0
$\begingroup$

Let $X$ be a continuous random variable with PDF

$$f_X(x)= \begin{cases} cx(1-x), & \text{$0<x<1$} \\ 0, & \text{elsewhere} \end{cases}$$ Find the median of $X$.

My question is how I am only given PDF, to calculate median, do I need to find CDF for it? And how?

$\endgroup$
2
  • 2
    $\begingroup$ Note that distribution is not uniform. $\endgroup$
    – herb steinberg
    Feb 11, 2020 at 5:06
  • $\begingroup$ @herbsteinberg fixed $\endgroup$
    – xyz12354
    Feb 11, 2020 at 5:46

2 Answers 2

Reset to default
3
$\begingroup$

You do not need to calculate anything.

The distribution is symmetric about $x=\frac 12$. (Draw it.)

So, median and mean are the same and are equal to $\frac 12$.

$\endgroup$
2
$\begingroup$

To get CDF $F_X(1)=1=c\int_0^1x(1-x)dx =\frac{c}{6}$, so $c=6$

Next $F_X(x)=6\int_0^xu(1-u)du=3x^2-2x^3$.

To get median $3x^2-2x^3=\frac{1}{2}$ and solve for $x$ in the interval $[0,1]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.