Let $Q(x)$ denote the complementary cumulative distribution function of a standard normal distribution (see here). Given $ 0 \leq a \leq b$, and define the $Q$-scale average of $a$ and $b$ as a $c$ satisfying $$1/2 Q(a) + 1/2 Q(b) = Q(c).$$
A `naive' way to calculate $c$ is $c = Q^{-1}[1/2Q(a) + 1/2 Q(b)]$. (Routines for calculating $Q(x)$ and $Q^{-1}(x)$ in a stable and accurate fashion are available in most standard numerical calculation libraries.)
However, I feel that the above formation kills the precision. Obviously $c \in (a, b)$. Moreover, $c$ is `very close' to $a$. Therefore, it makes sense to write $c = a + \Delta(a, b)$ and calculate $\Delta(a, b)$ instead. Nevertheless, I've so far been unable to find a formulae for $\Delta(a, b)$.
I wonder if someone else has encountered this problem before?