Percentile

Example 1

Consider the ordered list {15, 20, 35, 40, 50}, which contains 5 data values. What are the 5th, 30th, 40th, 50th and 100th percentiles of this list using the nearest-rank method?

Percentile P	Number in list N	Ordinal rank n	Number from the ordered list that has that rank	Percentile value	Notes
5th	5	${\displaystyle \left\lceil {\frac {5}{100}}\times 5\right\rceil =\lceil 0.25\rceil =1}$	the first number in the ordered list, which is 15	15	15 is the smallest element of the list; 0% of the data is strictly less than 15, and 20% of the data is less than or equal to 15.
30th	5	${\displaystyle \left\lceil {\frac {30}{100}}\times 5\right\rceil =\lceil 1.5\rceil =2}$	the 2nd number in the ordered list, which is 20	20	20 is an element of the ordered list.
40th	5	${\displaystyle \left\lceil {\frac {40}{100}}\times 5\right\rceil =\lceil 2.0\rceil =2}$	the 2nd number in the ordered list, which is 20	20	In this example, it is the same as the 30th percentile.
50th	5	${\displaystyle \left\lceil {\frac {50}{100}}\times 5\right\rceil =\lceil 2.5\rceil =3}$	the 3rd number in the ordered list, which is 35	35	35 is an element of the ordered list.
100th	5	${\displaystyle \left\lceil {\frac {100}{100}}\times 5\right\rceil =\lceil 5\rceil =5}$	the last number in the ordered list, which is 50	50	The 100th percentile is defined to be the largest value in the list, which is 50.

So the 5th, 30th, 40th, 50th and 100th percentiles of the ordered list {15, 20, 35, 40, 50} using the nearest-rank method are {15, 20, 20, 35, 50}.

Example 2

Consider an ordered population of 10 data values {3, 6, 7, 8, 8, 10, 13, 15, 16, 20}. What are the 25th, 50th, 75th and 100th percentiles of this list using the nearest-rank method?

Percentile P	Number in list N	Ordinal rank n	Number from the ordered list that has that rank	Percentile value	Notes
25th	10	${\displaystyle \left\lceil {\frac {25}{100}}\times 10\right\rceil =\lceil 2.5\rceil =3}$	the 3rd number in the ordered list, which is 7	7	7 is an element of the list.
50th	10	${\displaystyle \left\lceil {\frac {50}{100}}\times 10\right\rceil =\lceil 5.0\rceil =5}$	the 5th number in the ordered list, which is 8	8	8 is an element of the list.
75th	10	${\displaystyle \left\lceil {\frac {75}{100}}\times 10\right\rceil =\lceil 7.5\rceil =8}$	the 8th number in the ordered list, which is 15	15	15 is an element of the list.
100th	10	Last	20, which is the last number in the ordered list	20	The 100th percentile is defined to be the largest value in the list, which is 20.

So the 25th, 50th, 75th and 100th percentiles of the ordered list {3, 6, 7, 8, 8, 10, 13, 15, 16, 20} using the nearest-rank method are {7, 8, 15, 20}.

Example 3

Consider an ordered population of 11 data values {3, 6, 7, 8, 8, 9, 10, 13, 15, 16, 20}. What are the 25th, 50th, 75th and 100th percentiles of this list using the nearest-rank method?

Percentile P	Number in list N	Ordinal rank n	Number from the ordered list that has that rank	Percentile value	Notes
25th	11	${\displaystyle \left\lceil {\frac {25}{100}}\times 11\right\rceil =\lceil 2.75\rceil =3}$	the 3rd number in the ordered list, which is 7	7	7 is an element of the list.
50th	11	${\displaystyle \left\lceil {\frac {50}{100}}\times 11\right\rceil =\lceil 5.50\rceil =6}$	the 6th number in the ordered list, which is 9	9	9 is an element of the list.
75th	11	${\displaystyle \left\lceil {\frac {75}{100}}\times 11\right\rceil =\lceil 8.25\rceil =9}$	the 9th number in the ordered list, which is 15	15	15 is an element of the list.
100th	11	Last	20, which is the last number in the ordered list	20	The 100th percentile is defined to be the largest value in the list, which is 20.

So the 25th, 50th, 75th and 100th percentiles of the ordered list {3, 6, 7, 8, 8, 9, 10, 13, 15, 16, 20} using the nearest-rank method are {7, 9, 15, 20}.

All of the following variants have the following in common. Given the order statistics

${\displaystyle \{v_{i},i=1,2,...,N:v_{i+1}\geq v_{i},\forall i=1,2,...N-1\},}$

we seek a linear interpolation function that passes through the points ${\displaystyle (v_{i},i)}$. This is simply accomplished by

${\displaystyle v(x)=v_{\lfloor x\rfloor }+(x\%1)(v_{\lfloor x\rfloor +1}-v_{\lfloor x\rfloor }),\forall x\in [1,N]:v(i)=v_{i}{\text{, for }}i=1,2,\ldots ,N,}$

where ${\displaystyle \lfloor x\rfloor }$ uses the floor function to represent the integral part of positive ${\displaystyle x}$, whereas ${\displaystyle x\%1}$ uses the mod function to represent its fractional part (the remainder after division by 1). (Note that, though at the endpoint ${\displaystyle x=N}$, ${\displaystyle v_{\lfloor x\rfloor +1}}$ is undefined, it does not need to be because it is multiplied by ${\displaystyle x\%1=0}$.) As we can see, ${\displaystyle x}$ is the continuous version of the subscript ${\displaystyle i}$, linearly interpolating ${\displaystyle v}$ between adjacent nodes.

There are two ways in which the variant approaches differ. The first is in the linear relationship between the rank ${\displaystyle x}$, the percent rank ${\displaystyle P=100p}$, and a constant that is a function of the sample size ${\displaystyle N}$:

${\displaystyle x=f(p,N)=(N+c_{1})p+c_{2}.}$

There is the additional requirement that the midpoint of the range ${\displaystyle (1,N)}$, corresponding to the median, occur at ${\displaystyle p=0.5}$:

${\displaystyle f(0.5,N)={\frac {N+c_{1}}{2}}+c_{2}={\frac {N+1}{2}}\therefore 2c_{2}+c_{1}=1,}$

and our revised function now has just one degree of freedom, looking like this:

${\displaystyle x=f(p,N)=(N+1-2C)p+C.}$

The second way in which the variants differ is in the definition of the function near the margins of the ${\displaystyle [0,1]}$ range of ${\displaystyle p}$: ${\displaystyle f(p,N)}$ should produce, or be forced to produce, a result in the range ${\displaystyle [1,N]}$, which may mean the absence of a one-to-one correspondence in the wider region. One author has suggested a choice of ${\displaystyle C={\textstyle {\frac {1}{2}}}(1+\xi )}$ where ${\displaystyle \xi }$ is the shape of the Generalized extreme value distribution which is the extreme value limit of the sampled distribution.

The result of using each of the three variants on the ordered list {15, 20, 35, 40, 50}

(Sources: Matlab "prctile" function,[8][9])

${\displaystyle x=f(p)={\begin{cases}Np+{\frac {1}{2}},\forall p\in \left[p_{1},p_{N}\right],\\1,\forall p\in \left[0,p_{1}\right],\\N,\forall p\in \left[p_{N},1\right].\end{cases}},}$

where

${\displaystyle p_{i}={\frac {1}{N}}\left(i-{\frac {1}{2}}\right),i\in [1,N]\cap \mathbb {N} }$

${\displaystyle \therefore p_{1}={\frac {1}{2N}},p_{N}={\frac {2N-1}{2N}}.}$

Furthermore, let

${\displaystyle P_{i}=100p_{i}.}$

The inverse relationship is restricted to a narrower region:

${\displaystyle p={\frac {1}{N}}\left(x-{\frac {1}{2}}\right),x\in (1,N)\cap \mathbb {R} .}$

Consider the ordered list {15, 20, 35, 40, 50}, which contains five data values. What are the 5th, 30th, 40th and 95th percentiles of this list using the Linear Interpolation Between Closest Ranks method? First, we calculate the percent rank for each list value.

List value ${\displaystyle v_{i}}$	Position of that value in the ordered list ${\displaystyle i}$	Number of values ${\displaystyle N}$	Calculation of percent rank	Percent rank, ${\displaystyle P_{i}}$	Notes
15	1	5	${\displaystyle {\frac {100}{5}}\left(1-{\frac {1}{2}}\right)=10.}$	10
20	2	5	${\displaystyle {\frac {100}{5}}\left(2-{\frac {1}{2}}\right)=30.}$	30
35	3	5	${\displaystyle {\frac {100}{5}}\left(3-{\frac {1}{2}}\right)=50.}$	50
40	4	5	${\displaystyle {\frac {100}{5}}\left(4-{\frac {1}{2}}\right)=70.}$	70
50	5	5	${\displaystyle {\frac {100}{5}}\left(5-{\frac {1}{2}}\right)=90.}$	90

Then we take those percent ranks and calculate the percentile values as follows:

Percent rank ${\displaystyle P}$	Number of values ${\displaystyle N}$	Is ${\displaystyle P<P_{1}}$?	Is ${\displaystyle P>P_{n}}$?	Is there a percent rank equal to ${\displaystyle P}$?	What do we use for percentile value?	Percentile value ${\displaystyle v(f(p))}$	Notes
5	5	Yes	No	No	We see that P=5, which is less than the first percent rank p1=10, so use the first list value v1, which is 15	15	15 is a member of the ordered list
30	5	No	No	Yes	We see that P=30 is the same as the second percent rank p2=30, so use the second list value v2, which is 20	20	20 is a member of the ordered list
40	5	No	No	No	We see that P=40 is between percent rank p2=30 and p3=50, so we take k=2, k+1=3, P=40, pk=p2=30, vk=v2=20, vk+1=v3=35, N=5. Given those values we can then calculate v as follows: ${\displaystyle v=20+5\times {\frac {40-30}{100}}(35-20)=27.5}$	27.5	27.5 is not a member of the ordered list
95	5	No	Yes	No	We see that P=95, which is greater than the last percent rank pN=90, so use the last list value, which is 50	50	50 is a member of the ordered list

So the 5th, 30th, 40th and 95th percentiles of the ordered list {15, 20, 35, 40, 50} using the Linear Interpolation Between Closest Ranks method are {15, 20, 27.5, 50}

(Source: Some software packages, including NumPy[10] and Microsoft Excel[6] (up to and including version 2013 by means of the PERCENTILE.INC function). Noted as an alternative by NIST[2])

${\displaystyle x=f(p,N)=p(N-1)+1{\text{, }}p\in [0,1]}$

${\displaystyle \therefore p={\frac {x-1}{N-1}}{\text{, }}x\in [1,N].}$

Note that the ${\displaystyle x\leftrightarrow p}$ relationship is one-to-one for ${\displaystyle p\in [0,1]}$, the only one of the three variants with this property; hence the "INC" suffix, for inclusive, on the Excel function.

Example 1:

Consider the ordered list {15, 20, 35, 40, 50}, which contains five data values. What is the 40th percentile of this list using this variant method?

First we calculate the rank of the 40th percentile:

${\displaystyle x={\frac {40}{100}}(5-1)+1=2.6}$

So, x=2.6, which gives us ${\displaystyle \lfloor x\rfloor =2}$ and ${\displaystyle x\%1=0.6}$. So, the value of the 40th percentile is

${\displaystyle v(2.6)=v_{2}+0.6(v_{3}-v_{2})=20+0.6(35-20)=29.}$

Example 2:

Consider the ordered list {1,2,3,4} which contains four data values. What is the 75th percentile of this list using the Microsoft Excel method?

First we calculate the rank of the 75th percentile as follows:

${\displaystyle x={\frac {75}{100}}(4-1)+1=3.25}$

So, x=3.25, which gives us an integral part of 3 and a fractional part of 0.25. So, the value of the 75th percentile is

${\displaystyle v(3.25)=v_{3}+0.25(v_{4}-v_{3})=3+0.25(4-3)=3.25.}$

(The primary variant recommended by NIST.[2] Adopted by Microsoft Excel since 2010 by means of PERCENTIL.EXC function. However, as the "EXC" suffix indicates, the Excel version excludes both endpoints of the range of p, i.e., ${\displaystyle p\in (0,1)}$, whereas the "INC" version, the second variant, does not; in fact, any number smaller than 1/(N+1) is also excluded and would cause an error.)

${\displaystyle x=f(p,N)={\begin{cases}1{\text{, }}p\in \left[0,{\frac {1}{N+1}}\right]\\p(N+1){\text{, }}p\in \left({\frac {1}{N+1}},{\frac {N}{N+1}}\right)\\N{\text{, }}p\in \left[{\frac {N}{N+1}},1\right]\end{cases}}.}$

The inverse is restricted to a narrower region:

${\displaystyle p={\frac {x}{N+1}}{\text{, }}x\in (0,N).}$

Consider the ordered list {15, 20, 35, 40, 50}, which contains five data values. What is the 40th percentile of this list using the NIST method?

First we calculate the rank of the 40th percentile as follows:

${\displaystyle x={\frac {40}{100}}(5+1)=2.4}$

So x=2.4, which gives us ${\displaystyle \lfloor x\rfloor =2}$ and ${\displaystyle x\%1=0.4}$. So the value of the 40th percentile is calculated as:

${\displaystyle v(2.4)=v_{2}+0.4(v_{3}-v_{2})=20+0.4(35-20)=26}$

So the value of the 40th percentile of the ordered list {15, 20, 35, 40, 50} using this variant method is 26.