Change the Normalization property of the histogram object then get the appropriate element of the Values property of that object.
x = randn(10000,1);
h = histogram(x)
Since the default Normalization method is 'count', this will tell you that there are 133 elements of x that fall into bin 10. [Since I used rng default, you should get the exact same random numbers in x as I did and so generate the exact same histogram.]
h.Normalization = 'probability';
Now h.Values(10) is 0.0133 which makes sense: 133 / 10000 (the total number of points) = 0.0133.
If you wanted to get the same information without actually bringing up the plot, the histcounts function also lets you specify a 'Normalization' method.
And I'd guess that histogram you showed was created with something more like 900 data points than 90. According to the Y limits each of the 5 central bars contain more than 90 elements, assuming you're using the default 'count' Normalization. Still not Big Data, but bigger.