Let X be a continuous random variable with probability density function f(x)=4x^3 for 0≤x≤1 and f(x)=0 otherwise. Find the expected value E[X] and variance Var(X).
Answer
E[X]=4/5, Var(X)=2/75
The correct answer is E[X]=4/5, Var(X)=2/75 — Var(X)=E[X^2]−(E[X])^2 = 2/3 − (4/5)^2 = 2/3 − 16/25 = (50−48)/75 = 2/75.
Step-by-step solution
- Check normalisation: ∫0^1 4x^3 dx = 4[ x^4/4 ]0^1 =1.
- E[X] = ∫0^1 x·4x^3 dx = ∫0^1 4x^4 dx = 4[x^5/5]0^1 = 4/5.
- E[X^2] = ∫0^1 x^2·4x^3 dx = ∫0^1 4x^5 dx = 4[x^6/6]0^1 = 2/3.
- Var(X)=E[X^2]−(E[X])^2 = 2/3 − (4/5)^2 = 2/3 − 16/25 = (50−48)/75 = 2/75.
Common mistake: Mis‑computing the integral for E[X] or E[X^2] by wrong powers.
Concept tested: Expectation and variance of continuous variable
Published · How this solution was prepared and checked
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