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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?

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

2 Answers 2

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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 $\frac 12$.

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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 $[0,1]$.

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