Percentiles Calculator for ungrouped data with examples

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.

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

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

where $n$ is the total number of observations.

Percentiles calculator for ungrouped data

Use this calculator to find the percentiles for ungrouped (raw) data.

Percentile Calculator
Enter the X Values (Separated by comma,)
Which Percentile? (Between 1 to 99)
Results
Number of Obs. (n):
Ascending order of X values :
Required Percentile : P{{index}}

How to calculate percentiles for ungrouped data?

Step 1 - Enter the $x$ values separated by commas

Step 2 - Enter the nuber between 1 to 99 (inclusive)

Step 3 - Click on "Calculate" button to get percentile for ungrouped data

Step 4 - Gives the output as number of observations $n$

Step 5 - Gives the output as ascending order data

Step 6 - Gives the required percentile

Percentile for ungrouped data 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:

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

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:

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

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:

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

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

Percentile for ungrouped data 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 $n = 25$.

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

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

where $n$ is the total number of obs.s.

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:

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

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:

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

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:

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

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

Percenstiles for ungrouped data Example 3

Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

65,76,64,73,74,80,71,68,66,
81,79,75,70,62,83,63,77,78.

a. the upper diastolic BP for the lowest 30 % of the patients,

b. the lower diastolic BP for the upper 30 % of the patients.

Solution

a. The upper diastolic blood pressure for the lowest $30$ % of the patients is $P_{30}$.

$P_{30}$ can be computed as follows:

$$ \begin{aligned} P_{30} &=\text{Value of }\bigg(\dfrac{30(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{30(18+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(5.7\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}\\ &\quad +0.7 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=66+0.7\big(68 -66\big)\\ &=67.4 \text{ mmHg} \end{aligned} $$

Thus, the upper value of diastolic blood pressure for the lower $30$ % of the patients is $67.4$ mmHg.

b. The lower diastolic blood pressure for the upper $30$ % of the patients is $P_{70}$.

$P_{70}$ can be computed as follows:

$$ \begin{aligned} P_{70} &=\text{Value of }\bigg(\dfrac{70(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{70(18+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(13.3\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(13\big)^{th} \text{ obs.}\\ &\quad +0.3 \big(\text{Value of } \big(14\big)^{th}\text{ obs.}-\text{Value of }\big(13\big)^{th} \text{ obs.}\big)\\ &=77+0.3\big(78 -77\big)\\ &=77.3 \text{ mmHg} \end{aligned} $$

Thus, the lower value of diastolic blood pressure for the upper $30$ % of the patients is $27.5$ mmHg.

Percentiles for ungrouped data Example 4

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137, 
137, 138, 141, 143, 144, 146, 147, 152, 154, 161 

a. the maximum height for the lower 20 % of the children,

b. the minimum height for the upper 20 % of the children.

Solution

a. The maximum height for the lower $20$ % of the children is $P_{20}$.

$P_{20}$ can be computed as follows:

$$ \begin{aligned} P_{20} &=\text{Value of }\bigg(\dfrac{20(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{20(20+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(4.2\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}\\ &\quad +0.2 \big(\text{Value of } \big(5\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=132+0.2\big(132 -132\big)\\ &=132 \text{ cm} \end{aligned} $$

Thus, the maximum value of height for the lower $20$ % of the children is $132$ cm.

b. The minimum height for the upper $20$ % of the children is $P_{80}$.

$P_{80}$ can be computed as follows:

$$ \begin{aligned} P_{80} &=\text{Value of }\bigg(\dfrac{80(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{80(20+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(16.8\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(16\big)^{th} \text{ obs.}\\ &\quad +0.8 \big(\text{Value of } \big(17\big)^{th}\text{ obs.}-\text{Value of }\big(16\big)^{th} \text{ obs.}\big)\\ &=146+0.8\big(147 -146\big)\\ &=146.8 \text{ cm} \end{aligned} $$

Thus, the minimum height for the upper $20$ % of the children is $27.5$ cm.

Percentiles for ungrouped data Example 5

The rice production (in Kg) of 10 acres is given as: 1120, 1240, 1320, 1040, 1080, 1720, 1600, 1470, 1750, and 1885.

a. the maximum rice production for the lower 35 % of the plot,

b. the minimum rice production for the upper 15 % of the plot.

Solution

a. The maximum rice production for the lower $35$ % of the plot is $P_{35}$.

$P_{35}$ can be computed as follows:

$$ \begin{aligned} P_{35} &=\text{Value of }\bigg(\dfrac{35(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{35(10+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(3.85\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(3\big)^{th} \text{ obs.}\\ &\quad +0.85 \big(\text{Value of } \big(4\big)^{th}\text{ obs.}-\text{Value of }\big(3\big)^{th} \text{ obs.}\big)\\ &=1120+0.85\big(1240 -1120\big)\\ &=1222 \text{ Kg} \end{aligned} $$

Thus, the maximum value of rice production for the lower $35$ % of the plot is $132$ Kg.

b. The minimum rice production for the upper $15$ % of the plot is $P_{85}$.

$P_{85}$ can be computed as follows:

$$ \begin{aligned} P_{85} &=\text{Value of }\bigg(\dfrac{85(n+1)}{100}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{85(10+1)}{100}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(9.35\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(9\big)^{th} \text{ obs.}\\ &\quad +0.35 \big(\text{Value of } \big(10\big)^{th}\text{ obs.}-\text{Value of }\big(9\big)^{th} \text{ obs.}\big)\\ &=1750+0.35\big(1885 -1750\big)\\ &=1797.25 \text{ Kg} \end{aligned} $$

Thus, the minimum rice production for the upper $15$ % of the plot is $27.5$ Kg.

Conclusion

In this tutorial, you learned about formula for Percentiles for ungrouped data and how to calculate Percentiles for ungrouped data. You also learned about how to solve numerical problems based on Percentiles for ungrouped data.

To learn more about other descriptive statistics measures, please refer to the following tutorials:

Descriptive Statistics

Let me know in the comments if you have any questions on Percentiles calculator for ungrouped data with examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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