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I'm trying to find the median of $f(x) = 4xe^{-2x}$.

So far, I've tried solving for $q_{50}$ by plugging it into an integral and setting it equal to 0.5 like so: $\int_{0}^{q_{50}} 4xe^{-2x} dx = 0.5$. I eventually get to $-2q_{50}e^{-2q_{50}} - e^{-2q_{50}} + 1 = 0.5$. Unfortunately, at this point, I have been unable to solve for $q_{50}$.

Is there something I've done wrong up to this point or another method that I could be using instead to find the median? Thanks for the help!

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  • $\begingroup$ Looks right to me. Solving it numerically, as here yields $q_{50}\approx .839173$ $\endgroup$
    – lulu
    Dec 12, 2021 at 22:25
  • $\begingroup$ here it is in DESMOS $\endgroup$
    – WW1
    Dec 12, 2021 at 22:30

2 Answers 2

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Letting $x=q_{50}$, as @Vítězslav Štembera answered, you want to solve for $x$ the equation $$ (2 x+1)\,e^{-2 x}=k \quad\implies\quad(2x+1)\,e^{-(2 x+1)}=\frac k e$$ The only explicit solution of it is given by $$x=-\frac{1}{2} \left(1+W_{-1}\left(-\frac{k}{e}\right)\right)$$ where $W_{-1}(.)$ is the second branch of Lambert function.

If you cannot use Lambert function, only numerical methods would give the solution.

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Your equation \begin{align} -2q_{50}e^{-2q_{50}} - e^{-2q_{50}} + 1 = 0.5 \end{align} i.e. \begin{align} (2q_{50}+1)e^{-2q_{50}}= 0.5 \end{align} is correct, however it is trascendental and must be solved numerically. Using MAPLE for example you can find $q_{50}\approx 0.839173495$.

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