 Solution

For the Pole forward-backward asymmetry for charged leptons, integrating the normalized distribution over the two hemispheres cancels the even term in the numerator. Since ${\int }_{-1}^1(1+z^2)dz=8/3$ and ${\int }_0^{1}zdz-{\int }_{-1}^{0}zdz=1$,Within this single-$Z$, tree-level model, a nonzero asymmetry requires unequal left/right coupling magnitudes, demonstrating parity violation in the vertices. For $g_L=g_R$ the odd angular term vanishes and $A_{FB}=0$; equal magnitudes with opposite signs also give zero. This inference is process-specific: a forward-backward asymmetry is not universally a parity-odd observable, since reversing both beam and outgoing momenta leaves their relative Angle unchanged. Purely electromagnetic Bhabha scattering is a concrete parity-conserving process with a forward-peaked, asymmetric distribution.

For the supplied ${\sin }^2{\theta }_W=0.23153$,The measurement differs by $0.0188-0.01620=0.00260$, or $0.79$ times its quoted uncertainty of $0.0033$, so it is consistent with the prediction. Statistical fluctuations alone preclude exact equality; photon exchange, initial-state radiation and electroweak radiative corrections also modify the idealized pure-resonance prediction and provide physical reasons for small differences.