= Solution
The energy average must be taken over the atoms that actually arrive, not over the bulk population. Weighting the collision flux by $mv^2/2$ gives
$$
\langle K\rangle_{\rm hit}=\frac{(m/2)nA\pi I_5}{nA\pi I_3}=\frac m2\frac{I_5}{I_3},\qquad \boxed{\langle K\rangle_{\rm hit}=2k_BT.}
$$
The <wall-impact speed distribution> is proportional to $v^3e^{-\alpha v^2}$ rather than the bulk $v^2e^{-\alpha v^2}$. Faster atoms are overrepresented among arrivals. Equivalently, the mean normal <kinetic energy> of the arriving flux is $k_BT$, while each tangential component still contributes $k_BT/2$.
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