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I am using a four parameter logistic function $y = a + \dfrac{b-a}{1+e^{c(d-x)}}$. I would like to know how one finds the value $x$ that corresponds to $5\%$ (or $n\%$) cumulative density.

My intuition would be that I find the value between the lower and upper asymptotes that corresponds to an increase of $n\%$. Substituting this value for $y$ and solving the equation for x would give me the value $x$ at which there was a cumulative increase of $n\%$ from the lower asymptote. Is this correct?

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There's no such thing as a cumulative density. The word "cumulative" contradicts the word "density". See – Michael Hardy Jun 7 '13 at 16:09

Yes, the procedure you described is correct (except for the misuse of the word "density", pointed out by Michael Hardy). As $x\to-\infty$, $y\to a$ (assuming $c>0$), and as $x\to+\infty$, $y\to b$. Therefore, you want to solve the equation $y=a+0.05(b-a)$, or $y=a+(n/100)(b-a)$ in general. This amounts to $$\frac{1}{1+e^{c(d-x)}} = \frac{n}{100}$$ $$ e^{c(d-x)} = \frac{100}{n}-1$$ $$ c(d-x) = \ln\left(\frac{100}{n}-1\right)$$ and you can take it from there.

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