$X$ is a uniformly distributed random variable that takes values between zero and one. Find the value of $E( X^3)$.

$X$ is a uniformly distributed random variable that takes values between zero and one. The value of $E( X^3)$ will be 

(A) 0 

(B) $\dfrac{1}{8}$ 

(C) $\dfrac{1}{4}$ 

(D) $\dfrac{1}{2}$ 

Answer: (C) $\dfrac{1}{4}$ 

Explanation: 

Here, $X$ is a uniformly distributed random variable that takes values between 0 and 1. 

Therefore, its p.d.f. is $f(x) = \dfrac{1}{1 - 0} = 1, \mbox{for } 0 < x < 1.$ 

Hence,  $E(X^3) = \int\limits_{0}^{1}x^3f(x)dx = \int\limits_{0}^{1}x^3dx = \dfrac{1}{4}$ 

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