How to find the mean of a histogram without the mean function?

Can anyone give an example code on how to find the average/mean value of an histogram without using the mean or sd function, but rather making using of the bin width?

Answer 1

[Image Analyst]

What about a for loop summing up the values then dividing by the number of items you summed?

data = rand(100);
trueMean = mean(data, 'all') % ~0.5

trueMean = 0.4962

trueSD = std(data(:))		% ~0.29

trueSD = 0.2902

% Take the histogram
h = histogram(data, 10)

h = 
  Histogram with properties:

             Data: [100×100 double]
           Values: [1014 1064 996 1029 915 1029 983 961 1010 999]
          NumBins: 10
         BinEdges: [0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1]
         BinWidth: 0.1000
        BinLimits: [0 1]
    Normalization: 'count'
        FaceColor: 'auto'
        EdgeColor: [0 0 0]

  Show all properties

counts = h.Values;
binCenters = ([h.BinEdges(1:end-1) + h.BinEdges(2:end)])/2;
% for loop to sum data instead of using mean() and std().
theSum = 0;
numBins = numel(binCenters);
for k = 1 : numBins
	sumInThisBin = counts(k) * binCenters(k);
	theSum = theSum + sumInThisBin;
end
nMinus1 = sum(counts) - 1;
theMean = theSum / nMinus1

theMean = 0.4965

% Compute variance
theSumsSquared = 0;
for k = 1 : numBins
	sumInThisBin = counts(k) * (binCenters(k) - theMean);
	theSumsSquared = theSumsSquared + sumInThisBin ^ 2;
end
theVariance = theSumsSquared / (sum(counts)-1);
stdDev = sqrt(theVariance)

stdDev = 9.1852

I think there is an error with the variance computation but I'll let you find it.

[Gurpreet Kaur]

Thanks! I believe this was the fix to the variance computation.

theSumsSquared = 0;
for k = 1 : numBins
	sumInThisBin = counts(k) * (binCenters(k) - theMean)^2;
	theSumsSquared = theSumsSquared + sumInThisBin;
end
theVariance = theSumsSquared / (sum(counts)-1);
stdDev = sqrt(theVariance)