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# Octiles Calculator for grouped data with examples

## Octiles for grouped data

Octiles are the values of arranged data which divide whole data into eight equal parts. They are 7 in numbers namely $O_1,O_2, \cdots, O_7$. Here $O_1$ is first octile, $O_2$ is second octile, $O_3$ is third octile and so on.

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

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where,

• $N$ is total number of observations.

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

$O_i=l + \bigg(\dfrac{\dfrac{iN}{8} - F_<}{f}\bigg)\times h$; $i=1,2,\cdots,7$

where,

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

## Octiles Calculator for grouped data

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

Octile Calculator (Grouped Data)
Type of Freq. Dist. DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Which Octile? (Between 1 to 7)
Results
Number of Obs. (N):
Required Octile : O{{index}}

## How to find octiles 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 – Select the octile you want to caculate

Step 5 Click on "Calculate" for octile calculation

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

Step 7 – Calculate required octile

## Octiles 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 No. of Students
30 8
32 12
35 20
38 10
40 5

Calculate $O_1$ and $O_4$.

#### 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 $i^{th}$ octile is

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where $N$ is the total number of observations.

Second octile $O_2$

 \begin{aligned} O_{2} &=\bigg(\dfrac{2(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(55)}{8}\bigg)^{th}\text{ value}\\ &=\big(13.75\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $13.75$ is $20$. The corresponding value of $X$ is the $2^{nd}$ octile. That is, $O_2 =32$ minutes.

Fourth octile $O_4$

 \begin{aligned} O_{4} &=\bigg(\dfrac{4(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{4(55)}{8}\bigg)^{th}\text{ value}\\ &=\big(27.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $27.5$ is $40$. The corresponding value of $X$ is the $4^{th}$ octile. That is, $O_4 =35$ minutes.

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

Using octiles calculate

a. the maximum time spent on the internet by lower 25 % of the students,

b. the minimum time spent on the internet by upper 25 % 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 25 % of the students is second octile $O_2$.

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

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where $N$ is the total number of observations.

 \begin{aligned} O_{2} &=\bigg(\dfrac{2(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(56)}{8}\bigg)^{th}\text{ value}\\ &=\big(14\big)^{th}\text{ value} \end{aligned}
The cumulative frequency just greater than or equal to $14$ is $15$, the corresponding class $12.5-15.5$ is the $2^{nd}$ octile class.

Thus

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

The second octile $O_2$ can be computed as follows:

 \begin{aligned} O_2 &= l + \bigg(\frac{\frac{2(N)}{8} - F_<}{f}\bigg)\times h\\ &= 12.5 + \bigg(\frac{\frac{2*56}{8} - 3}{12}\bigg)\times 3\\ &= 12.5 + \bigg(\frac{14 - 3}{12}\bigg)\times 3\\ &= 12.5 + \big(0.9167\big)\times 3\\ &= 12.5 + 2.75\\ &= 15.25 \text{ minutes} \end{aligned}
The maximum time spent on the internet by lower $25$ % of the students is second octile $O_2 = 15.25$ minutes.

b. The minimum time spent on the internet by upper 25 % of the students is sixth octile $O_6$.

 \begin{aligned} O_{6} &=\bigg(\dfrac{6(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{6(56)}{8}\bigg)^{th}\text{ value}\\ &=\big(42\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $42$ is $54$, the corresponding class $18.5-21.5$ is the $6^{th}$ octile class.

Thus

• $l = 18.5$, the lower limit of the $6^{th}$ octile class
• $N=56$, total number of observations
• $f =24$, frequency of the $6^{th}$ octile class
• $F_< = 30$, cumulative frequency of the class previous to $6^{th}$ octile class
• $h =3$, the class width

The sixth octile $O_6$ can be computed as follows:

 \begin{aligned} O_6 &= l + \bigg(\frac{\frac{6(N)}{8} - F_<}{f}\bigg)\times h\\ &= 18.5 + \bigg(\frac{\frac{6*56}{8} - 30}{24}\bigg)\times 3\\ &= 18.5 + \bigg(\frac{42 - 30}{24}\bigg)\times 3\\ &= 18.5 + \big(0.5\big)\times 3\\ &= 18.5 + 1.5\\ &= 20 \text{ minutes} \end{aligned}
The minimum time spent on the internet by upper $25$ % of the students is sixth octile $O_6 = 20$ minutes.

## Octiles for grouped data Example 3

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

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

Calculate third and fifth octiles.

#### Solution

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

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

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where $N$ is the total number of observations.

 \begin{aligned} O_{3} &=\bigg(\dfrac{3(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{3(45)}{8}\bigg)^{th}\text{ value}\\ &=\big(16.875\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $16.875$ is $26$, the corresponding class $30-40$ is the $3^{nd}$ octile class.

Thus

• $l = 30$, the lower limit of the $3^{nd}$ octile class
• $N=45$, total number of observations
• $f =12$, frequency of the $3^{nd}$ octile class
• $F_< = 14$, cumulative frequency of the class previous to $3^{nd}$ octile class
• $h =10$, the class width

The third octile $O_3$ can be computed as follows:

 \begin{aligned} O_3 &= l + \bigg(\frac{\frac{3(N)}{8} - F_<}{f}\bigg)\times h\\ &= 30 + \bigg(\frac{\frac{3*45}{8} - 14}{12}\bigg)\times 10\\ &= 30 + \bigg(\frac{16.875 - 14}{12}\bigg)\times 10\\ &= 30 + \big(0.2396\big)\times 10\\ &= 30 + 2.3958\\ &= 32.3958 \text{ Scores} \end{aligned}

The maximum score for lower $37.5$ % of the students is third octile $O_3 = 32.3958$ Scores.

 \begin{aligned} O_{5} &=\bigg(\dfrac{5(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{5(45)}{8}\bigg)^{th}\text{ value}\\ &=\big(28.125\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $28.125$ is $36$, the corresponding class $40-50$ is the $5^{th}$ octile class.

Thus

• $l = 40$, the lower limit of the $5^{th}$ octile class
• $N=45$, total number of observations
• $f =10$, frequency of the $5^{th}$ octile class
• $F_< = 26$, cumulative frequency of the class previous to $5^{th}$ octile class
• $h =10$, the class width

The fifth octile $O_5$ can be computed as follows:

 \begin{aligned} O_5 &= l + \bigg(\frac{\frac{5(N)}{8} - F_<}{f}\bigg)\times h\\ &= 40 + \bigg(\frac{\frac{5*45}{8} - 26}{10}\bigg)\times 10\\ &= 40 + \bigg(\frac{28.125 - 26}{10}\bigg)\times 10\\ &= 40 + \big(0.2125\big)\times 10\\ &= 40 + 2.125\\ &= 42.125 \text{ Scores} \end{aligned}
The maximum score for lower $62.5$ % of the students is fifth octile $O_5 = 42.125$ Scores.

## Octiles 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 first and second octiles for the above frequency distribution.

#### 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

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

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where $N$ is the total number of observations.

 \begin{aligned} O_{1} &=\bigg(\dfrac{1(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(60)}{8}\bigg)^{th}\text{ value}\\ &=\big(7.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $7.5$ is $19$, the corresponding class $10.25-10.75$ is the $1^{nd}$ octile class.

Thus

• $l = 10.25$, the lower limit of the $1^{nd}$ octile class
• $N=60$, total number of observations
• $f =12$, frequency of the $1^{nd}$ octile class
• $F_< = 7$, cumulative frequency of the class previous to $1^{nd}$ octile class
• $h =0.5$, the class width

The first octile $O_1$ can be computed as follows:

 \begin{aligned} O_1 &= l + \bigg(\frac{\frac{1(N)}{8} - F_<}{f}\bigg)\times h\\ &= 10.25 + \bigg(\frac{\frac{1*60}{8} - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{7.5 - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \big(0.0417\big)\times 0.5\\ &= 10.25 + 0.0208\\ &= 10.2708 \text{ tons} \end{aligned}

The maximum load for lower $12.5$ % of the cables is first octile $O_1 = 10.2708$ tons.

 \begin{aligned} O_{2} &=\bigg(\dfrac{2(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{2(60)}{8}\bigg)^{th}\text{ value}\\ &=\big(15\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $15$ is $19$, the corresponding class $10.25-10.75$ is the $2^{th}$ octile class.

Thus

• $l = 10.25$, the lower limit of the $2^{th}$ octile class
• $N=60$, total number of observations
• $f =12$, frequency of the $2^{th}$ octile class
• $F_< = 7$, cumulative frequency of the class previous to $2^{th}$ octile class
• $h =0.5$, the class width

The second octile $O_2$ can be computed as follows:

 \begin{aligned} O_2 &= l + \bigg(\frac{\frac{2(N)}{8} - F_<}{f}\bigg)\times h\\ &= 10.25 + \bigg(\frac{\frac{2*60}{8} - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \bigg(\frac{15 - 7}{12}\bigg)\times 0.5\\ &= 10.25 + \big(0.6667\big)\times 0.5\\ &= 10.25 + 0.3333\\ &= 10.5833 \text{ tons} \end{aligned}
The maximum load for lower $25$ % of the cable is second octile $O_2 = 10.5833$ tons.

## Octiles 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 first, and seventh octile for the above data.

#### 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

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

$O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$

where $N$ is the total number of observations.

 \begin{aligned} O_{1} &=\bigg(\dfrac{1(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{1(100)}{8}\bigg)^{th}\text{ value}\\ &=\big(12.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $12.5$ is $18$, the corresponding class $6-8$ is the $1^{nd}$ octile class.

Thus

• $l = 6$, the lower limit of the $1^{nd}$ octile class
• $N=100$, total number of observations
• $f =14$, frequency of the $1^{nd}$ octile class
• $F_< = 4$, cumulative frequency of the class previous to $1^{nd}$ octile class
• $h =2$, the class width

The first octile $O_1$ can be computed as follows:

 \begin{aligned} O_1 &= l + \bigg(\frac{\frac{1(N)}{8} - F_<}{f}\bigg)\times h\\ &= 6 + \bigg(\frac{\frac{1*100}{8} - 4}{14}\bigg)\times 2\\ &= 6 + \bigg(\frac{12.5 - 4}{14}\bigg)\times 2\\ &= 6 + \big(0.6071\big)\times 2\\ &= 6 + 1.2143\\ &= 7.2143 \text{ ('00 grams)} \end{aligned}

The maximum weight of lower $12.5$ % of the pumpkins is first octile $O_1 = 7.2143$ ('00 grams).

 \begin{aligned} O_{7} &=\bigg(\dfrac{7(N)}{8}\bigg)^{th}\text{ value}\\ &= \bigg(\dfrac{7(100)}{8}\bigg)^{th}\text{ value}\\ &=\big(87.5\big)^{th}\text{ value} \end{aligned}

The cumulative frequency just greater than or equal to $87.5$ is $100$, the corresponding class $12-14$ is the $7^{th}$ octile class.

Thus

• $l = 12$, the lower limit of the $7^{th}$ octile class
• $N=100$, total number of observations
• $f =20$, frequency of the $7^{th}$ octile class
• $F_< = 80$, cumulative frequency of the class previous to $7^{th}$ octile class
• $h =2$, the class width

The seventh octile $O_7$ can be computed as follows:

 \begin{aligned} O_7 &= l + \bigg(\frac{\frac{7(N)}{8} - F_<}{f}\bigg)\times h\\ &= 12 + \bigg(\frac{\frac{7*100}{8} - 80}{20}\bigg)\times 2\\ &= 12 + \bigg(\frac{87.5 - 80}{20}\bigg)\times 2\\ &= 12 + \big(0.375\big)\times 2\\ &= 12 + 0.75\\ &= 12.75 \text{ ('00 grams)} \end{aligned}
The maximum weight of lower $87.5$ % of the students is seventh octile $O_7 = 12.75$ ('00 grams).

## Conclusion

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