There is a well-known method for finding the mean of a normal distribution (given its variance) given the probability below a certain endpoint by normalizing the distribution: $$X \sim N(\mu, \sigma^2)$$ $$P(X<a) = 0.8$$ $$P(Z<\frac{a-\mu}{\sigma}) = 0.8$$ $$\frac{a-\mu}{\sigma} = 0.841...$$ after which, provided $\sigma$ and $a$, one can solve for $\mu$.
However, what if one is given two endpoints, between which the probability lies? $$X \sim N(\mu, \sigma^2)$$ $$P(a<X<b) = 0.8$$ $$P(\frac{a-\mu}{\sigma}<Z<\frac{b-\mu}{\sigma}) = 0.8$$ $$P(Z<\frac{b-\mu}{\sigma}) - P(Z <\frac{a-\mu}{\sigma}) = 0.8$$
How would I continue to find the mean $\mu$ of this distribution provided the values of $\sigma$, $a$, and $b$?
Let's write $$ F(\mu):=\textstyle P(Z<\frac{b-\mu}{\sigma}) - P(Z <\frac{a-\mu}{\sigma})=\Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})\,. $$ The derivative of this w.r.t. $\mu$ is $$ F'(\mu)=\frac{\Phi'(\frac{a-\mu}{\sigma}) - \Phi'(\frac{b-\mu}{\sigma})}{\sigma} $$ When $\mu<a<b$ then $F'(\mu)>0\,$ and when $a<b<\mu$ then $F'(\mu)<0$.
This means that the equation $F(\mu)=c$ has typically two solutions $\mu$. Newton-Raphson should be stable enough to find them based on reasonable starting values.