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I am teaching an introductory unit on descriptive statistics. One topic is finding the median of a raw set of data.

I know the formula for finding the position of the median is $(n+1)/2$.

My trouble: I cannot think of a simple clear justification of this formula.

I understand that if I have an ascending (ordered) data set $X$ of size $n$, then the smallest value would be $x_1$, the value in position $i$ is $x_i$ and the largest data value would be $x_n$.

Since the numbering of the data values is arithmetic, I can think of the numbering as a uniformly distributed (discrete) set, and the mean of the indices would be the centroid of the set. The mean of the set of indices is the sum of the first $n$ integers, which is $n(n+1)/2$, divided by $n$, which results in $(n+1)/2$.

But I don't believe this is simple or clear for my students. Would someone provide a simpler justification for the formula $(n+1)/2$?

Assume my students have a working knowledge of Algebra 2/Precalculus.

Tim Thayer
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1 Answers1

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When $n$, the number of numbers in your set, is odd, the median is one of the numbers; half of the remaining $n-1$ numbers are below it. So the median is at position $(n-1)/2+1=(n+1)/2$ in the list that arranges the numbers in nondecreasing order.

Of course, when $n$ is even, the standard definition of the median (at this level, in primary or secondary school) is the average of the middle two terms, which are at positions $n/2$ and $n/2+1$ respectively.

symplectomorphic
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  • I believe my students will be able to follow these justifications (the even argument will be easier for them than the odd argument, but I'll get 'em there). Thanks! – Tim Thayer Aug 17 '17 at 02:30
  • "Of course" is a little strong here. If there are six numbers, anything between the third and fourth is a median. – Henry Jun 30 '21 at 13:29
  • @Henry: in college, yes, but not in secondary school, where the “the” in “the median” is taken literally for all data sets, at least in US schools. – symplectomorphic Jun 30 '21 at 13:55