Let $X$ be a continuous random variable with pdf $f (x) = 3 x^{2}$ for $0 l e x l e 1$ and $f (x) = 0$ otherwise. Find the expected value $E [X]$ and the variance $Var (X)$ .

medium 2 marks Class 12 · Mathematics · Probability

Answer

E[X] = 3/4, Var(X) = 3/80

The correct answer is E[X] = 3/4, Var(X) = 3/80 — Variance $Var (X) = E [X^{2}] - (E [X])^{2} = (3/5) - (3/4)^{2} = 3/5 - 9/16 = 48/80 - 45/80 = 3/80$ .

Step-by-step solution

Verify that $f (x)$ is a pdf: $in t_{0}^{1} 3 x^{2} d x = 3 [x^{3} /3]_{0}^{1} = 1$ .
Compute $E [X] = in t_{0}^{1} x c d o t 3 x^{2} d x = 3 in t_{0}^{1} x^{3} d x = 3 [x^{4} /4]_{0}^{1} = 3/4$ .
Compute $E [X^{2}] = in t_{0}^{1} x^{2} c d o t 3 x^{2} d x = 3 in t_{0}^{1} x^{4} d x = 3 [x^{5} /5]_{0}^{1} = 3/5$ .
Variance $Var (X) = E [X^{2}] - (E [X])^{2} = (3/5) - (3/4)^{2} = 3/5 - 9/16 = 48/80 - 45/80 = 3/80$ .

Common mistake: Missing the square of

x

when computing

E [X^{2}]

or mis‑integrating

Concept tested: Mean and variance of continuous distributions

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

← All Probability questions

Let X be a continuous random variable with pdf f(x)=3x2 for 0lexle1 and f(x)=0 otherwise. Find the expected value E[X] and the variance Var(X).

Step-by-step solution

Let $X$ be a continuous random variable with pdf $f (x) = 3 x^{2}$ for $0 l e x l e 1$ and $f (x) = 0$ otherwise. Find the expected value $E [X]$ and the variance $Var (X)$ .