Discussion
A continuous random variable X follows the probability law :
\(f(x) = E.x^{2};\ 0 \le x \le 1\)
What is the value of E ?
Answer: D
The value of E is 3.
A probability density must integrate to 1 over its range:
\[ \begin{aligned} \int_{0}^{1} E x^{2}\, dx &= 1 \\ E \left[ \frac{x^{3}}{3} \right]_{0}^{1} &= 1 \\ \frac{E}{3} &= 1 \Rightarrow E = 3 \end{aligned} \]
E = 1 forgets the division by 3 from integrating \(x^{2}\), giving a total probability of only \(\frac{1}{3}\). E = 2 is the constant for \(f(x) = Ex\), and E = 6 would make the total probability 2.
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