## Percentiles for ungrouped data

Percentiles are the values of arranged data which divide whole data into **hundred** equal parts. They are 9 in numbers namely `$P_1,P_2, \cdots, P_{99}$`

. Here `$P_1$`

is first percentile, `$P_2$`

is second percentile, `$P_3$`

is third percentile and so on.

## Formula

The formula for $i^{th}$ percentile is

$P_i =$ Value of $\bigg(\dfrac{i(n+1)}{100}\bigg)^{th}$ observation, $i=1,2,3,\cdots,99$

where $n$ is the total number of observations.

## Example 1

The test score of a sample of 20 students in a class are as follows:

20,30,21,29,10,17,18,15,27,25,16,15,19,22,13,17,14,18,12 and 9.

Find the value of `$P_{10}$`

, `$P_{20}$`

and `$P_{80}$`

.

### Solution

The sample size is $n = 20$.

The formula for $i^{th}$ percentile is

$P_i =$ Value of $\bigg(\dfrac{i(n+1)}{100}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 99$

where $n$ is the total number of observations.

**Arrange the data in ascending order**

9, 10, 12, 13, 14, 15, 15, 16, 17, 17, 18, 18, 19, 20, 21, 22, 25, 27, 29, 30

**Tenth percentile $P_{10}$**

The tenth percentile $P_{10}$ can be computed as follows:

`$$ \begin{aligned} P_{10} &=\text{Value of }\bigg(\dfrac{10(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{10(20+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(2.1\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}+0.1 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=10+0.1\big(12 -10\big)\\ &=10.2 \end{aligned} $$`

Thus, lower $10$ % of the students had test score less than or equal to $10.2$.

**Twentieth percentile $P_{20}$**

The twentieth percentile $P_{20}$ can be computed as follows:

`$$ \begin{aligned} P_{20} &=\text{Value of }\bigg(\dfrac{20(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{20(20+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(4.2\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}+0.2 \big(\text{Value of } \big(5\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=13+0.2\big(14 -13\big)\\ &=13.2 \end{aligned} $$`

Thus, lower $20$ % of the students had test score less than or equal to $13.2$.

**Eightieth percentile $P_{80}$**

The eightieth percentile $P_{80}$ can be computed as follows:

`$$ \begin{aligned} P_{80} &=\text{Value of }\bigg(\dfrac{80(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{80(20+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(16.8\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(16\big)^{th} \text{ obs.}+0.8 \big(\text{Value of } \big(17\big)^{th}\text{ obs.}-\text{Value of }\big(16\big)^{th} \text{ obs.}\big)\\ &=22+0.8\big(25 -22\big)\\ &=24.4 \end{aligned} $$`

Thus, lower $80$ % of the students had test score less than or equal to $24.4$.

## Example 2

The following data gives the hourly wage rates (in dollars) of 25 employees of a company.

20, 28, 30, 18, 27, 19, 22, 21, 24, 25, 18, 25, 20, 27, 24, 20, 23, 32, 20, 35, 22, 26, 25, 28, 31.

a. the upper wage rate for the lowest 15 % of the employees,

b. the upper wage rate for the lowest 45 % of the employees,

c. the lower wage rate for the upper 25 % of the employees.

### Solution

The sample size is $n = 25$.

**Arrange the data in ascending order**

18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 23, 24, 24, 25, 25, 25, 26, 27, 27, 28, 28, 30, 31, 32, 35

a. The upper wage rate for the lowest $15$ % of the employees is $P_{15}$.

The fifteenth percentile $P_{15}$ can be computed as follows:

`$$ \begin{aligned} P_{15} &=\text{Value of }\bigg(\dfrac{15(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{15(25+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(3.9\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(3\big)^{th} \text{ obs.}+0.9 \big(\text{Value of } \big(4\big)^{th}\text{ obs.}-\text{Value of }\big(3\big)^{th} \text{ obs.}\big)\\ &=19+0.9\big(20 -19\big)\\ &=19.9 \text{ dollars} \end{aligned} $$`

Thus, the upper value of hourly wage rate for the lower $15$ % of the employees is $19.9$ dollars.

b. The upper wage rate for the lowest $45$ % of the employees is $P_{45}$.

The fourty fifth percentile $P_{45}$ can be computed as follows:

`$$ \begin{aligned} P_{45} &=\text{Value of }\bigg(\dfrac{45(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{45(25+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(11.7\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(11\big)^{th} \text{ obs.}+0.7 \big(\text{Value of } \big(12\big)^{th}\text{ obs.}-\text{Value of }\big(11\big)^{th} \text{ obs.}\big)\\ &=23+0.7\big(24 -23\big)\\ &=23.7 \text{ dollars} \end{aligned} $$`

Thus, the upper value of hourly wage rate for the lower $45$ % of the employees is $23.7$ dollars.

c. The lower wage rate for the upper $25$ % of the employees is $P_{75}$.

The seventy fifth percentile $P_{75}$ can be computed as follows:

`$$ \begin{aligned} P_{75} &=\text{Value of }\bigg(\dfrac{75(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{75(25+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(19.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(19\big)^{th} \text{ obs.}+0.5 \big(\text{Value of } \big(20\big)^{th}\text{ obs.}-\text{Value of }\big(19\big)^{th} \text{ obs.}\big)\\ &=27+0.5\big(28 -27\big)\\ &=27.5 \text{ dollars} \end{aligned} $$`

Thus, the lower value of hourly wage rate for the upper $25$ % of the employees is $27.5$ dollars.