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Here is a link that describes the formula to find the mode of grouped data.

Here is a link that gives a graphical method to finding the mode of grouped data.

Question: Can someone please explain how the formula corresponds with the graphical method? The formula is an interpolation but I am not able to see how it expresses the idea of the graphical method. If it helps, I understand the formula for calculating the mean of grouped data.

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I have not bothered to check the math at the link you gave but from the figure the mode is the intersection of the two diagonal lines. The end points of both the lines is known so all you need to do is to find out the intersection of those two lines to get the mode.

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    $\begingroup$ You are correct: the formula is an elementary plane geometry exercise. The lines create three triangles at the left, top, and right. The left and right ones are similar, whence the estimated mode partitions its bin in the same proportion as the bases of these triangles. In the notation of the formula, one base is fm-f1 and the other is fm-f2, so the fraction of the way to go across the bin equals (fm-f1) / (sum of the two bases). The interpolated value is obtained by starting at the left endpoint L and adding this fraction of the bin width h. $\endgroup$
    – whuber ♦
    Commented Nov 4, 2010 at 2:32
  • $\begingroup$ @Srikant: Thanks. I know it is only an estimate but is there any reason why the mode must be at the intersection of the two diagonal lines and not somewhere else? @whuber: Thanks. Your comments was very helpful in understanding the answer. $\endgroup$
    – Sara
    Commented Nov 4, 2010 at 9:55
  • $\begingroup$ @Sara The mode in general depends on how the data are grouped, so the best you can hope for is some reasonable estimate. I suspect this formula may originally have been derived by assuming numbers were drawn independently from a unimodal distribution and then grouped somehow. On the average the interpolated value ought to depart less from the mode of the underlying distribution than some other estimate (such as the midpoint or one of the endpoints of the largest group). When you know more about the distribution--e.g., if it's Normal--then this interpolator is not necessarily best on average. $\endgroup$
    – whuber ♦
    Commented Nov 4, 2010 at 15:30

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