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.
