# Deciles Calculator for Ungrouped Data with Examples

## 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.

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

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

where,

• $n$ is the total number of observations.

## Deciles calculator for ungrouped data

Use this calculator to find the deciles for ungrouped (raw) data.

Decile Calculator
Enter the X Values (Separated by comma,)
Which Decile? (Between 1 to 9)
Results
Number of Obs. (n):
Ascending order of X values :
Required Decile : D{{index}}

## How to calculate Decile for ungrouped data?

Step 1 - Enter the $x$ values separated by commas

Step 2 - Enter the nuber between 1 to 9 (inclusive)

Step 3 - Click on "Calculate" button to get Decile for ungrouped data

Step 4 - Gives the output as number of observations $n$

Step 5 - Gives the output as ascending order data

Step 6 - Calculate the Decile for Ungrouped Data

## Deciles for ungrouped data 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 decile formula for $i^{th}$ decile is

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

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.}\\ &\quad +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.}\\ &\quad +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.}\\ &\quad +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.

## Deciles for ungrouped data 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 decile 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{ Obs.}\\ &=\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.}\\ &\quad +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.}\\ &\quad +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.}\\ &\quad +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.

## Deciles for ungrouped data Example 3

Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

65,76,64,73,74,80, 71, 68,66, 81, 79, 75, 70, 62, 83,63, 77, 78.

Find

a. the maximum Diastolic BP level for the lowest 10 % of the patients,
b. the minimum Diastolic BP level for the upper 10 % of the patients.

#### Solution

a. The maximum diastolic blood pressure 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{ Obs.}\\ &=\text{Value of }\bigg(\dfrac{1(18+1)}{10}\bigg)^{th} \text{ Obs.}\\ &= \text{Value of }\big(1.9\big)^{th} \text{ Obs.}\\ &= \text{Value of }\big(1\big)^{th} \text{ obs.}\\ &\quad +0.9 \big(\text{Value of } \big(2\big)^{th}\text{ obs.}-\text{Value of }\big(1\big)^{th} \text{ obs.}\big)\\ &=62+0.9\big(63 -62\big)\\ &=62.9 \text{ mmHg}. \end{aligned}

Thus, the maximum diastolic blood pressure for the loweer $10$ % of the patients is $62.9$ mmHg.

b. The minimum diastolic blood pressure for the upper 10% of the patients is $D_9$. (because above $D_9$ upper 10% patients lies)

The ninth 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(18+1)}{10}\bigg)^{th} \text{ Obs.}\\ &= \text{Value of }\big(17.1\big)^{th} \text{ Obs.}\\ &= \text{Value of }\big(17\big)^{th} \text{ obs.}\\ &\quad +0.1 \big(\text{Value of } \big(18\big)^{th}\text{ obs.}-\text{Value of }\big(17\big)^{th} \text{ obs.}\big)\\ &=81+0.1\big(83 -81\big)\\ &=81.2\text{ mmHg} \end{aligned}

Thus, the minimum diastolic blood pressure level for the upper $10$ % of the patients is $81.2$ mmHg.

## Deciles for ungrouped data Example 4

The following data are the heights, correct to the nearest centimeters, for a group of children:

126, 129, 129, 132, 132, 133, 133, 135, 136, 137,
137, 138, 141, 143, 144, 146, 147, 152, 154, 161 

a. the maximum height for the lower 30 % of the children,
b. the minimum height for the upper 20 % of the children.

#### Solution

a. The maximum height for the lowest 30% of the is children $D_3$.

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.}\\ &\quad +0.3 \big(\text{Value of } \big(7\big)^{th}\text{ obs.}-\text{Value of }\big(6\big)^{th} \text{ obs.}\big)\\ &=133+0.3\big(133 -133\big)\\ &=133 \text{ cm}. \end{aligned}

Thus, the maximum height for the lower $30$ % of the children is $13.2$ cm.

b. The minimum height for the upper 20% of the children is $D_8$. (because above $D_8$ upper 20% children 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.}\\ &\quad +0.8 \big(\text{Value of } \big(17\big)^{th}\text{ obs.}-\text{Value of }\big(16\big)^{th} \text{ obs.}\big)\\ &=146+0.8\big(147 -146\big)\\ &=146.8\text{ cm} \end{aligned}

Thus, the minimum height for the upper $20$ % of the children is $146.8$ cm.

## Deciles for ungrouped data Example 5

The rice production (in Kg) of 10 acres is given as: 1120, 1240, 1320, 1040, 1080, 1720, 1600, 1470, 1750, and 1885.

Find the second octile for the given data.

#### Solution

The minimum rice production for the upper 30% of the plots is $D_7$. (because above $D_7$ upper 30% plot lies)

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

 \begin{aligned} D_{7} &=\text{Value of }\bigg(\dfrac{7(n+1)}{10}\bigg)^{th} \text{ Obs.}\\ &=\text{Value of }\bigg(\dfrac{7(10+1)}{10}\bigg)^{th} \text{ Obs.}\\ &= \text{Value of }\big(7.7\big)^{th} \text{ Obs.}\\ &= \text{Value of }\big(7\big)^{th} \text{ obs.}\\ &\quad +0.7 \big(\text{Value of } \big(8\big)^{th}\text{ obs.}-\text{Value of }\big(7\big)^{th} \text{ obs.}\big)\\ &=1600+0.7\big(1720 -1600\big)\\ &=1684\text{ Kg} \end{aligned}

Thus, the minimum rice production for the upper $30$ % of the plots is $1684$ Kg.

## Conclusion

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