# Median of a continuous r.v.

Let X be a continuous r.v., with pdf $$f_X(x) = kx(1-x), 0 < x <1$$

I evaluated k = 6 and found the cdf $$F_x(x) = 3x^2 - 2x^3$$, but then I am asked to find the median. The equation $$F_x(x) = 1/2$$ brings me to $$-4x^3 + 6x^2 -1 = 0$$, for which I don't know any quick way to find its root (but I randomly plugged in $$x = 1/2$$ and got my answer). The question is, apart from the tedious cubic formula, is there another quick trick to solve general cubic equations, especially when I am not looking for integer solutions?

## 1 Answer

Cubic equations are indeed tedious.

I guess you are expected here to realize that the pdf is symmetric around $$x=1/2$$ : that is, $$f_X(x-\frac12)=f_X(x+\frac12)$$. Hence, the median is $$1/2$$, because $$\int_0^{1/2} f_X(x) dx =\int_{1/2}^1 f_X(x) dx$$

• I see now, but I didn't know that the graph was symmetric in advance, and I suspect that examining the function (find roots, first and second derivatives and their roots, potential asymptotes etc) is too farfetched for the purposes of this question. – JBuck Apr 19 at 21:17
• Which brings this question to my mind: Is every 2nd degree polynomial symmetric around its extremum? – JBuck Apr 19 at 21:19