Measure of Position

Percentiles are a measure of the relative standing of observation within a data. Percentiles divide a set of observations into 100 equal parts, and percentile scores are frequently used to report results from national standardized tests such as NAT, GAT, and GRE, etc.

The $p$th percentile is the value $Y_{(p)}$ in order statistic such that $p$ percent of the values are less than the value $Y_{(p)}$ and $(100-p)$ (100-p) percent of the values are greater $Y_{(p)}$. The 5th percentile is denoted by $P_5$, the 10th by $P_{10}$ and 95th by $P_{95}$.

Percentiles for the Ungrouped data

To calculate percentiles (a measure of the relative standing of an observation) for the ungrouped data, adopt the following procedure:

1. Order the observation
2. For the $m$th percentile, determine the product $\frac{m.n}{100}$. If $\frac{m.n}{100}$ is not an integer, round it up and find the corresponding ordered value and if $\frac{m.n}{100}$ is an integer, say k, then calculate the mean of the $K$th and $(k+1)$th ordered observations.

Example: For the following height data collected from students find the 10th and 95th percentiles. 91, 89, 88, 87, 89, 91, 87, 92, 90, 98, 95, 97, 96, 100, 101, 96, 98, 99, 98, 100, 102, 99, 101, 105, 103, 107, 105, 106, 107, 112.

Solution: The ordered observations of the data are 87, 87, 88, 89, 89, 90, 91, 91, 92, 95, 96, 96, 97, 98, 98, 98, 99, 99, 100, 100, 101, 101, 102, 103, 105, 105, 106, 107, 107, 112.

\[P_{10}= \frac{10 \times 30}{100}=3\]

So the 10th percentile i.e. $P_{10}$ is the 3rd observation in sorted data is 88, which means that 10 percent of the observations in the data set are less than 88.

\[P_{95}=\frac{95 \times 30}{100}=28.5\]

The 29th observation is our 95th Percnetile i.e., $P_{95}=107$

Percentiles for the Frequency Distribution Table (Grouped data)

The $m$th percentile (a measure of the relative standing of an observation) for the Frequency Distribution Table (grouped data) is

\[P_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)\]

Like median, $\frac{m.n}{100}$ is used to locate the $m$th percentile group.

$l$    is the lower class boundary of the class containing the $m$th percentile
$h$   is the width of the class containing $P_m$
$f$    is the frequency of the class containing
$n$   is the total number of frequencies $P_m$
$c$    is the cumulative frequency of the class immediately preceding the class containing $P_m$

Note that the 50th percentile is the median by definition as half of the values in the data are smaller than the median and half of the values are larger than the median. Similarly, the 25th and 75th percentiles are the lower ($Q_1$) and upper quartiles ($Q_3$) respectively. The quartiles, deciles, and percentiles are also called quantiles or fractiles.

Example: For the following grouped data compute $P_{10}$, $P_{25}$, $P_{50}$, and $P_{95}$ given below.Solution:

1. Locate the 10th percentile (lower deciles i.e. $D_1$)by $\frac{10 \times n}{100}=\frac{10 \times 3o}{100}=3$ observation.
   so, $P_{10}$ group is 85.5–90.5 containing the 3rd observation
   \begin{align*}
   P_{10}&=l+\frac{h}{f}\left(\frac{10 n}{100}-c\right)\\
   &=85.5+\frac{5}{6}(3-0)\\
   &=85.5+2.5=88
   \end{align*}
2. Locate the 25th percentile (lower quartiles i.e. $Q_1$)  by $\frac{10 \times n}{100}=\frac{25 \times 3o}{100}=7.5$ observation.
   so, $P_{25}$ group is 90.5–95.5 containing the 7.5th observation
   \begin{align*}
   P_{25}&=l+\frac{h}{f}\left(\frac{25 n}{100}-c\right)\\
   &=90.5+\frac{5}{4}(7.5-6)\\
   &=90.5+1.875=92.375
   \end{align*}
3. Locate the 50th percentile (Median i.e. 2nd quartiles, 5th deciles) by $\frac{50 \times n}{100}=\frac{50 \times 3o}{100}=15$ observation.
   so, P₅₀ group is 95.5–100.5 containing the 15th observation
   \begin{align*}
   P_{50}&=l+\frac{h}{f}\left(\frac{50 n}{100}-c\right)\\
   &=95.5+\frac{5}{10}(15-10)\\
   &=95.5+2.5=98
   \end{align*}
4. Locate the 95th percentile by $\frac{95 \times n}{100}=\frac{95 \times 30}{100}=28.5$th observation.
   so, $P_{95}$ group is 105.5–110.5 containing the 3rd observation
   \begin{align*}
   P_{95}&=l+\frac{h}{f}\left(\frac{95 n}{100}-c\right)\\
   &=105.5+\frac{5}{3}(28.5-26)\\
   &=105.5+4.1667=109.6667
   \end{align*}

The percentiles and quartiles may be read directly from the graphs of the cumulative frequency function.

Further Reading: https://en.wikipedia.org/wiki/Percentile

Drawing Graphs and Charts in R Language

The deciles are the values (nine in number) of the variable that divides an ordered (sorted, arranged) data set into ten equal parts so that each part represents $\frac{1}{10}$ of the sample or population and are denoted by $D_1, D_2, \cdots D_9$, where First decile ($D_1$) is the value of order statistics that exceed 1/10 of the observations and less than the remaining $\frac{9}{10}$. The $D_9$ (ninth decile) is the value in order statistic that exceeds $\frac{9}{10}$ of the observations and is less than $\frac{1}{10}$ remaining observations. Note that the fifth deciles are equal to the median. The deciles determine the values for 10%, 20%… and 90% of the data.

Calculating Deciles for Ungrouped Data

To calculate the decile for the ungrouped data, first order all observations according to the magnitudes of the values, then use the following formula for $m$th decile.

\[D_m= m \times \left( \frac{(n+1)}{10} \right) \mbox{th value; } \qquad \mbox{where} m=1,2,\cdots,9\]

Example: Calculate the 2nd and 8th deciles of the following ordered data 13, 13,13, 20, 26, 27, 31, 34, 34, 34, 35, 35, 36, 37, 38, 41, 41, 41, 45, 47, 47, 47, 50, 51, 53, 54, 56, 62, 67, 82.
Solution:
\begin{eqnarray*}
D_m &=&m \times \{\frac{(n+1)}{10} \} \mbox{th value}\\
&=& 2 \times \frac{30+1}{10}=6.2\\
\end{eqnarray*}

We have to locate the sixth value in the ordered array and then move 0.2 of the distance between the sixth and seventh values. i.e. the value of the 2nd decile can be calculated as
\[6 \mbox{th observation} + \{7 \mbox{th observation} – 6 \mbox{th observation} \}\times 0.2\]
as 6th observation is 27 and 7th observation is 31.
The second decile would be $27+\{31-27\} \times 0.2 = 27.8$

Similarly, $D_8$ can be calculated. $D_8=52.6$.

Calculating Deciles for Grouped Data

The following formula can calculate the $m$th decile for grouped data (in ascending order).

\[D_m=l+\frac{h}{f}\left(\frac{m.n}{10}-c\right)\]

where

$l$ = is the lower class boundary of the class containing $m$th deciles
$h$ = is the width of the class containing $m$th deciles
$f$ = is the frequency of the class containing $m$th deciles
$n$ = is the total number of frequencies
$c$ = is the cumulative frequency of the class preceding the class containing $m$th deciles

Example: Calculate the first and third decile(s) of the following grouped data

Solution: The Decile class for $D_1$ can be calculated from $\left(\frac{m.n}{10}-c\right) = \frac{1 \times 30}{10} = 3$rd observation. As 3rd observation lies in the first class (first group) so

\begin{eqnarray*}
D_m&=&l+\frac{h}{f}\left(\frac{m.n}{10}-c\right)\\
D_1&=&85.5+\frac{5}{6}\left(\frac{1\times30}{10}-0\right)\\
&=&88\\
\end{eqnarray*}

The Decile class for $D_7$ is 100.5—105.5 as $\frac{7 \times 30}{10}=21$th observation which is in fourth class (group).
\begin{eqnarray*}
D_m&=&l+\frac{h}{f}\left(\frac{m.n}{10}-c\right)\\
D_7&=&100.5+\frac{5}{6}\left(\frac{7\times30}{10}-20\right)\\
&=&101.333\\
\end{eqnarray*}

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