Finding the median given the PDF $f(x) = cx^2$.

Question

I'm new to stats and I facing problems in finding the median of a PDF.

I have to find the median of this PDF

$$ f(x) = \begin{cases} cx^2, & \text{if 0 $\le$ $x$ $\le$ 3} \\ 0, & \text{otherwise} \end{cases}$$

How I'm trying to solve it

First I found the value o $c$

$$\int_0^3 cx^2 dx = 1 \implies c = \frac{1}{26}$$

Second

Given that the median is a number $m$ such that $P(X\le m) =\frac{1}{2}$ I did:

$$\int_0^m\frac{1}{26}x^2dx = \frac{1}{2}$$

I have two questions concerning my attempt:

1 - What I'm doing is correct? As I said, I'm new to stats and I'm sure if my approach is the correct one to find the median of the PDF

2 - How can I can integrate from zero to $m$? I have $x$ as a parameter in my integral and $m$ as its upper bond. I'm confused in how to solve that.

Answer 1

1. You made an error. The setup is correct, but $c = 1/9$.

2. Why did you stop? Follow through next time. $$\frac12 = \int_0^m \frac{1}{9}x^2\,dx = \frac{1}{27}m^3-\frac{1}{27}(0)^3 = \frac{1}{27}m^3.$$ This gives $m = \sqrt[3]{27/2}.$

Answer 2

Your approach is correct, however $c=\frac{1}{9}$ and not $c=\frac{1}{26}$.

As for your second point, what you can do is find your antiderivative (with respect to $x$) and evaluate it from $0$ to $m$ like you would a definite integral, but set that equal to $\frac{1}{2}$. Then you can solve for $m$.