# Tagged: Measure of Position

## Quantiles or Fractiles

When the number of observations is sufficiently large, the principle by which a distribution is divided into two equal parts may be extended to divide the distribution into four, five, eight, ten, or hundred equal parts. Let us start learning about Quantiles or Fractiles.

#### Quartiles

These are the values that divide a distribution into four equal parts. There are three quartiles denoted by $Q_1, Q_2$, and $Q_3$. If $x_1,x_2,\cdots,x_n$ are $n$ observations on a variable $X$, and $x_{(1)}, x_{(2)}, \cdots, x_{(n)}$ is their array then $r$th quartile $Q_r$ is the values of $X$, such that $\frac{r}{4}$ of the observations is less than that value of $X$ and $\frac{4-r}{4}$ of the observations is greater. Thus $Q_1$ is the value of $X$ such that $\frac{1}{4}$ of the observations is less than the value of $X$ and $\frac{4-1}{4}$ of the observations is greater, the $Q_3$ is the value of $X$, such that $\frac{3}{4}$ of the observations is less than that of $X$ and $\frac{4-3}{4}$ of the observations is greater.

#### Deciles

These are the values that divide a distribution into ten equal parts. There are 9 deciles $D_1, D_2, \cdots, D_9$.

#### Percentiles

These are the values that divide a distribution into a hundred equal parts. There are 99 percentiles denoted as $P_1,P_2,\cdots, P_{99}$.

The median, quartiles, deciles, percentiles, and other partition values are collectively called quantiles or fractiles. All quantiles are percentages. For example, $P_{50}, Q_2$, and $D_5$ are also median.

\begin{align*}
Q_2 &= D_5 = P_{50}\\
Q_1 &= P_{25} = D_{2.5}\\
Q_3 &= P_{75}=D_{7.5}
\end{align*}
The $r$th quantile, $k$th decile, and $j$th percentile are located in the array by the following relation:
For ungrouped Date

\begin{align}
Q_r &=\frac{r(n+1)}{4}\text{th value in the distribution and } r=1,2,3\\
D_k &=\frac{k(n+1)}{10}\text{th value in the distribution and } k=1,2,\cdots, 9\\
P_j &=\frac{j(n+1)}{100}\text{th value in the distribution and } k=1,2,\cdots, 99
\end{align}
For grouped Data
\begin{align}
Q_r&= l+\frac{h}{f}\left(\frac{rn}{4}-c\right)\\
D_k&= l+\frac{h}{f}\left(\frac{kn}{10}-c\right)\\
P_j&= l+\frac{h}{f}\left(\frac{jn}{100}-c\right)
\end{align}

A procedure for obtaining percentile (quartiles, deciles) of a data set of size $n$ is as follows:

Step 1: Arrange the data in ascending/ descending order.
Step 2: Compute an index $i$ as follows: $i=\frac{p}{100} (n+1)$th (in case of odd observation).

• If $i$ is an integer, the $p$th percentile is the average of the $i$th and $(i+1)$th data values.
• if $i$ is not an integer then round $i$ up to the nearest integer and take the value at that position or use some mathematics to locate the value of percentile between $i$th and $(i+1)$th value.

Percentile Example:

Consider the following (sorted) data values: 380, 600, 690, 890, 1050, 1100, 1200, 1900, 890000.

For the $p=10$th percentile, $i=\frac{p}{100} (n+1) =\frac{10}{100} (9+1)= 1$. So the 10th percentile is the first sorted value or 380.

For the $p=75$ percentile, $i=\frac{p}{100} (n+1)= \frac{75}{100}(9+1) = 7.5$

To get the actual value we need to compute 7th value + (8th value – 7th value) $\times 0.5$. That is, $1200 + (1900-1200)\times 0.5 = 1200+350 = 1550$.

## Quartiles: Measure of relative standing of an observation within data

Like Percentile and Deciles, Quartiles is a type of Quantile, which is a measure of the relative standing of observation within the data set. Quartiles are the values are three points that divide the data into four equal parts each group comprising a quarter of the data (the first quartile $Q_1$, second quartile $Q_2$ (also median) and the third quartile $Q_3$) in the order statistics. The first quartile, (also known as the lower quartile) is the value of order statistic that exceeds 1/4 of the observations and less than the remaining 3/4 observations. The third quartile is known as upper quartile is the value in the order statistic that exceeds 3/4 of the observations and is less than remaining 1/4 observations, while the second quartile is the median.

## Quartiles for Ungrouped Data

For ungrouped data, the quartiles are calculated by splitting the order statistic at the median and then calculating the median of the two halves. If n is odd, the median can be included in both sides.

Example: Find the $Q_1, Q_2$ and $Q_3$ for the following ungrouped data set 88.03, 94.50, 94.90, 95.05, 84.60.Solution: We split the order statistic at the median and calculate the median of two halves. Since n is odd, we can include the median in both halves. The order statistic is 84.60, 88.03, 94.50, 94.90, 95.05.

\begin{align*}
Q_2&=median=Y_{(\frac{n+1}{2})}=Y_{(3)}\\
&=94.50  (\text{the third observation})\\
Q_1&=\text{Median of the first three value}=Y_{(\frac{3+1}{2})}\\&=Y_{(2)}=88.03 (\text{the second observation})\\
Q_3&=\text{Median of the last three values}=Y_{(\frac{3+5}{2})}\\
&=Y_{(4)}=94.90 (\text{the forth observation})
\end{align*}

## Quartiles for Grouped Data

For the grouped data (in ascending order) the quartiles are calculated as:
\begin{align*}
Q_1&=l+\frac{h}{f}(\frac{n}{4}-c)\\
Q_2&=l+\frac{h}{f}(\frac{2n}{4}-c)\\
Q_3&=l+\frac{h}{f}(\frac{3n}{4}-c)
\end{align*}
where
l    is the lower class boundary of the class containing the $Q_1,Q_2$ or $Q_3$.
h    is the width of the class containing the $Q_1,Q_2$ or $Q_3$.
f    is the frequency of the class containing the $Q_1,Q_2$ or $Q_3$.
c    is the cumulative frequency of the class immediately preceding to the class containing $Q_1,Q_2$ or $Q_3, \left[\frac{n}{4},\frac{2n}{4} \text{or} \frac{3n}{4}\right]$ are used to locate $Q_1,Q_2$ or $Q_3$ group.

Example: Find the quartiles for the following grouped data Solution: To locate the class containing $Q_1$, find $\frac{n}{4}$th observation which is here $\frac{30}{4}$th observation i.e. 7.5th observation. Note that 7.5th observation falls in the group ($Q_1$ group) 90.5–95.5.
\begin{align*}
Q_1&=l+\frac{h}{f}(\frac{n}{4}-c)\\
&=90.5+\frac{5}{4}(7.5-6)=90.3750
\end{align*}

For $Q_2$, the $\frac{2n}{4}$th observation=$\frac{2 \times 30}{4}$th observation = 15th observation falls in the group 95.5–100.5.
\begin{align*}
Q_2&=l+\frac{h}{f}(\frac{2n}{4}-c)\\
&=95.5+\frac{5}{10}(15-10)=98
\end{align*}

For $Q_3$, the $\frac{3n}{4}$th observation=$\frac{3\times 30}{4}$th = 22.5th observation. So
\begin{align*}
Q_3&=l+\frac{h}{f}(\frac{3n}{4}-c)\\
&=100.5+\frac{5}{6}(22.5-20)=102.5833
\end{align*}

Reference:

## Percentiles: Measure of relative standing of an observation within data

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, etc.

The pth 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) 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 mth 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 Kth 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  P10 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.

$P_{95}=\frac{95 \times 30}{100}=28.5$

29th observation is our 95th percentile i.e. P95=107.

## Percentiles for the Grouped data

The mth percentile (a measure of the relative standing of an observation) for 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 mth percentile group.

l     is the lower class boundary of the class containing the mth percentile
h   is the width of the class containing Pm
f    is the frequency of the class containing
n   is the total number of frequencies Pm
c    is the cumulative frequency of the class immediately preceding to the class containing Pm

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, 25th and 75th percentiles are the lower (Q1) and upper quartiles (Q3) respectively. The quartiles, deciles, and percentiles are also called quantiles or fractiles.

Example: For the following grouped data compute P10 , P25 , P50 , and P95 given below.Solution:

1. Locate the 10th percentile (lower deciles i.e. D1)by $\frac{10 \times n}{100}=\frac{10 \times 3o}{100}=3$ observation.
so, P10 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. Q1)  by $\frac{10 \times n}{100}=\frac{25 \times 3o}{100}=7.5$ observation.
so, P25 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 $\frac{95 \times n}{100}=\frac{95 \times 3o}{100}=28.5$th observation.
so, P95 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 cumulative frequency function.

## The deciles: Measure of Position

The deciles are the values (nine in numbers) of the variable that divide an ordered (sorted, arranged) data set into ten equal parts so that each part represents 1/10 of the sample or population, and are denoted by $D_1, D_2, \cdots D_9$, where First decile (D1) is the value of order statistics that exceed 1/10 of the observations and less than the remaining 9/10 and the D9 (ninth decile) is the value in order statistic that exceeds 9/10 of the observations and is less than 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 for Ungrouped Data

To calculate deciles for the ungrouped data, first order the all observation according to the magnitudes of the values, then use the following formula for mth decile.

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

Example: Calculate 2nd and 8th deciles of 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 have to more 0.2 of the distance between the sixth and seventh values. i.e. the value of 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 can be calculated. D8 = 52.6.

## Calculating for Grouped Data

The mth decile for grouped data (in ascending order) can be calculated from the following formula.

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

where

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

Example: Calculate the first and third deciles of the following grouped data Solution: Deciles class for D1 can be calculated from $\left(\frac{m.n}{10}-c\right) = \frac{1 \times 30}{10} = 3$rd observation. As 3rd observation lie in 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*}

Deciles class for D7 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*}