If we have a normal distribution of N(10,2) and we are asked on what is the proportion of values betwen 7 and 8 we can calculate this by:
F(8) = NORM.DIST(8,10,2,TRUE) = 15.9%
F(7) = NORM.DIST(7,10,2,TRUE) = 6.7%
F(8) - F(7) = 9.2%
Is there a way to calculate F(x) and F(y) so that F(x) - F(y) = 10%?
Let $X \sim \mathcal N(\mu, \sigma^2)$
Then $P(X\leq x+x')-P(X\leq x)=F(x+x')-F(x)=c$
In your case $c=0.1$. For a given $x$ the equality can be arranged to $F(x+x')=c+F(x)$.
$c+F(x)=r$ is a real value between $0$ and $1$.
$F(x+x')=r$
$x+x'=F^{-1}(r)$
$x'=F^{-1}(r)-x$
For a given $x$ the value of $x'$ is $F^{-1}(r)-x$