Deciles for grouped data

Deciles are the values of arranged data which divide whole data 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, $D_3$ is third decile and so on.

Formula

For discrete frequency distribution, 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 total number of observations.

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

$D_i=l + \bigg(\dfrac{\dfrac{iN}{10} - F_<}{f}\bigg)\times h$; $i=1,2,\cdots, 9$

where,

• $l$ is 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
• $h$ 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 $D_1$ 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

Deciles

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.

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 $2^{st}$ decile class.

Thus

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