Question

In a continuous distribution, whose probability density function is given by
f(x) = 3/4x(2 – x) , 0 < x < 2. Find the expected value of x.

Answer 1

we know the relation between Probability density function and expected value of x .
e.g., E(x) = $\int\limits^a_b {xf(x)} \, dx$ where interval [a, b] is domain of f(x) , E(x) is the expected value of x and f(x) is probability density function.

Now, given, g(x) = -3x(2 - x)/4 , 0 < x < 2
So, E(x) = $\bold{ \int\limits^2_0{x[\frac{3}{4}x(2-x)]} \, dx }$
= 3/4 ∫₀²[2x - x² ] dx
= 3/4 [x² - x³/3]₀²
= 3/4 [4 - 8/3 ]
= 3/4 × 4/3
= 1

Hence, expected value of x = 1