Deciles Calculator for grouped data with examples

Deciles for grouped data

Deciles are the values which divide whole distribution into ten equal parts. They are 9 in numbers namely $D_1, D_2, \cdots, D_9$. Here $D_1$ is first decile, $D_2$ is second decile and so on.

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

$$ \begin{aligned} D_i=l + \bigg(\frac{\frac{iN}{10} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned} $$

where,

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

Deciles Calculator for grouped data

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

Deciles Calculator (Grouped Data)
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Which Decile? (Between 1 to 9)
Results
Number of Obs. (N):
Required Decile : D{{index}}

How to find deciles 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 decile number between 1 to 9.

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

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

Step 7 - Gives required decile

Deciles for 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 30 32 35 38 40
No. of students 8 12 20 10 5

Calculate $D_1$ and $D_6$.

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 $i^{th}$ deciles is

$D_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

First Decile $D_1$

$$ \begin{aligned} D_{1} &=\bigg(\dfrac{1(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(55)}{10}\bigg)^{th}\text{ value}\\ &=\big(5.5\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $5.5$ is $8$. The corresponding value of $X$ is the $1^{st}$ decile. That is, $D_1 =30$ minutes.

Thus, $10$ % of the students spent less than or equal to $30$ minutes.

Sixth Decile $D_6$

$$ \begin{aligned} D_{6} &=\bigg(\dfrac{6(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{6(55)}{10}\bigg)^{th}\text{ value}\\ &=\big(33\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $33$ is $40$. The corresponding value of $X$ is the $6^{th}$ decile. That is, $D_6 =35$ minutes.

Thus, $60$ % of the students spent less than or equal to $35$ minutes.

Deciles for grouped data Example 2

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 ($x$) 10-12 13-15 16-18 19-21 22-24
No. of students ($f$) 3 12 15 24 2

Calculate

a. the maximum time spent on the internet by lower 20 % of the students,
b. the maximum time spent on the internet by lower 50 % of the students,
c. the minimum time spent on the internet by upper 30 % 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 time spent on the internet by lower 20 % of the students is second decile $D_2$.

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

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

$$ \begin{aligned} D_{2} &=\bigg(\dfrac{2(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(11.2\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $11.2$ is $15$, the corresponding class $12.5-15.5$ is the $2^{nd}$ decile class.

Thus

  • $l = 12.5$, the lower limit of the $2^{nd}$ decile class
  • $N=56$, total number of observations
  • $f =12$, frequency of the $2^{nd}$ decile class
  • $F_< = 3$, cumulative frequency of the class previous to $2^{nd}$ decile class
  • $h =3$, the class width

The second decile $D_2$ can be computed as follows:

$$ \begin{aligned} D_2 &= l + \bigg(\frac{\frac{2(N)}{10} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{2*56}{10} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{11.2 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.6833\big)\times 3\\ &= 12.5 + 2.05\\ &= 14.55 \text{ minutes} \end{aligned} $$

The maximum time spent on the internet by lower $20$ % of the students is second decile $D_2 = 14.55$ minutes.

b. The maximum time spent on the internet by lower 50 % of the students is fifth decile $D_5$.

$$ \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(28\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $28$ is $30$, the corresponding class $15.5-18.5$ is the $5^{th}$ decile class.

Thus

  • $l = 15.5$, the lower limit of the $5^{th}$ decile class
  • $N=56$, total number of observations
  • $f =15$, frequency of the $5^{th}$ decile class
  • $F_< = 15$, cumulative frequency of the class previous to $5^{th}$ decile class
  • $h =3$, the class width

The fifth decile $D_5$ can be computed as follows:

$$ \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 15.5 + \bigg(\frac{\frac{5*56}{10} - 15}{15}\bigg)\times 3\\ &= 15.5 + \bigg(\frac{28 - 15}{15}\bigg)\times 3\\ &= 15.5 + \big(0.8667\big)\times 3\\ &= 15.5 + 2.6\\ &= 18.1 \text{ minutes} \end{aligned} $$

The maximum time spent on the internet by lower $50$ % of the students is fifth decile $D_5 = 18.1$ minutes.

c. The minimum time spent on the internet by upper 30 % of the students is seventh decile $D_7$.

$$ \begin{aligned} D_{7} &=\bigg(\dfrac{7(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{7(56)}{10}\bigg)^{th}\text{ value}\\ &=\big(39.2\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $39.2$ is $54$, the corresponding class $18.5-21.5$ is the $7^{th}$ decile class.

Thus

  • $l = 18.5$, the lower limit of the $7^{th}$ decile class
  • $N=56$, total number of observations
  • $f =24$, frequency of the $7^{th}$ decile class
  • $F_< = 30$, cumulative frequency of the class previous to $7^{th}$ decile class
  • $h =3$, the class width

The seventh decile $D_7$ can be computed as follows:

$$ \begin{aligned} D_7 &= l + \bigg(\frac{\frac{7(N)}{10} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{7*56}{10} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{39.2 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.3833\big)\times 3\\ &= 18.5 + 1.15\\ &= 19.65 \text{ minutes} \end{aligned} $$

The minimum time spent on the internet by upper $30$ % of the students is seventh decile $D_7 = 19.65$ minutes.

Deciles for grouped data Example 3

The Scores of students in a Math test is given in the table below :

Class Interval 10-20 20-30 30-40 40-50 50-60 60-70
Frequency ($f$) 6 8 12 10 5 4

Find

a. the maximum score for lower 20% of the students.
b. the minimum score for upper 10% of the students.

Solution

Let $X$ denote the scores in Math Test.

Class Interval Class Boundries $f_i$ $cf$
10-20 10-20 6 6
20-30 20-30 8 14
30-40 30-40 12 26
40-50 40-50 10 36
50-60 50-60 5 41
60-70 60-70 4 45
Total 45

a. The maximum marks score by lower 20 % of the students is second decile $D_2$.

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

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

$$ \begin{aligned} D_{2} &=\bigg(\dfrac{2(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(45)}{10}\bigg)^{th}\text{ value}\\ &=\big(9\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $9$ is $14$, the corresponding class $20-30$ is the $2^{nd}$ decile class.

Thus

  • $l = 20$, the lower limit of the $2^{nd}$ decile class
  • $N=45$, total number of observations
  • $f =8$, frequency of the $2^{nd}$ decile class
  • $F_< = 6$, cumulative frequency of the class previous to $2^{nd}$ decile class
  • $h =10$, the class width

The second decile $D_2$ can be computed as follows:

$$ \begin{aligned} D_2 &= l + \bigg(\frac{\frac{2(N)}{10} - F_<}{f}\bigg)\times h\\ &= 20 + \bigg(\frac{\frac{2*45}{10} - 6}{8}\bigg)\times 10\\ &= 20 + \bigg(\frac{9 - 6}{8}\bigg)\times 10\\ &= 20 + \big(0.375\big)\times 10\\ &= 20 + 3.75\\ &= 23.75 \text{ Scores} \end{aligned} $$

The maximum score by lower $20$ % of the students is second decile $D_2 = 23.75$ Scores.

c. The minimum score by upper 10 % of the students is ninth decile $D_9$.

$$ \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(45)}{10}\bigg)^{th}\text{ value}\\ &=\big(40.5\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $40.5$ is $41$, the corresponding class $50-60$ is the $9^{th}$ decile class.

Thus

  • $l = 50$, the lower limit of the $9^{th}$ decile class
  • $N=45$, total number of observations
  • $f =5$, frequency of the $9^{th}$ decile class
  • $F_< = 36$, cumulative frequency of the class previous to $9^{th}$ decile class
  • $h =10$, the class width

The ninth decile $D_9$ can be computed as follows:

$$ \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 50 + \bigg(\frac{\frac{9*45}{10} - 36}{5}\bigg)\times 10\\ &= 50 + \bigg(\frac{40.5 - 36}{5}\bigg)\times 10\\ &= 50 + \big(0.9\big)\times 10\\ &= 50 + 9\\ &= 59 \text{ Scores} \end{aligned} $$

The minimum score by upper $10$ % of the students is ninth decile $D_9 = 59$ Scores.

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 fifth, sixth and ninth decile for the above frequency distribution.

Solution

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

Decile formula for grouped data

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

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

Fifth Decile

$$ \begin{aligned} D_{5} &=\bigg(\dfrac{5(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $30$ is $36$, the corresponding class $10.75-11.25$ is the $5^{th}$ decile class.

Thus

  • $l = 10.75$, the lower limit of the $5^{th}$ decile class
  • $N=60$, total number of observations
  • $f =17$, frequency of the $5^{th}$ decile class
  • $F_< = 19$, cumulative frequency of the class previous to $5^{th}$ decile class
  • $h =0.5$, the class width

The fifth $D_5$ can be computed as follows:

$$ \begin{aligned} D_5 &= l + \bigg(\frac{\frac{5(N)}{10} - F_<}{f}\bigg)\times h\\ &= 10.75 + \bigg(\frac{\frac{5*60}{10} - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \bigg(\frac{30 - 19}{17}\bigg)\times 0.5\\ &= 10.75 + \big(0.6471\big)\times 0.5\\ &= 10.75 + 0.3235\\ &= 11.0735 \text{ tons} \end{aligned} $$

$50$ % of the cables have maximum load less than $D_5 = 11.0735$ tons.

Sixth Decile

$$ \begin{aligned} D_{6} &=\bigg(\dfrac{6(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{6(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(36\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $36$ is $50$, the corresponding class $11.25-11.75$ is the $6^{th}$ decile class.

Thus

  • $l = 11.25$, the lower limit of the $6^{th}$ decile class
  • $N=60$, total number of observations
  • $f =14$, frequency of the $6^{th}$ decile class
  • $F_< = 36$, cumulative frequency of the class previous to $6^{th}$ decile class
  • $h =0.5$, the class width

The sixth decile $D_6$ can be computed as follows:

$$ \begin{aligned} D_6 &= l + \bigg(\frac{\frac{6(N)}{10} - F_<}{f}\bigg)\times h\\ &= 11.25 + \bigg(\frac{\frac{6*60}{10} - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \bigg(\frac{36 - 36}{14}\bigg)\times 0.5\\ &= 11.25 + \big(0\big)\times 0.5\\ &= 11.25 + 0\\ &= 11.25 \text{ tons} \end{aligned} $$

$60$ % of the pumpkins weigh less than $D_6 = 11.25$ tons.

Ninth Decile

$$ \begin{aligned} D_{9} &=\bigg(\dfrac{9(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{9(60)}{10}\bigg)^{th}\text{ value}\\ &=\big(54\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $54$ is $56$, the corresponding class $11.75-12.25$ is the $9^{th}$ decile class.

Thus

  • $l = 11.75$, the lower limit of the $9^{th}$ decile class
  • $N=60$, total number of observations
  • $f =6$, frequency of the $9^{th}$ decile class
  • $F_< = 50$, cumulative frequency of the class previous to $9^{th}$ decile class
  • $h =0.5$, the class width

The ninth decile $D_9$ can be computed as follows:

$$ \begin{aligned} D_9 &= l + \bigg(\frac{\frac{9(N)}{10} - F_<}{f}\bigg)\times h\\ &= 11.75 + \bigg(\frac{\frac{9*60}{10} - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \bigg(\frac{54 - 50}{6}\bigg)\times 0.5\\ &= 11.75 + \big(0.6667\big)\times 0.5\\ &= 11.75 + 0.3333\\ &= 12.0833 \text{ tons} \end{aligned} $$

$90$ % of the pumpkins weigh less than $D_9 = 12.0833$ tons.

Deciles 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 third, fourth and eigth decile.

Solution

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

Decile formula for grouped data

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

$D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$

where $N$ is the total number of observations.

Third Decile

$$ \begin{aligned} D_{3} &=\bigg(\dfrac{3(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(30\big)^{th}\text{ value} \end{aligned} $$
The cumulative frequency just greater than or equal to $30$ is $52$, the corresponding class $8-10$ is the $3^{rd}$ decile class.

Thus

  • $l = 8$, the lower limit of the $3^{rd}$ decile class
  • $N=100$, total number of observations
  • $f =34$, frequency of the $3^{rd}$ decile class
  • $F_< = 18$, cumulative frequency of the class previous to $3^{rd}$ decile class
  • $h =2$, the class width

The third $D_3$ can be computed as follows:

$$ \begin{aligned} D_3 &= l + \bigg(\frac{\frac{3(N)}{10} - F_<}{f}\bigg)\times h\\ &= 8 + \bigg(\frac{\frac{3*100}{10} - 18}{34}\bigg)\times 2\\ &= 8 + \bigg(\frac{30 - 18}{34}\bigg)\times 2\\ &= 8 + \big(0.3529\big)\times 2\\ &= 8 + 0.7059\\ &= 8.7059 \text{ ('00 grams)} \end{aligned} $$

$30$ % of the pumpkins weigh less than $D_3 = 8.7059$ ('00 grams).

Fourth Decile

$$ \begin{aligned} D_{4} &=\bigg(\dfrac{4(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{4(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(40\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $40$ is $52$, the corresponding class $8-10$ is the $4^{th}$ decile class.

Thus

  • $l = 8$, the lower limit of the $4^{th}$ decile class
  • $N=100$, total number of observations
  • $f =34$, frequency of the $4^{th}$ decile class
  • $F_< = 18$, cumulative frequency of the class previous to $4^{th}$ decile class
  • $h =2$, the class width

The fourth decile $D_4$ can be computed as follows:

$$ \begin{aligned} D_4 &= l + \bigg(\frac{\frac{4(N)}{10} - F_<}{f}\bigg)\times h\\ &= 8 + \bigg(\frac{\frac{4*100}{10} - 18}{34}\bigg)\times 2\\ &= 8 + \bigg(\frac{40 - 18}{34}\bigg)\times 2\\ &= 8 + \big(0.6471\big)\times 2\\ &= 8 + 1.2941\\ &= 9.2941 \text{ ('00 grams)} \end{aligned} $$

Maximum weight for lower $40$ % of the pumpkins is $D_4 = 9.2941$ ('00 grams).

Eigth Decile

$$ \begin{aligned} D_{8} &=\bigg(\dfrac{8(N)}{10}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{8(100)}{10}\bigg)^{th}\text{ value}\\ &=\big(80\big)^{th}\text{ value} \end{aligned} $$

The cumulative frequency just greater than or equal to $80$ is $100$, the corresponding class $12-14$ is the $8^{th}$ decile class.

Thus

  • $l = 12$, the lower limit of the $8^{th}$ decile class
  • $N=100$, total number of observations
  • $f =20$, frequency of the $8^{th}$ decile class
  • $F_< = 80$, cumulative frequency of the class previous to $8^{th}$ decile class
  • $h =2$, the class width

The seventh decile $D_8$ can be computed as follows:

$$ \begin{aligned} D_8 &= l + \bigg(\frac{\frac{8(N)}{10} - F_<}{f}\bigg)\times h\\ &= 12 + \bigg(\frac{\frac{8*100}{10} - 80}{20}\bigg)\times 2\\ &= 12 + \bigg(\frac{80 - 80}{20}\bigg)\times 2\\ &= 12 + \big(0\big)\times 2\\ &= 12 + 0\\ &= 12 \text{ ('00 grams)} \end{aligned} $$

Maximum weight for lower $80$ % of the pumpkins is $D_8 = 12$ ('00 grams).

Conclusion

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

Leave a Comment