You can also see the shape of a distribution based on a boxplot, although some features such as gaps are not visible. See the graphic below with a boxplot displaying the same data as the histogram below it for each image. See how "uniform" and "bimodal" look very similar? If a boxplot is your only graphical display, likely the best you could say is "mostly symmetric" if the boxplot has a somewhat symmetric appearance. Even though there is some ambiguity in shape, boxplots do reveal a skew pretty well.

Determine the shape of the distribution for each boxplot below.

a) [Graphs generated by this script: setBorder(15); initPicture(1,5,-3,4);axes(1,100,1,null,null,1,'off');text([2.975,-3],"data");line([1.03,2],[1.03,4]); rect([2.43,2],[3.42,4]); line([3.01,2],[3.01,4]);line([4.98,2],[4.98,4]); line([1.03,3],[2.43,3]); line([3.42,3],[4.98,3]);]

b) [Graphs generated by this script: setBorder(15); initPicture(1,6,-3,4);axes(1,100,1,null,null,1,'off');text([2.98,-3],"data");line([1.08,2],[1.08,4]); rect([2.99,2],[4.72,4]); line([3.96,2],[3.96,4]);line([5.04,2],[5.04,4]); line([1.08,3],[2.99,3]); line([4.72,3],[5.04,3]);]

c) [Graphs generated by this script: setBorder(15); initPicture(1,6,-3,4);axes(1,100,1,null,null,1,'off');text([2.98,-3],"data");line([1.07,2],[1.07,4]); rect([1.29,2],[2.9,4]); line([2.07,2],[2.07,4]);line([5.03,2],[5.03,4]); line([1.07,3],[1.29,3]); line([2.9,3],[5.03,3]);]

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