# 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 for Grouped Data
Type of Frequency Distribution 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 Observation (N):
Required Decile : D{{index}}

## How to use deciles calculator 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 - Calculate 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 decile formula for group data $i^{th}$ 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 decile 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:

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, we have covered deciles formula for grouped data and how to calculate deciles for grouped data. You also learnt about how to solve numerical problems based on deciles for grouped data.