Best method of quantifying probability of new datum belonging to either of two distanced normal distributions? 
I have two samples A and B from two separate normally distributed populations. The population mean of B is higher than that of A, but both are unknown. My aim is to find a threshold value between the two distributions such that a new datum can be assigned to A if its value falls below this central point, and to B if it lies above, with a certain level of accuracy. I would like to obtain the percentage chance of a datum from population B being incorrectly assigned to A and vice versa.
I have solved for the point at which the Z-scores of the two samples are equal, and found the percentage that lie above and below this point for both distributions. As I am not terribly well-versed in statistics, I am left wondering if this is the correct approach. For example, whether Student's T-distribution should instead be used, or another method altogether.
I have provided the below information, for the sake of interest.
$$
\bar{A} = 103.72,  s_A = 8.62 \\
\bar{B} = 161.17, s_B = 13.62
$$
 A: Here the aim "is to find a threshold value between the two distributions such that a new datum can be assigned to $A$ if its value falls below this central point, and to $B$ if it lies above, with a certain level of accuracy".
Suppose we measure the accuracy as (probability of wrong assignment for data in $A$) + (probability of wrong assignment for data in $B$).
Then we are looking for a threshold value $t$ to minimize
$$P[A>t\ |\ A\sim N(m_A,s_A)] + P[B<t\ |\ B\sim N(m_B,s_B)]$$
The derivative of this with respect to $t$ should be 0:
$$\frac{-e^{-(t-m_A)^2/(2s_A^2)}}{\sqrt{2\pi} s_A}
+\frac{e^{-(t-m_B)^2/(2s_B^2)}}{\sqrt{2\pi} s_B} = 0$$
This can be solved analytically with some algebra and the quadratic formula:
$$(t-m_A)^2/(2s_A^2) + \ln s_A=
(t-m_B)^2/(2s_B^2) + \ln s_B$$
$$t = \frac{b\pm\sqrt{b^2-ac}}{a},\text{ where}$$
$$a=\frac{1}{s_A^2}-\frac{1}{s_B^2},\ \ 
b=\frac{m_A}{s_A^2}-\frac{m_B}{s_B^2},\ \
c=\frac{m_A^2}{s_A^2}-\frac{m_B^2}{s_B^2}+\ln\left(\frac{s_A^2}{s_B^2}\right)$$
For the particular numerical values in the question, this gives $a=0.00807$, $b=0.527$, $c=3.84$, and $t=126.9$ as the option in between $m_A$ and $m_B$. The measure of accuracy is $0.95\%$.
For other ways of measuring accuracy we would get other values of $t$; this is one way of getting a reasonable value.
