In statistics, we usually deal with the large amount of numerical data. There are various concepts and formulas in statistics which are widely applicable in different researches and surveys. Quartile is also one of the very useful topic in statistics. The quartiles are defined as the set of values having three points that divide the data set into four equal parts. Quartiles are the values that divide a list of numbers into three quarters. The middle part computes the central point of a distribution. It indicates the data which are nearest to the central point. Also, the lower part indicates just about half of the information set falls under the median and the upper portion tells about the half that falls over the median. Overall, the quartiles represent the distribution or dispersion of the data.

In the median, a usual measure of dispersion can be found from the lesser and higher quartile. Besides mean and median, there are other measures which divide a series into certain equal parts. In the case of a median, 50% of the values lie above it and an equal number below it, dividing the series into equal parts. Partitional values are mainly divided into three ways:

- Quartiles
- Deciles
- Percentiles

Let us go ahead in this article below and understand more about quartiles and methods for finding them.

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Quartiles divide the whole series into 4 equal parts. So there are 3 quartiles namely first Quartile denoted by Q_{1}, second Quartile denoted by Q_{2} and third Quartile denoted by Q_{3}. Second Quartile is nothing but Median. Since it denotes the position of the item in the series, it is a positional average.

Whenever we find quartiles, we have to arrange the data in the ascending order.

Here, we will learn how to solve quartiles with the help of the quartiles formula.

### Upper Quartiles

Upper quartile is nothing but the third quartile. The upper quartile (Q_{3}) is the median of the upper half of the data set. Where as the lower quartile is nothing but the first quartile. The second quartile is also known as median.

**Quartiles in individual series**

When there are n number of items, Quartiles are given by

$Q_{1}$ = $\frac{n+1^{th}}{4}$ item

$Q_{2}$ = $\frac{n+1^{th}}{2}$ item

$Q_{3}$ = 3$\frac{n+1^{th}}{4}$ item

Formula for the rth Quartile:

$Q_r$ = $l_1 + \frac{r(\frac{N}{4}) - c}{f}(l_2 - l_1)$

where, $Q_r$ = rth quartile.

$l_1$ = the lower limit of the quartile class.

$l_2$ = the upper limit of the quartile class.

f = the frequency of the quartile class.

c = cumulative frequency of the class preceding the quartile class.

### Finding Quartiles

Let us see with the help of examples, how to find the quartiles:

### Solved Examples

**Question 1: **Find the quartiles of the following data: 3, 5, 6, 7, 9, 22, 33.

** Solution: **

Here the numbers are arranged in the increasing order, n =**7**

**Lower quartile, **$Q_{1}$ = $\frac{n+1^{th}}{4}$ item

= $\frac{(7+1)}{4}$ item

= 2^{nd }item =** 5**

**Median,** $Q_{2}$ = $\frac{n+1^{th}}{2}$ item = $\frac{(7+1)}{2}$ item = 4^{th} item = **7**

Upper Quartile,$Q_{3}$ = 3 $\frac{n+1^{th}}{4}$ item = $\frac{3(7+1)}{4}$ item = 6^{th} item = **22**

**Question 2: **Find the Quartiles of the following marks:-

21, 12, 36, 15, 25, 34, 25, 34

** Solution: **

First we have to arrange the numbers in the ascending order.

12, 15, 21, 25, 25, 34, 34, 36

n = 8

**Lower Quartile, **$Q_{1}$ = $\frac{n+1^{th}}{4}$ item = $\frac{8+1}{4}$ item = 2.25^{th} item

= 2^{nd} item + 0.25(3^{rd} item - 2^{nd} item)

=15 + 0.25(21 - 15) = 15 + 0.25(6) = 16.5

**Second Quartile,** $Q_{2}$ = $\frac{n+1^{th}}{2}$ item

= $\frac{8+1}{2}$ item = 4.5^{th} item

= 4^{th} item + 0.5(5^{th} item - 4^{th} item)

=15 + 0.5(25 - 15)

= 15 + 0.5(10) = 20

**Third Quartile**, $Q_{3}$ = 3$\frac{n+1^{th}}{4}$ item

= $\frac{3(8+1)}{4}$ item

= 6.75th item

= 6^{th} item + 0.75(7^{th} item - 6^{th} item)

= 34 + 0.75(34 - 34)

= 34 + 0.75(0)

= 34

Half the distance between the third and first quartile is called Quartile Deviation. It is also known as Semi Inter quartile range.

Quartile deviation = $\frac{Q_{3}-Q_{1}}{2}$

where, Q_{1} is the first quartile and Q_{3} is the third quartile of the series.

The difference between the Upper and Lower Quartile is called inter quartile range. Interquartile range (IQR), also called the midspread, is a measure of statistical dispersion, being equal to the difference between the upper and lower quartiles. The interquartile range is a measure of variability, based on dividing a data set into quartiles.

Inter quartile range = Q_{3} - Q_{1}where, Q_{1} is the first quartile and Q_{3} is the third quartile of the series._{}
→ Read More
The income quartile method has been adopted as the most objective method of comparing change in the income profile of a community over time. The income quartile method assumes an even distribution within each income group. To examine the nature of income distribution household were divided into income quartiles. The major characteristics of households falling in different income quartiles. Households are divided into eight income classes, 4 income quartiles in each of rural and urban areas. Factor incomes in agriculture and non agriculture are distributed to each quartile on the basis of a income survey data.

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N.

Then $Q_{1}$ = Size of $\frac{N+1^{th}}{4}$ item

$Q_{2}$ = Size of $\frac{N+1^{th}}{2}$ item

$Q_{3}$ = Size of $3$$\frac{N+1^{th}}{4}$ item

### Solved Examples

**Question 1: **Find the Quartiles of the following marks

** Solution: **

Here, N = 18

**First Quartile $Q_{1}$ **= Size of $\frac{N+1^{th}}{4}$ item

= Size of $\frac{19}{4}^{th}$ item

= Size of 4.75^{th} item = **2**

Second Quartiles $Q_{2}$ = Size of $\frac{N+1^{th}}{2}$ item

= Size of $\frac{19}{2}^{th}$ item

= Size of 9.5^{th} item =** 6**

Third Quartile, $Q_{3}$= Size of $3\frac{N+1^{th}}{4}$ item

= Size of $\frac{3(19)}{4}^th$ item

= Size of 14.25^{th} item =**10**

**Question 2: **Find the Quartiles of the following data

** Solution: **

Here, N = 40

Lower Quartile(Q_{1}) = Size of $\frac{N+1^{th}}{4}$ item

= Size of $\frac{40}{4}$ item

= Size of 10^{th} item = **5**

Second Quartile, Q_{2} = Size of $\frac{N+1^{th}}{2}$ item

= Size of $\frac{40}{2^{th}}$ item

= Size of 20^{th} item = **12**

Upper Quartile, Q_{3} = Size of 3 $\frac{N+1^{th}}{4}$ item

= Size of item $\frac{3(40)}{4^{th}}$

= Size of 30^{th} item = **15**

In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.

$Q_{1}$ = l + $\frac{(\frac{N}{4}-cf)c}{f}$

$Q_{2}$ = l + $\frac{(\frac{N}{2}-cf)c}{f}$

$Q_{3}$ = l + $\frac{(\frac{3N}{4}-cf)c}{f}$

Where,

l = lower limit of the Q_{1}, Q_{2 }and Q_{3} classes respectively.

cf = cumulative frequency of the class just preceding the corresponding classes.

F = frequency of the Q_{1}, Q_{2} and Q_{3} classes respectively and

c = class width of the corresponding classes.

### Solved Example

**Question: **Find the Quartiles of the following data:-

** Solution: **

Here, N = 15

**Step 1:**

$\frac{N}{4}$ = $\frac{15}{4}$ = 3.75 which lies in 10-20.

Q_{1} class = 10-20

l = 10

cf = 0, c = 10

f = 4

Therefore** Lower Quartile Q**_{1} = l + $\frac{(\frac{N}{4}-cf)c}{f}$

= 10 + $\frac{(3.75-0)10}{4}$

= 10 + 9.38 =**19.38**

**Step 2:**

$\frac{N}{2}$ = $\frac{15}{2}$ = 7.5 which lies in 30-40.

Q_{2 } class = 30-40

l = 30

cf = 7 c = 10

f = 2

Therefore**Second Quartile, Q**_{2} = l + $\frac{(\frac{N}{2}-cf)c}{f}$

= 30 + $\frac{(7.5-7)10}{2}$

= 30 + 2.5 =** 32.5**

**Step 3:**

$\frac{3N}{4}$ = $\frac{3(15)}{4}$ = 11.25 which lies in 50-60.

Q_{2} = 50-60

l = 50

cf = 10 c = 10

f = 5

Therefore**Upper Quartile Q**_{3} = l + $\frac{(\frac{3N}{4}-cf)c}{f}$

= 50 + $\frac{(11.25-10)10}{5}$

= 50 + 2.5 =**52.5**

Deciles divide the whole series into 10 equal parts. So there are 9 quartiles namely first decile denoted by D_{1}, second decile denoted by D_{2} …........ and 9th decile denoted by D_{9}. 5th decile is nothing but Median. Since it denotes the position of the item in the series, it is a positional average. Whenever we find deciles we have to arrange the data in the ascending order.

### Deciles in Individual Series

When there are n numbers of items,

The r^{th} decile are given by D_{r } = $r\frac{n+1^{th}}{10}$ item

### Deciles in Discrete Series

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N, then

r^{th} decile, D_{r} = Size of $r\frac{N+1^{th}}{10}$ item

### Deciles in Continuous Series

In the case of continuous series we find the cumulative frequency first and then use the interpolation formula. The r^{th} decile is given by:

D_{r} = l + $\frac{(\frac{rN}{10}-cf)c}{f}$

Where, l is the lower limit of the D_{r} class.

cf is the cumulative frequency of the class just preceding D_{r} class.

f is the frequency and c is the class width of the D_{r} class.

→ Read More Percentiles divide the whole series into 100 equal parts. So there are 99 quartiles namely first percentile denoted by P_{1}, second percentile denoted by P_{2} ….... and 99^{th} percentile denoted by P_{99}. 50^{th} percentile is nothing but Median. Since it denotes the position of the item in the series, it is a positional average. Whenever we find Percentiles we have to arrange the data in the ascending order.

### Percentiles in Individual Series

When there are n number of items, The k^{th} percentile are given by

$P_{k}$ = $k\frac{n+1^{th}}{100}$ item

### Percentiles in Discrete Series

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N, then

k^{th} percentile, P_{k} = Size of $k\frac{N+1^{th}}{100}$ item

### Percentiles in Continuous Series

In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.

The k^{th} percentile is given by P_{k} = l + $\frac{(\frac{kN}{100}-cf)c}{f}$

Where,

l is the lower limit of the P_{k} class.

cf is the cumulative frequency of the class just preceding P_{k} class.

f is the frequency and c is the class width of the P_{k} class.

→ Read More### Solved Examples

**Question 1: **Determine the median, lesser quartile, higher quartile, inter-quartile
and range of the given information of the following sequence.

9, 40, 11, 35, 8, 13, 24, 7, 25, 5, 20, 18

** Solution: **

**Step 1:**

Arrange the data in ascending order:

Given: 5, 7, 8, 9, 11, 13, 18, 20, 24, 25, 35, 40

Step 2:

Lower quartile = `(8 + 9)/2` = 8.5

Step 3:

Median = `(13+18)/2` = 15.5

Step 4:

Upper quartile = `(24+25)/2` = 24.5

Step 5:

Inter quartile range = Upper quartile – lower quartile

= 24.5 – 8.5

= 16

Step 6:

Range = largest value – smallest value

= 40 – 5

= 35

**Answer:** Inter-quartile range = 16

**Question 2: **Estimate the median, lower quartile, upper quartile, inter-quartile and
range of the given information of the following sequence.

45, 23, 11, 42, 78, 21, 12, 27, 34, 52, 99, 10

** Solution: **

**Step 1: **Arrange the data in ascending order:

Given data: 10, 11, 12, 21, 23, 27, 34, 42, 45, 52, 78, 99

Step 2:

Lower quartile = `(12+21)/2` =16.5

Step 3:

Median = `(27+34)/2` = 30.5

Step 4:

Upper quartile = `(45+52)/2` = 48.5

Step 5:

Inter-quartile range = Upper quartile – lower quartile

= 48.5 – 30.5

= 18

Step 6: To find the range

Range = largest value – smallest value

= 99 – 10

= 89

Given below are some of the practice problems on Quartiles.

### Practice Problems

**Question 1: **

**Question 2: **Evaluate the median, lower quartile, upper quartile, inter-quartile and
range of the given information of the following sequence.

2, 8, 10, 12, 6, 4

Whenever we find quartiles, we have to arrange the data in the ascending order.

Here, we will learn how to solve quartiles with the help of the quartiles formula.

Upper quartile is nothing but the third quartile. The upper quartile (Q

When there are n number of items, Quartiles are given by

$Q_{1}$ = $\frac{n+1^{th}}{4}$ item

$Q_{2}$ = $\frac{n+1^{th}}{2}$ item

$Q_{3}$ = 3$\frac{n+1^{th}}{4}$ item

Formula for the rth Quartile:

$Q_r$ = $l_1 + \frac{r(\frac{N}{4}) - c}{f}(l_2 - l_1)$

where, $Q_r$ = rth quartile.

$l_1$ = the lower limit of the quartile class.

$l_2$ = the upper limit of the quartile class.

f = the frequency of the quartile class.

c = cumulative frequency of the class preceding the quartile class.

Let us see with the help of examples, how to find the quartiles:

Here the numbers are arranged in the increasing order, n =

= $\frac{(7+1)}{4}$ item

= 2

Upper Quartile,

21, 12, 36, 15, 25, 34, 25, 34

First we have to arrange the numbers in the ascending order.

12, 15, 21, 25, 25, 34, 34, 36

n = 8

= 2

=15 + 0.25(21 - 15) = 15 + 0.25(6) = 16.5

= $\frac{8+1}{2}$ item = 4.5

= 4

=15 + 0.5(25 - 15)

= 15 + 0.5(10) = 20

= $\frac{3(8+1)}{4}$ item

= 6.75th item

= 6

= 34 + 0.75(34 - 34)

= 34 + 0.75(0)

= 34

Half the distance between the third and first quartile is called Quartile Deviation. It is also known as Semi Inter quartile range.

Quartile deviation = $\frac{Q_{3}-Q_{1}}{2}$

where, Q

The difference between the Upper and Lower Quartile is called inter quartile range. Interquartile range (IQR), also called the midspread, is a measure of statistical dispersion, being equal to the difference between the upper and lower quartiles. The interquartile range is a measure of variability, based on dividing a data set into quartiles.

Inter quartile range = Q

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N.

Then $Q_{1}$ = Size of $\frac{N+1^{th}}{4}$ item

$Q_{2}$ = Size of $\frac{N+1^{th}}{2}$ item

$Q_{3}$ = Size of $3$$\frac{N+1^{th}}{4}$ item

X |
2 |
4 | 6 | 8 | 10 |

F | 5 | 4 | 3 | 2 | 4 |

X |
F |
Cumulative Frequency (c.f.) |

2 |
5 |
5 |

4 |
4 |
5 + 4 = 9 |

6 |
3 |
9 + 3 = 12 |

8 |
2 |
12 + 2 = 14 |

10 |
4 |
14 + 4 = 18 |

Here, N = 18

= Size of $\frac{19}{4}^{th}$ item

= Size of 4.75

Second Quartiles $Q_{2}$

= Size of $\frac{19}{2}^{th}$ item

= Size of 9.5

Third Quartile, $Q_{3}$

= Size of $\frac{3(19)}{4}^th$ item

= Size of 14.25

X |
5 | 10 | 12 | 15 | 28 |

F | 12 | 4 | 6 | 8 | 10 |

X |
F | Cumulative Frequency |

5 |
12 |
12 |

10 |
4 |
12 + 4 = 16 |

12 |
6 |
16 + 6 = 22 |

15 |
8 |
22 + 8 = 30 |

28 |
10 |
30 + 10 = 40 |

Here, N = 40

Lower Quartile(Q

= Size of $\frac{40}{4}$ item

= Size of 10

Second Quartile, Q

= Size of $\frac{40}{2^{th}}$ item

= Size of 20

Upper Quartile, Q

= Size of item $\frac{3(40)}{4^{th}}$

= Size of 30

$Q_{1}$ = l + $\frac{(\frac{N}{4}-cf)c}{f}$

$Q_{2}$ = l + $\frac{(\frac{N}{2}-cf)c}{f}$

$Q_{3}$ = l + $\frac{(\frac{3N}{4}-cf)c}{f}$

Where,

l = lower limit of the Q

cf = cumulative frequency of the class just preceding the corresponding classes.

F = frequency of the Q

c = class width of the corresponding classes.

Class |
F |

10-20 |
4 |

20-30 |
3 |

30-40 |
2 |

40-50 |
1 |

50-60 |
5 |

Class |
F |
Cumulative Frequency |

10-20 |
4 |
4 |

20-30 |
3 |
4 + 3 = 7 |

30-40 |
2 |
7 + 2 = 9 |

40-50 |
1 |
9 + 1 = 10 |

50-60 |
5 |
10 + 1 = 15 |

Here, N = 15

$\frac{N}{4}$ = $\frac{15}{4}$ = 3.75 which lies in 10-20.

Q

l = 10

cf = 0, c = 10

f = 4

Therefore

= 10 + $\frac{(3.75-0)10}{4}$

= 10 + 9.38 =

$\frac{N}{2}$ = $\frac{15}{2}$ = 7.5 which lies in 30-40.

Q

l = 30

cf = 7 c = 10

f = 2

Therefore

= 30 + $\frac{(7.5-7)10}{2}$

= 30 + 2.5 =

Q

l = 50

cf = 10 c = 10

f = 5

Therefore

= 50 + $\frac{(11.25-10)10}{5}$

= 50 + 2.5 =

The r

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N, then

r

In the case of continuous series we find the cumulative frequency first and then use the interpolation formula. The r

D

Where, l is the lower limit of the D

cf is the cumulative frequency of the class just preceding D

f is the frequency and c is the class width of the D

→ Read More Percentiles divide the whole series into 100 equal parts. So there are 99 quartiles namely first percentile denoted by P

$P_{k}$ = $k\frac{n+1^{th}}{100}$ item

In the case of discrete series, we first find the cumulative frequency. The last cumulative frequency will be N, then

k

In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.

The k

Where,

l is the lower limit of the P

cf is the cumulative frequency of the class just preceding P

f is the frequency and c is the class width of the P

→ Read More

Given below are some of the examples on quartiles.

9, 40, 11, 35, 8, 13, 24, 7, 25, 5, 20, 18

Arrange the data in ascending order:

Given: 5, 7, 8, 9, 11, 13, 18, 20, 24, 25, 35, 40

Step 2:

Lower quartile = `(8 + 9)/2` = 8.5

Step 3:

Median = `(13+18)/2` = 15.5

Step 4:

Upper quartile = `(24+25)/2` = 24.5

Step 5:

Inter quartile range = Upper quartile – lower quartile

= 24.5 – 8.5

= 16

Step 6:

Range = largest value – smallest value

= 40 – 5

= 35

45, 23, 11, 42, 78, 21, 12, 27, 34, 52, 99, 10

Given data: 10, 11, 12, 21, 23, 27, 34, 42, 45, 52, 78, 99

Step 2:

Lower quartile = `(12+21)/2` =16.5

Step 3:

Median = `(27+34)/2` = 30.5

Step 4:

Upper quartile = `(45+52)/2` = 48.5

Step 5:

Inter-quartile range = Upper quartile – lower quartile

= 48.5 – 30.5

= 18

Step 6:

Range = largest value – smallest value

= 99 – 10

= 89

Evaluate the median, lower quartile, upper quartile, inter-quartile and range of the given information of the following sequence.

60, 50, 20,30,10,402, 8, 10, 12, 6, 4

More topics in Quartiles | |

Interquartile Range | Percentile |

First Quartile | Deciles |

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