Median Formula

Is the 50th percentile also the median?

Where \(\displaystyle n =\) number of list. \(\displaystyle P\) = percentile

\(\displaystyle \dfrac{P}{100}(n + 1) =\) rank

\(\displaystyle \dfrac{50}{100}(n + 1) =\) median

How about in this situation?

Find the median of 1, 2, 3, 4, 5

\(\displaystyle \dfrac{50}{100} (5 + 1) = 3\) median

Jason76 said:

How about in this situation?

Find the median of 1, 2, 3, 4, 5

\(\displaystyle \dfrac{50}{100} (5 + 1) = 3\) median

Click to expand...

You have a wrong idea about percentiles. Medians and percentiles are about the number of instances, not the numeric value of the instances.

1, 2, 3, 4, 7. Median is still 3.

Quite obviously \(\displaystyle \dfrac{50}{100} * (7 + 1) = \dfrac{1}{2} * 8 = 4 \ne 3.\)

Median, quartile, etc are about counting an ordered set, not adding anything.

Given numbers \(\displaystyle a_1\), \(\displaystyle a_2\), ..., \(\displaystyle a_n\). Then the midrange is \(\displaystyle \dfrac{a_1+ a_n}{2}\).

The mean is \(\displaystyle \dfrac{a_2+ a_2+ \cdot\cdot\cdot+ a_n}{n}\).

The median is \(\displaystyle a_{(n+1)/2}\), the middle number, if n is odd. If n is even, the median is half way between the two middle numbers, \(\displaystyle \dfrac{a_{n/2}+ a_{(n+2)/2}}{2}\)

IF the numbers form an arithmetic sequence so that, as in your example, there is the same difference between all consecutive pairs of numbers, then "mean" ,"midrange", and "median" are the same.