## 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.

## Formula

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

## 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 $O_1$ 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.

## 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 |

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.