# Calculating the Mean, Median, and Mode of Continuous Random Variable

Given a continuous random variable $x$ with CDF of $x^3$ for $0\le x\le 1$ (and $0$ for $x \lt 0$ and $1$ for $x \gt 1$, rank the median, mode and mean.

My attempt: To find the mean, I first found the PDF to be $3x^2$. I then took $\int_0^1 x(3x^2) \,dx = \frac{3}{4}$

For the median, I set the CDF of $x^3$ equal to $\frac{1}{2}$ which is $\left(\frac{1}{2}\right)^\frac{1}{3}$

Finally, for the mode, I think that this would be the highest value of the PDF of $3x^2$ on the interval of $0 \le x \le 1$ , which should be $1$ but I am unsure if this reasoning is correct.

I am struggling to understand the concept of mode for continuous random variables since the probability of any individual point is $0$. I am unsure of whether my reasoning which is mostly carried over from discrete variables is applicable.

• Here the mode is just the PDF's maximum. Jun 25, 2017 at 20:33

In this case, our domain is the closed interval $[0,1]$, so the pdf $3x^{2}$ takes on a maximal value at either a critical point or at the endpoints $0,1$. The only critical point is $0,$ and $3x^{2}_{x=0} = 0$. Since $3x^{2}>0$ at $x=1$, your answer is correct: The mode is $1$.