* Percentiles* are

*measure of relative standing*of an observation within a data.

*divides a set of observations into 100 equal parts, and*

**Percentiles***are frequently used to report results from national standardized tests such as NAT, GAT etc.*

**percentile scores**The *p*th * percentile *is the value

*Y*in order statistic such that

_{(p) }*p*percent of the values are less than the value

*Y*

_{(p) }and

*(100-p)*percent of the values are greater

*Y*. The 5th

_{(p) }*is denoted by*

**percentile***P*, the 10th by

_{5 }*P*and 95th by

_{10}*P*.

_{95 }**Percentiles for the ungrouped data**

To calculate * percentiles (measure of relative standing of an observation) *for the

*ungrouped data*, adopt the following procedure

- Order the observation
- For the
*m*thdetermine 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**percentile,***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***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.*

**percentiles.****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*is 3rd observation in sorted data is 88, means that 10 percent of the observations in data set are less than 88.

_{10 }\[P_{95}=\frac{95 \times 30}{100}=28.5\]

**95th****percentile**i.e.

*P*=107.

_{95}**Percentiles for the Grouped data**

The *m*th * percentile* (

*) for grouped data is*

**measure of relative standing of an observation**\[P_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)\]

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 to the class containing

*P*

_{m}Note that 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

*and*

**25th***are the*

**75th percentiles***respectively. The*

**lower (Q**_{1}) and upper quartiles (Q_{3})

*quartiles,**and*

**deciles***are also called*

**percentiles***quantiles*or

*fractiles.*

**Example:** For the following grouped data compute *P _{10 }*,

*P*,

_{25 }*P5*, and

_{0 }*P*given below.

_{95}**Solution:**

- Locate the
(lower**10th percentile**i.e.**deciles***D*)by $\frac{10 \times n}{100}=\frac{10 \times 3o}{100}=3$ observation._{1}

so,*P*group is 85.5–90.5 containing the 3rd observation_{10}

\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*} - Locate the
(lower**25th percentile**i.e.*quartiles**Q*) by $\frac{10 \times n}{100}=\frac{25 \times 3o}{100}=7.5$ observation._{1}

so,*P*group is 90.5–95.5 containing the 7.5th observation_{25}

\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*} - Locate the 50th percentile (Median i.e. 2nd
, 5th*quartiles*by $\frac{50 \times n}{100}=\frac{50 \times 3o}{100}=15$ observation.**deciles)**

so,*P*group is 95.5–100.5 containing the 15th observation_{50}

\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*} - Locate the
by $\frac{95 \times n}{100}=\frac{95 \times 3o}{100}=28.5$th observation.**95th percentile**

so,*P*group is 105.5–110.5 containing the 3rd observation_{95}

\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

**may be read directly from the graphs of cumulative frequency function.**

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

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