Hello SX,
Ordinarily to find the cdfs you would have to use numerical integration. In this case for the normal distributions, the cdf function is available. Then you can interpolate using the cdf as the independent variable. Here is an example. In the plot you get a wide minimum which you might expect.
mu1 = 1;
mu2 = 2;
sig1 = 1;
sig2 = 3;
w1 = .3;
w2 = .7;
c = .9; % confidence span, there is probably a better name for this
x = -20:.00001:20;
cdf = w1*normcdf(x,mu1,sig1) +w2*normcdf(x,mu2,sig2);
cdn = linspace(min(cdf),max(cdf)-c,1e4);
xdn = interp1(cdf,x,cdn);
cup = linspace(min(cdf)+c,max(cdf),1e4);
xup = interp1(cdf,x,cup);
figure(1); grid on
plot(xup-xdn)
[x0 ind] = min(xup-xdn);
xdn(ind) % lower end of confidence interval
xup(ind) % upper end of confidence interval
cdn(ind) % lower cdf value
cup(ind) % upper cdf value
% ans = -2.4087
% ans = 6.3858
% ans = 0.0497
% ans = 0.9497
D = xup(ind)-xdn(ind) % the result
cup(ind)-cdn(ind) % check, should be c = confidence span
% D = 8.7945
% ans = 0.9000
% try a different case, get a larger confidence interval
xtest = interp1(cdf,x,[.07 .97]);
Dtest = diff(xtest)
% xtest = -1.8602 7.1554
% Dtest = 9.0156
