## Deciles for ungrouped 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

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

$D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 9$

where,

- $n$ is the total number of observations.

## Example 1

The marks obtained by a sample of 20 students in a class test are as follows:

20,30,21,29,10,17,18,15,27,25,16,15,19,22,13,17,14,18,12 and 9.

Find

a. the upper marks for the lowest 20 % of the students,,

b. the upper marks for the lowest 50 % of the students,

c. the lower marks for the upper 10 % of the students.

### Solution

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

$D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 9$

where $n$ is the total number of observations.

**Arrange the data in ascending order**

9, 10, 12, 13, 14, 15, 15, 16, 17, 17, 18, 18, 19, 20, 21, 22, 25, 27, 29, 30

a. The upper marks for the lowest 20 % of the students is $D_2$.

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

`$$ \begin{aligned} D_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{2(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(4.2\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(4\big)^{th} \text{ obs.}+0.2 \big(\text{Value of } \big(5\big)^{th}\text{ obs.}-\text{Value of }\big(4\big)^{th} \text{ obs.}\big)\\ &=13+0.2\big(14 -13\big)\\ &=13.2 \text{ marks} \end{aligned} $$`

Thus, the upper limit of marks obtained by the students for the lowest $20$ % of the students is $13.2$ marks.

b. The upper marks for the lowest 50 % of the students is $D_5$.

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

`$$ \begin{aligned} D_{5} &=\text{Value of }\bigg(\dfrac{5(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{5(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(10.5\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(10\big)^{th} \text{ obs.}+0.5 \big(\text{Value of } \big(11\big)^{th}\text{ obs.}-\text{Value of }\big(10\big)^{th} \text{ obs.}\big)\\ &=17+0.5\big(18 -17\big)\\ &=17.5 \text{ marks} \end{aligned} $$`

Thus, the upper limit of marks obtained by the students for the lowest $50$ % of the students is $17.5$ marks.

c. The lower marks for the upper 10 % of the students is $D_9$.

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

`$$ \begin{aligned} D_{9} &=\text{Value of }\bigg(\dfrac{9(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{9(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(18.9\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(18\big)^{th} \text{ obs.}+0.9 \big(\text{Value of } \big(19\big)^{th}\text{ obs.}-\text{Value of }\big(18\big)^{th} \text{ obs.}\big)\\ &=27+0.9\big(29 -27\big)\\ &=28.8 \text{ marks} \end{aligned} $$`

Thus, the lower marks obtained by the students for the upper $10$ % of the students is $28.8$ marks.

## Example 2

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75,89,72,78,87, 85, 73, 75, 97, 87, 84, 76,73,79,99,86,83,76,78,73.

Find

a. the upper limit of Blood sugar level for the lowest 10 % of the patients,

b. the upper limit of Blood sugar level for the lowest 30 % of the patients,

c. the lower limit of Blood sugar level for the upper 20 % of the patients.

### Solution

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

$D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,\cdots, 9$

where $n$ is the total number of observations.

**Arrange the data in ascending order**

72, 73, 73, 73, 75, 75, 76, 76, 78, 78, 79, 80, 82, 83, 84, 85, 86, 87, 97, 99

a. The upper limit of blood sugar level for the lowest 10% of the patients is $D_1$.

The first decile $D_1$ can be computed as follows:

`$$ \begin{aligned} D_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{10}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(2.1\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(2\big)^{th} \text{ obs.}+0.1 \big(\text{Value of } \big(3\big)^{th}\text{ obs.}-\text{Value of }\big(2\big)^{th} \text{ obs.}\big)\\ &=73+0.1\big(73 -73\big)\\ &=73 \text{ mg/dl} \end{aligned} $$`

Thus, the upper limit of blood sugar level for the lowest $10$ % of the patients is $73$ mg/dl.

b. The upper limit of blood sugar level for the lowest 30% of the patients is $D_3$.

The third decile $D_3$ can be computed as follows:

`$$ \begin{aligned} D_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{3(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(6.3\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(6\big)^{th} \text{ obs.}+0.3 \big(\text{Value of } \big(7\big)^{th}\text{ obs.}-\text{Value of }\big(6\big)^{th} \text{ obs.}\big)\\ &=75+0.3\big(76 -75\big)\\ &=75.3 \text{ mg/dl}. \end{aligned} $$`

Thus, the upper limit of blood sugar level for the lowest $30$ % of the patients is $75.3$ mg/dl.

c. The lower limit of blood sugar level for the upper 20% of the patients is $D_8$. (because above $D_8$ upper 20% patients lies)

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

`$$ \begin{aligned} D_{8} &=\text{Value of }\bigg(\dfrac{8(n+1)}{10}\bigg)^{th} \text{ obs.}\\ &=\text{Value of }\bigg(\dfrac{8(20+1)}{10}\bigg)^{th} \text{ obs.}\\ &= \text{Value of }\big(16.8\big)^{th} \text{ obs.}\\ &= \text{Value of }\big(16\big)^{th} \text{ obs.}+0.8 \big(\text{Value of } \big(17\big)^{th}\text{ obs.}-\text{Value of }\big(16\big)^{th} \text{ obs.}\big)\\ &=85+0.8\big(86 -85\big)\\ &=85.8\text{ mg/dl} \end{aligned} $$`

Thus, the lower limit of blood sugar level level for the upper $20$ % of the patients is $85.8$ mg/dl.