Let 𝑋X be a continuous random variable with the following PDF:𝑓𝑋(𝑥)={𝑥1(4𝑥2+3𝑥−2)0<𝑥≤10otherwisef X (x)={ x 1 (4x 2 +3x−2)0 0<x≤1otherwise Find the value of 9𝐸[𝑋].E[X].Enter your answer correct to two decimals accuracy.
Question
Let 𝑋X be a continuous random variable with the following PDF:𝑓𝑋(𝑥)={𝑥1(4𝑥2+3𝑥−2)0<𝑥≤10otherwisef X (x)={ x 1 (4x 2 +3x−2)0 0<x≤1otherwise Find the value of 9𝐸[𝑋].E[X].Enter your answer correct to two decimals accuracy.
Solution
The expected value E[X] of a continuous random variable X with probability density function f(x) is given by the integral of x*f(x) over the range of X. In this case, the range of X is 0 < x ≤ 1.
So, we need to compute the integral ∫x*f(x) dx from 0 to 1.
Given f(x) = x(4x^2 + 3x - 2), we have:
E[X] = ∫x*f(x) dx from 0 to 1 = ∫x^2(4x^2 + 3x - 2) dx from 0 to 1 = ∫(4x^4 + 3x^3 - 2x^2) dx from 0 to 1 = [x^5 + (3/4)x^4 - (2/3)x^3] from 0 to 1 = 1 + 3/4 - 2/3 - 0 = 1.08 (rounded to two decimal places)
Therefore, 9E[X] = 9 * 1.08 = 9.72 (rounded to two decimal places).
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