My data has numbers such as 70.35, 70.39, 70.1. None of the numbers are exactly the same. How do I find the mode?
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$\begingroup$ Hint: Arrange the numbers from lowest to highest. The most frequently occurring number (irrespective of whether it has decimals or not) will be the mode. (This is subject to some minor caveats though, as you will see in online discussions about the mode.) $\endgroup$– Jose Arnaldo Bebita DrisCommented Jul 23, 2021 at 3:01
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2$\begingroup$ If no two numbers are the same, then there is no mode. If you need to have a mode, round the numbers to the nearest integer. $\endgroup$– Gerry MyersonCommented Jul 23, 2021 at 3:02
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3 Answers
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The mode is the value occurring with the highest frequency in a data set. If there is a "tie" (with two or more values having equal highest frequencies) you can have more than one mode - bimodal, multimodal etc. However, if every data value occurs exactly one, you generally don't consider that multimodal, you usually just say there's no mode.
In your data set, there is no mode. However, it is possible to transform the data set by truncation, rounding or grouping. For instance if we denoted your (single) variable $x$ and assigned non-overlapping classes like $70.0<x \leq 70.1, 70.1<x \leq 70.2,70.2<x \leq 70.3, 70.3<x \leq 70.4$ then you could argue that your data set has modal class $70.3<x \leq 70.4$.
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The modes of a dataset are those values with the greatest frequency of occurrence. This definition allows that there may be more than one mode, when several tie for the position of most frequent.
When all values in the dataset occur with a frequency of "exactly once", they are all modes.
answered Jul 23, 2021 at 3:05
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You can fit a probability density to your data. For example, see this article https://en.wikipedia.org/wiki/Kernel_density_estimation or use an in-built method in Python (scipy.stats or sckit-learn). The mode can then be defined as the point where you have a peak in the estimated probability density. This method will work if you have a large enough dataset to do a fit.
