]
The expected value of x i
.e E(X) can now be estimated by taking the integral:
[tex]E(X) = \int ^{\infty}_{\infty} x f(x) \ dx[/tex]
[tex]E(X) = \int ^{1}_{- \infty} x 0 dx + \int^2_1 \ x \dfrac{1}{2}\ dx + \int ^{\infty}_2 \ x0dx[/tex]
[tex]E(X) = \int ^{2}_{1} x \dfrac{1}{2} dx[/tex]
[tex]E(X) = \dfrac{1}{2}[\dfrac{x^2}{2}]^2_1[/tex]
[tex]E(X) = \dfrac{1}{2}[\dfrac{4}{2}-\dfrac{1}{2}][/tex]
[tex]E(X) = \dfrac{1}{2} \times [\dfrac{3}{2}][/tex]
[tex]\mathbf{E(X) = \dfrac{3}{4}}[/tex]
