Trying to find a histogram half-sum point

I'm trying to find the x coordinate in a histogram that describes the point where the sum of the values is exactly half (basically like the half-life in an exponential decay graph).

My code that doesn't quite work: (avoiding the use of loops)

h = hist(dataset);
% ex: h = [3 45 1 4 6 3 4 6 8 23 6 3 44];
n = 1:1:length(h);
a(n) = (sum(h(1:n)) <= sum(h)/2); % Returns a Boolean matrix which identifies where the half-sum point is
x = find(a); % Returns the non-zero indices of a, which can be used as an accessing array 
fprintf('Angular Resolution Metric is at %f', x)

The problem is that (sum(h(1:n)) < sum(h)/2) isn't working... any tips?

(Also, I realize that it should give me something like [1 1 1 1 1 0 0 0 0 0 0 0] ...etc. so I'd just reference the maximum x value to get the point)

edit:

The following code works, but I'd like to avoid using loops, so if anyone figures out how to write this w/out loops that'd be good to know:

n=1;
while n < length(h)
    if sum(h(1:n)) < sum(h)/2
        n=n+1;
    else
        x=n;
        fprintf('Angular Resolution Metric is at %f \n', x)
        break
    end
end

Answer 1

[Paulo Silva]

h = [3 45 1 4 6 3 4 6 8 23 6 3 44]
x=sum(cumsum(h) < sum(h)/2) 
fprintf('Angular Resolution Metric is at %d \n', x)

[Lena Heffern]

Gracias!

Answer 2

[Bjorn Gustavsson]

This slight modification seems to work as I expect it would:

   sum_h = sum(h)
   cumsum(a)
   a = cumsum(h) <= sum_h/2

This produces:

   sum_h =
           156
   cumsum_h = 3  48  49  53  59  62  66  72  80 103 109 112  156
   a =        1   1   1   1   1   1   1   1   0   0   0   0    0

HTH