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.
