I came across this problem asking the lower quartile of the ungrouped data. My answer is 3, but other references say it should be 2.5. Here's the data:

1, 1, 2, 2, 3, 3, 4, 4, 6, 7, 8, 10, 11, 14, 15, 20, 22

What do you think?

  • 2
    $\begingroup$ $Q_1 = 2.5$ is indeed correct. How do you define the lower quartile? $\endgroup$ – Brian Apr 10 at 1:09
  • $\begingroup$ At least half a dozen definitions of 'quantile' (specifically, 'quartile') are in common use. They can give noticeably different answers for datasets as small as yours, especially when there are ties. The default method in R statistical software gives 3 for your 17 observations.That result happens to match @translocations's rule. $\endgroup$ – BruceET Apr 11 at 7:40

From a purely mathematical point of view, the first quartile $Q_1$ is the $25$th percentile and represents a cutoff, that should have the following two properties:

  • at most $25\%$ of the data is less than $Q_1$ and
  • at most $75\%$ of the data is greater than $Q_1$

So, $\color{blue}{Q_1 =3}$.

Note, that $\color{red}{2.5}$ does not satisfy the above given conditions: $\frac{13}{17} \approx 76.5\%$ of the data items are larger than $2.5$. Hence, $2.5$ does not quality as a quartile.

Nevertheless, some sources will continue to report the result $2.5$, because they may use a method, where one determines the lower and upper quartile as the median of a so-called lower or upper "data half".

The result then depends on whether one "includes" the median of the whole data set in such a "data half" or not and whether the data items near the quartile postition are equal or not. (Just note that if the data set were $112\color{blue}{3}33...$ the result would be $3$)

So, in the case that the data items aren't equal around the quartile position, then

  • If the median is included in the "data halves", then the median method is incorrect, if the data size is $4k+3$.
  • If the median is excluded in the "data halves", then the median method is incorrect, if the data size is $4k+1$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.