I am aware of the definition for outliers in a Box-Whisker-Plot:

  • ๐‘ž1โˆ’๐‘˜ร—IQR
  • ๐‘ž3+๐‘˜ร—IQR

with IQR := ๐‘ž3โˆ’๐‘ž1

The question at hand is: (Maximum) How many outliers are possible in a Box-Whisker-Plot for a sample of 66 observations (numerical variable)?

I just don't see the logic behind it. May be you have any hint.


1 Answer 1


Your IQR will cover the middle 33 observations. Set all of those equal to zero. Then the IQR is zero, and any nonzero point will be an outlier. Thus, half of the points can be "outliers" by your definition.

(There is some funkiness because of the even number of observations, but there are many different ways to calculate empirical IQR, and this is the gist. Because of the even number of observations, it seems that the answer is 32 points instead of 33, and that the general solution is going to be: $$\left \lfloor{\dfrac{n - 1}{2}}\right \rfloor .$$

(That is, round down the $x.5$ number for an even $n$, or take the integer value for an odd $n$.)

x <- c(rep(-100, 16), rep(0, 34), rep(100, 16)
y <- c(rep(-100, 17), rep(0, 33), rep(100, 17)
boxplot(x, y)
  • $\begingroup$ Tukey's original construction of a boxplot used hinges rather than quartiles. With 66 data values, the hinges are the 17th highest and 17th lowest, leaving at most (17-1)+(17-1)=32 points beyond the box. The ordinal 17 arises because the median is the (66+1)/2 = 33h position (the mean of the 33rd highest and 33rd lowest values) and the hinges are the medians of the top 33 and bottom 33 values. I give a little more detail at stats.stackexchange.com/a/134239/919. The boxplot is a visual representation of the 5-letter summary: stats.stackexchange.com/a/87096/919. $\endgroup$
    – whuber
    Commented Jun 8, 2021 at 16:21

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