Logistic Scoring Correction

Question

"Consider the logistic curve $f(x)=\frac{1}{1+e^{-bx}}, -1 \leq x \leq 1$. We wish to use this curve to make a scoring correction formula $g(x)$ for an $n$ item test. The domain and range are both $[0,n]$, with $g(0)=0$ and $g(n)=n$. With $b=2$ and $n=10$ the graph looks like

Find an explicit formula for this function for an $n$ item test."

I have tried at least 20 different equations similar to the given logistic curve, and I cannot get it to match the given graph for the given values. Any insight would be greatly appreciated!

Answer 1

Two desired curve is the composition of three functions:

1. linear transformations of the interval $[0,n]$ onto $[-1,1]$. That is, $$x = \frac{2t}{n} -1$$
2. the given logistic function $ y = 1/(1+e^{-bx})$.
3. linear transformation of the range $[1/(1+e^b), 1/(1+e^{-b})]$ onto $[0,n]$. That is, $$u = n\frac{1+e^{-b}}{e^b-e^{-b}}((1+e^b)y-1)$$

As a result, you have $u$ as a function of $t$. It's a bit complicated, but it works: