Percentiles Calculator for grouped data with examples

Percentiles for grouped data

Percentiles are the values which divide whole distribution into hundred equal parts. They are 99 in numbers namely $P_1, P_2, \cdots, P_{99}$. Here $P_1$ is first percentile, $P_2$ is second percentile and so on.

For discrete frequency distribution, the formula for $i^{th}$ percentile is

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where,

  • $N$ is total number of observations.

For continuous frequency distribution, the formula for $i^{th}$ percentile is

$P_i=l + \bigg(\frac{\frac{iN}{100} - F_<}{f}\bigg)\times h; \quad i=1,2,\cdots,99$

where,

  • $l :$ the lower limit of the $i^{th}$ percentile class
  • $N=\sum f :$ total number of observations
  • $f :$ frequency of the $i^{th}$ percentile class
  • $F_< :$ cumulative frequency of the class previous to $i^{th}$ percentile class
  • $h :$ the class width

Percentiles Calculator for grouped data

Use this calculator to find the Percentiles for grouped (frequency distribution) data.

Percentiles Calculator (Grouped Data)
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Which Percentile? (Between 1 to 99)
Results
Number of Obs. (N):
Required Percentile : P{{index}}

How to find percentile for grouped data?

Step 1 - Select type of frequency distribution (Discrete or continuous)

Step 2 - Enter the Range or classes (X) seperated by comma (,)

Step 3 - Enter the Frequencies (f) seperated by comma

Step 4 - Enter the require percentile number between 1 to 99.

Step 5 - Click on "Calculate" button for percentile calculation

Step 6 - Gives output as number of observation (N)

Step 7 - Gives required percentile

Percentile of Grouped data Example 1

A librarian keeps the records about the amount of time spent (in minutes) in a library by college students. Data is as follows:

Time spent No. of Students
30 8
32 12
35 20
38 10
40 5

Calculate $P_{15}$ and $P_{40}$.

Solution

$x_i$ $f_i$ $cf$
30 8 8
32 12 20
35 20 40
38 10 50
40 5 55
Total 55

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

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

Fiftieth percentile $P_{15}$

$$
\begin{aligned}
P_{15} &=\bigg(\dfrac{15(N)}{100}\bigg)^{th}\text{ value}\
&= \bigg(\dfrac{15(55)}{100}\bigg)^{th}\text{ value}\
&=\big(8.25\big)^{th}\text{ value}
\end{aligned}
$$

The cumulative frequency just greater than or equal to $8.25$ is $20$. The corresponding value of $X$ is the $15^{th}$ percentile. That is, $P_{15} =32$ minutes.

Fortieth percentile $P_{40}$

$$ \begin{aligned} P_{40} &=\bigg(\dfrac{40(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{40(55)}{100}\bigg)^{th}\text{ value}\\ &=\big(22\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $22$ is $40$. The corresponding value of $X$ is the $40^{th}$ percentile. That is, $P_{40} =35$ minutes.

Percentile of Grouped data Example 2

The following table gives a frequency distribution of weight (in pounds) of 57 children at a day care center.

Weight 10-19 20-29 30-39 40-49 50-59 60-69 70-79
children 5 19 10 13 4 4 2

Calculate

a. the maximum weight of lower 30 % of the children,

b. the minimum weight of upper 30 % of the children,

c. the limits for the weight of middle 40 % of the children.

Solution

Let $X$ denote the weight of children at a day care center.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries $f_i$ $cf$
10-20 9.5-20.5 5 5
20-30 19.5-30.5 19 24
30-40 29.5-40.5 10 34
40-50 39.5-50.5 13 47
50-60 49.5-60.5 4 51
60-70 59.5-70.5 4 55
10-20 9.5-20.5 2 57
Total 57

a. The maximum weight of lower $30$ % of the children is $P_{30}$.

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

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

$$ \begin{aligned} P_{30} &=\bigg(\dfrac{30(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{30(57)}{100}\bigg)^{th}\text{ value}\\ &=\big(17.1\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $17.1$ is $24$, the corresponding class $19.5-30.5$ is the $30^{th}$ percentile class.

Thus

  • $l = 19.5$, the lower limit of the $30^{th}$ percentile class
  • $N=57$, total number of observations
  • $f =19$, frequency of the $30^{th}$ percentile class
  • $F_< = 5$, cumulative frequency of the class previous to $30^{th}$ percentile class
  • $h =10$, the class width

The thirtieth percentile $P_{30}$ can be computed as follows:

$$ \begin{aligned} P_{30} &= l + \bigg(\frac{\frac{30(N)}{100} - F_<}{f}\bigg)\times h\\ &= 19.5 + \bigg(\frac{\frac{30*57}{100} - 5}{19}\bigg)\times 10\\ &= 19.5 + \bigg(\frac{17.1 - 5}{19}\bigg)\times 10\\ &= 19.5 + \big(0.6368\big)\times 10\\ &= 19.5 + 6.3684\\ &= 25.8684 \text{ pounds} \end{aligned} $$

The maximum weight of lower $30$ % of the children is $P_{30}= 25.8684$ pounds.

b. The minimum weight of upper $30$ % of the children is $P_{70}$.

$$ \begin{aligned} P_{70} &=\bigg(\dfrac{70(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{70(57)}{100}\bigg)^{th}\text{ value}\\ &=\big(39.9\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $39.9$ is $47$, the corresponding class $39.5-50.5$ is the $70^{th}$ percentile class.

Thus

  • $l = 39.5$, the lower limit of the $70^{th}$ percentile class
  • $N=57$, total number of observations
  • $f =13$, frequency of the $70^{th}$ percentile class
  • $F_< = 34$, cumulative frequency of the class previous to $70^{th}$ percentile class
  • $h =10$, the class width

The seventieth percentile $P_{70}$ can be computed as follows:

$$ \begin{aligned} P_{70} &= l + \bigg(\frac{\frac{70(N)}{100} - F_<}{f}\bigg)\times h\\ &= 39.5 + \bigg(\frac{\frac{70*57}{100} - 34}{13}\bigg)\times 10\\ &= 39.5 + \bigg(\frac{39.9 - 34}{13}\bigg)\times 10\\ &= 39.5 + \big(0.4538\big)\times 10\\ &= 39.5 + 4.5385\\ &= 44.0385 \text{ pounds} \end{aligned} $$

The minimum weight of upper $30$ % of the children is $P_{30}= 44.0385$ pounds.

c. The limits for the weight of middle $40$ % of the children is $P_{30}$ and $P_{70}$.

Thus the limits for the weight of middle 40 % of the children is $P_{30} = 25.8684$ pounds and $P_{70}= 44.0385$ pounds.

Percentile for grouped data Example 3

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students.

Time spent on Internet ($x$) No. of Students ($f$)
10-12 3
13-15 12
16-18 15
19-21 24
22-24 2

Using percentile calculate

a. the maximum time spent on the internet by lower 35 % of the students,

b. the minimum time spent on the internet by upper 15 % of the students.

Solution

Let $X$ denote the amount of time (in minutes) spent on the internet.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval Class Boundries $f_i$ $cf$
10-12 9.5-12.5 3 3
13-15 12.5-15.5 12 15
16-18 15.5-18.5 15 30
19-21 18.5-21.5 24 54
22-24 21.5-24.5 2 56
Total 56

a. The maximum weight of lower $35$ % of the children is $P_{35}$.

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

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

$$ \begin{aligned} P_{35} &=\bigg(\dfrac{35(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{35(56)}{100}\bigg)^{th}\text{ value}\\ &=\big(19.6\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $19.6$ is $30$, the corresponding class $15.5-18.5$ is the $35^{th}$ percentile class.

Thus

  • $l = 15.5$, the lower limit of the $35^{th}$ percentile class
  • $N=56$, total number of observations
  • $f =15$, frequency of the $35^{th}$ percentile class
  • $F_< = 15$, cumulative frequency of the class previous to $35^{th}$ percentile class
  • $h =3$, the class width

The thirty fifth percentile $P_{35}$ can be computed as follows:

$$ \begin{aligned} P_{35} &= l + \bigg(\frac{\frac{35(N)}{100} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{35*56}{100} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{19.6 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.3067\big)\times 3\\ &= 15.5 + 0.92\\ &= 16.42 \text{ minutes} \end{aligned} $$

The maximum time spend by lower $35$ % of the students is $P_{35}= 16.42$ minutes.

b. The minimum time spend by upper $15$ % of the children is $P_{85}$.

$$ \begin{aligned} P_{85} &=\bigg(\dfrac{85(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{85(56)}{100}\bigg)^{th}\text{ value}\\ &=\big(47.6\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $47.6$ is $54$, the corresponding class $18.5-21.5$ is the $85^{th}$ percentile class.

Thus

  • $l = 18.5$, the lower limit of the $85^{th}$ percentile class
  • $N=56$, total number of observations
  • $f =24$, frequency of the $85^{th}$ percentile class
  • $F_< = 30$, cumulative frequency of the class previous to $85^{th}$ percentile class
  • $h =3$, the class width

The eighty fifth percentile $P_{85}$ can be computed as follows:

$$ \begin{aligned} P_{85} &= l + \bigg(\frac{\frac{85(N)}{100} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{85*56}{100} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{47.6 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.7333\big)\times 3\\ &= 18.5 + 2.2\\ &= 20.7 \text{ minutes} \end{aligned} $$

The minimum time spend by upper $15$ % of the children is $P_{85}= 20.7$ minutes.

Deciles for grouped data Example 4

The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:

Maximum load No. of Cables
9.25-9.75 2
9.75-10.25 5
10.25-10.75 12
10.75-11.25 17
11.25-11.75 14
11.75-12.25 6
12.25-12.75 3
12.75-13.25 1

Compute

a. maximum load for lower 40 percent of the cables,
b. minimum load for the upper 16 percent of the cables.

Solution

Class Boundries $f_i$ $cf$
9.25-9.75 2 2
9.75-10.25 5 7
10.25-10.75 12 19
10.75-11.25 17 36
11.25-11.75 14 50
11.75-12.25 6 56
12.25-12.75 3 59
12.75-13.25 1 60
Total 60

a. The maximum load of lower $40$ % of the cables is $P_{40}$.

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

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

$$ \begin{aligned} P_{40} &=\bigg(\dfrac{40(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{40(60)}{100}\bigg)^{th}\text{ value}\\ &=\big(24\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $24$ is $36$, the corresponding class $10.75-11.25$ is the $40^{th}$ percentile class.

Thus

  • $l = 10.75$, the lower limit of the $40^{th}$ percentile class
  • $N=60$, total number of observations
  • $f =17$, frequency of the $40^{th}$ percentile class
  • $F_< = 19$, cumulative frequency of the class previous to $40^{th}$ percentile class
  • $h =0.5$, the class width

The fortieth percentile $P_{40}$ can be computed as follows:

$$ \begin{aligned} P_{40} &= l + \bigg(\frac{\frac{40(N)}{100} - F_<}{f}\bigg)\times h\\ &= 10.75 + \bigg(\frac{\frac{40*60}{100} - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{24 - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \big(0.2941\big)\times 0.5\\ &= 10.75 + 0.1471\\ &= 10.8971 \text{ tons} \end{aligned} $$

The maximum load by lower $40$ % of the cables is $P_{40}= 10.8971$ tons.

b. The minimum load for upper $16$ % of the cables is $P_{84}$.

$$ \begin{aligned} P_{84} &=\bigg(\dfrac{84(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{84(60)}{100}\bigg)^{th}\text{ value}\\ &=\big(50.4\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $50.4$ is $56$, the corresponding class $11.75-12.25$ is the $84^{th}$ percentile class.

Thus

  • $l = 11.75$, the lower limit of the $84^{th}$ percentile class
  • $N=60$, total number of observations
  • $f =6$, frequency of the $84^{th}$ percentile class
  • $F_< = 50$, cumulative frequency of the class previous to $84^{th}$ percentile class
  • $h =0.5$, the class width

The eighty fourth percentile $P_{84}$ can be computed as follows:

$$ \begin{aligned} P_{84} &= l + \bigg(\frac{\frac{84(N)}{100} - F_<}{f}\bigg)\times h\\ &= 11.75 + \bigg(\frac{\frac{84*60}{100} - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \bigg(\frac{50.4 - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \big(0.0667\big)\times 0.5\\ &= 11.75 + 0.0333\\ &= 11.7833 \text{ tons} \end{aligned} $$

The minimum load for upper $16$ % of the cable is $P_{84}= 11.7833$ tons.

Percentiles for grouped data Example 5

Following table shows the weight of 100 pumpkin produced from a farm :

Weight ('00 grams) Frequency
$4 \leq x < 6$ 4
$6 \leq x < 8$ 14
$8 \leq x < 10$ 34
$10 \leq x < 12$ 28
$12 \leq x < 14$ 20

Calculate

a. minimum weight for upper 9 percent of the pumpkins,
b. maximum weight for lower 24 percent of the pumpkins.

Solution

Class Boundries $f_i$ $cf$
4-6 4 4
6-8 14 18
8-10 34 52
10-12 28 80
12-14 20 100
Total 100

a. The minimum weight for upper $9$ % of the pumpkins is $P_{91}$.

$$ \begin{aligned} P_{91} &=\bigg(\dfrac{91(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{91(100)}{100}\bigg)^{th}\text{ value}\\ &=\big(91\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $91$ is $100$, the corresponding class $12-14$ is the $91^{th}$ percentile class.

Thus

  • $l = 12$, the lower limit of the $91^{th}$ percentile class
  • $N=100$, total number of observations
  • $f =20$, frequency of the $91^{th}$ percentile class
  • $F_< = 80$, cumulative frequency of the class previous to $91^{th}$ percentile class
  • $h =2$, the class width

The ninety first percentile $P_{91}$ can be computed as follows:

$$ \begin{aligned} P_{91} &= l + \bigg(\frac{\frac{91(N)}{100} - F_<}{f}\bigg)\times h\\ &= 12 + \bigg(\frac{\frac{91*100}{100} - 80}{20}\bigg)\times 2\\ &= 12 + \bigg(\frac{91 - 80}{20}\bigg)\times 2\\ &= 12 + \big(0.55\big)\times 2\\ &= 12 + 1.1\\ &= 13.1 \text{ ('00 grams)} \end{aligned} $$

The minimum weight for upper $9$ % of the pumpkins is $P_{91}= 13.1$ ('00 grams).

b. The maximum weight for lower $24$ % of the pumpkins is $P_{24}$.

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

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

$$ \begin{aligned} P_{24} &=\bigg(\dfrac{24(N)}{100}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{24(100)}{100}\bigg)^{th}\text{ value}\\ &=\big(24\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $24$ is $52$, the corresponding class $8-10$ is the $24^{th}$ percentile class.

Thus

  • $l = 8$, the lower limit of the $24^{th}$ percentile class
  • $N=100$, total number of observations
  • $f =34$, frequency of the $24^{th}$ percentile class
  • $F_< = 18$, cumulative frequency of the class previous to $24^{th}$ percentile class
  • $h =2$, the class width

The twenty fourth percentile $P_{24}$ can be computed as follows:

$$ \begin{aligned} P_{24} &= l + \bigg(\frac{\frac{24(N)}{100} - F_<}{f}\bigg)\times h\\ &= 8 + \bigg(\frac{\frac{24*100}{100} - 18}{34}\bigg)\times 2\\ &= 8 + \bigg(\frac{24 - 18}{34}\bigg)\times 2\\ &= 8 + \big(0.1765\big)\times 2\\ &= 8 + 0.3529\\ &= 8.3529 \text{ ('00 grams)} \end{aligned} $$

The maximum weight for lower $24$ % of the pumpkins is $P_{24}= 8.3529$ ('00 grams).

Conclusion

In this tutorial, you learned about formula for percentile for grouped data and how to calculate percentile for grouped data. You also learned about how to solve numerical problems based on percentile for grouped 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 Percentile calculator for grouped 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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