= Solution
Integrating the distribution in <4/d>, use $\int_{-1}^1(1+z^2)\,dz=8/3$ and $\int_0^1z\,dz-\int_{-1}^0z\,dz=1$. Therefore
$$
\boxed{A_{\rm fb}=\frac34\left(\frac{a-b}{a+b}\right)^2.}
$$
Subtract backgrounds and correct the measured forward and backward yields for <detector acceptance>, charge misidentification and selection efficiency; radiative effects and other mediators must be removed or included in a fit. With $t=\sqrt{4A_{\rm fb}/3}$ and $r=|g_l/g_r|$, inversion gives
$$
\boxed{r=\sqrt{\frac{1+t}{1-t}}\quad\text{or}\quad r=\sqrt{\frac{1-t}{1+t}}.}
$$
A single unpolarized measurement determines the ratio up to $r\leftrightarrow1/r$. A polarized-beam observable or another signed chiral asymmetry is needed to choose which coupling is larger. Values outside $0\le A_{\rm fb}\le3/4$ would contradict this ideal universal single-mediator model, apart from experimental uncertainties.
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