Partition Value | Quartiles, Deciles and Percentiles

Partition values are statistical measures that divide a dataset into equal parts to help in understanding the distribution and spread of data by indicating where certain percentages of the data fall. The most commonly used partition values are quartiles, deciles, and percentiles.

Quartiles, deciles, and Percentiles of partition values represent various perspectives on the same subject. To put it another way, these are values that partition the same collection of observations in several ways. As a result, it can divide these into many equal parts.

• Each quartile splits the total area into 4 equal parts (0.25 each).
• Each decile splits the total area into 10 equal parts (0.10 each).
• Each percentile splits the total area into 100 equal parts (0.01 each).

Quartiles

Quartiles divide a dataset into four equal parts, each containing 25% of the data.

The three quartiles are:

• Q1 (First Quartile/ Lower Quartile): 25% of the data fall below this value.
• Q2 (Second Quartile / Median): 50% of the data fall below this value.
• Q3 (Third Quartile / Upper Quartile): 75% of the data fall below this value.

The general equation to find the position of the Quartile is:

Q_k=\frac{k(N+1)}{4},for \:k=1,2, and \: 3.

When we put the values we get,

Q_{1}=[\frac{N+1}{4}]^{th}~item

Q_{2}=[\frac{N+1}{2}]^{th}~item

Q_{3}=[\frac{3(N+1)}{4}]^{th}~item

where, n is the total number of observations, Q₁ is First Quartile, Q₂ is Second Quartile, and Q₃ is Third Quartile.

Example 1: 

Calculate the lower and upper quartiles of the following weights in the family: 25, 17, 32, 11, 40, 35, 13, 5, and 46.

Solution: 

Age (in years)	Number of Employee	Cumulative
500 - 600	10	10
600 - 700	12	22(m)
700 - 800	16 (f)	38
800 - 900	14	52
900 - 1000	8	60

First of all organise the numbers in ascending order. 5, 11, 13, 17, 25, 32, 35, 40, 46

Lower quartile, Q_{1}=[\frac{N+1}{4}]^{th}~item

Q_{1}=[\frac{9+1}{4}]^{th}~item

Q₁ = 2.5^th term

As per the quartile formula;

Q₁ = 2^nd term + 0.5(3^rd term - 2^nd term)

Q₁ = 11 + 0.5(13 - 11) = 12

Q₁ = 12

Upper Quartile, Q_{3}=[\frac{3(N+1)}{4}]^{th}~item

Q_{3}=[\frac{3(9+1)}{4}]^{th}~item

Q₃ = 7.5^th item

Q₃ = 7^th term + 0.5(8^th term - 7^th term)

Q₃ = 35 + 0.5(40 - 35) = 37.5

Q₃ = 37.5

Example 2:

Calculate Q1 and Q3 for the data related to the age in years of 99 members in a housing society.

Solution:

Age (in years)	Number of Members	Cumulative Frequency
10	20	20
18	5	25
25	10	35
35	30	65
40	20	85
45	14	99

Q_{1}=[\frac{N+1}{4}]^{th}~item

Q_{1}=[\frac{99+1}{4}]^{th}~item

Q₁ = 25^th item

Now, the 25^th item falls under the cumulative frequency of 25 and the age against this cf value is 18.

Q₁ = 18 years

Q_{3}=[\frac{3(N+1)}{4}]^{th}~item

Q_{3}=[\frac{3(99+1)}{4}]^{th}~item

Q₃ = 75^th item

Now, the 75^th item falls under the cumulative frequency of 85 and the age against this cf value is 40.

Q₃ = 40 years 

Example 3:

Determine the quartiles Q₁ and Q₃ for the company's salaries listed below.

Solution:

Salaries(per day in ₹)	Number of Employee	Cumulative Frequency
500 - 600	10	10(m1)
600 - 700	12(f1)	22
700 - 800	16	38(m2)
800 - 900	14(f2)	52
900 - 1000	8	60

Q_{1}~Class=\frac{N}{4}

Q_{1}~Class=\frac{60}{4}

= 15^th item

Now, the 15^th item falls under the cumulative frequency 22 and the salary against this cf value lies in the group 600-700.

Q_{1}=l_{1}+\frac{\frac{N}{4}-m_{1}}{f_{1}}\times{c_{1}}

Q_{1}=600+\frac{\frac{60}{4}-10}{12}\times{100}

Q₁ = ₹641.67

Q_{3}~Class=\frac{3N}{4}

Q_{3}~Class=\frac{180}{4}

Q₃ = 45^th item

Now, the 45^th item falls under the cumulative frequency 52 and the salary against this cf value lies in the group 800-900.

Q_{3}=l_{1}+\frac{\frac{3N}{4}-m_{3}}{f_{3}}\times{c_{3}}

Q_{3}=800+\frac{\frac{180}{4}-38}{14}\times{100}

Q_{3}=800+\frac{7}{14}\times{100}

Q_{3}=800+50

Deciles

The deciles involve dividing a dataset into ten equal parts based on numerical values, each containing 10% of the data.

The three quartiles are:

• D1 (First Decile): 10% of the data.
• D5 (Fifth Decile / Median): 50% of the data fall below this value.
• D9 (Ninth Decile): 90% of the data fall below this value.

The general equation to find the position of the Quartile is:

D_k=\frac{k(N+1)}{10},for \:k=1,2, \dots 9.

When substituting for each, we get,

D_{1}=[\frac{N+1}{10}]^{th}~item

D_{2}=[\frac{2(N+1)}{10}]^{th}~item

\dots

D_{9}=[\frac{9(N+1)}{10}]^{th}~item

where, n is the total number of observations, D₁ is First Decile, D₂ is Second Decile, ... D₉ is Ninth Quartile.

Example 1: Calculate the D₁ and D₅ from the following weights in a family: 25, 17, 32, 11, 40, 35, 13, 5, and 46.

Solution: 

First of all, organise the numbers in ascending order.

5, 11, 13, 17, 25, 32, 35, 40, 46

D_{1}=[\frac{N+1}{10}]^{th}~item

D_{1}=[\frac{9+1}{10}]^{th}~item

D₁ = 1^st item = 5

D_{5}=[\frac{5(N+1)}{10}]^{th}~item

D_{1}=[\frac{5(9+1)}{10}]^{th}~item

D₅ = 5^th item = 25

Example 2: Calculate D₂ and D₆ for the data related to the age (in years) of 99 members in a housing society.

Solution:

Age (in years)	Number of Members	Cumulative Frequency
10	20	20
18	5	25
25	10	35
35	30	65
40	20	85
45	14	99

D_{2}=[\frac{2(N+1)}{10}]^{th}~item

D_{2}=[\frac{2(99+1)}{10}]^{th}~item

D₂ = 20^th item

Now, the 20^th item falls under the cumulative frequency of 25 and the age against this cf value is 18.

D₂ = 18 years

Similarly D_{6}=[\frac{6(N+1)}{10}]^{th}~item

D_{6}=[\frac{6(99+1)}{10}]^{th}~item

D₆ = 60^th item

Now, the 60^th item falls under the cumulative frequency of 65 and the age against this cf value is 35.

D₆ = 35 years

Example 3: Determine D₄ for the company's salary listed below.

Solution:

Salaries(per day in ₹)	Number of Employee	Cumulative Frequency
500 - 600	10	10
600 - 700	12	22(m)
700 - 800	16(f)	38
800 - 900	14	52
900 - 1000	8	60

In case N is an even number, the following formula is used:

D_{4}=[\frac{4N}{10}]^{th}~item

D_{4}=[\frac{4(60)}{10}]^{th}~item

D₄ = 24^th item

Now, the 24^th item falls under the cumulative frequency 22 and the salary against this cf value lies in the group 700-800.

D_{4}=l+\frac{\frac{4(N)}{10}-m}{f}\times{c}

D_{4}=700+\frac{\frac{4(60)}{10}-22}{16}\times{100}

D₄ =  ₹712.5

Percentiles

Centiles are another term for percentiles. Percentiles divide a dataset into 100 equal parts, with each percentile representing the value below which a certain percentage of the data falls. These percentiles are commonly denoted as P1, P2, P3, ..., P99.₁, P₂, P₃,..P₉₉.

For example:

• P1: 1% of the data is less than or equal to this value.
• P50: 50% of the data is less than or equal to this value (also called the median).
• P90: 90% of the data is less than or equal to this value (commonly used in performance benchmarking).

The Three Quartiles (special percentiles)

• P25: 25th percentile (also known as Q1) – 25% of the data is below this value.
• P50: 50th percentile (also known as Q2 or the median) – 50% of the data is below this value.
• P75: 75th percentile (also known as Q3) – 75% of the data is below this value.

The general equation to find the position of the Quartile is:

P_k=\frac{k(N+1)}{100},for \:k=1,2, \dots 99.

When substituting for each, we get,

P_{1}=[\frac{N+1}{100}]^{th}~item

P_{2}=[\frac{2(N+1)}{100}]^{th}~item

\dots

D_{99}=[\frac{99(N+1)}{100}]^{th}~item

where, n is the total number of observations, P₁ is First Percentile, P₂ is Second Percentile, ... P₉₉ is Ninety Ninth Percentile.

Example 1: Calculate the P₂₀ and P₉₀ from the following weights in the family: 25, 17, 32, 11, 40, 35, 13, 5, and 46.

Solution: 

First of all, organise the numbers in ascending order.

5, 11, 13, 17, 25, 32, 35, 40, 46

P_{20}=[\frac{20(N+1)}{100}]^{th}~item

P_{20}=[\frac{20(9+1)}{100}]^{th}~item

P₂₀ = 2^nd item

P₂₀ = 11

P_{90}=[\frac{90(N+1)}{100}]^{th}~item

P_{90}=[\frac{90(9+1)}{100}]^{th}~item

P₉₀ = 9^th item

P₉₀ = 40

Example 2: Calculate P₁₀ and P₇₅ for the data related to the age (in years) of 99 members in a housing society.

Solution:

P_{10}=[\frac{10(N+1)}{100}]^{th}~item

P_{10}=[\frac{10(99+1)}{100}]^{th}~item

P₁₀ = 10^th item

Now, the 10^th item falls under the cumulative frequency of 20 and the age against this cf value is 10.

P₁₀ = 10 years

P_{75}=[\frac{75(N+1)}{100}]^{th}~item

P_{75}=[\frac{75(99+1)}{100}]^{th}~item

P₇₅ = 75^th item

Now, the 75^th item falls under the cumulative frequency of 85 and the age against this cf value is 40.

P₇₅ = 40 years

Example 3: Determine the value of P₅₀ for the company's salary listed below.

Solution:

Salaries(per day in ₹)	Number of Employee
500 - 600	10
600 - 700	12
700 - 800	16
800 - 900	14
900 - 1000	8

In case N is an even number, the following formula is used:

P_{50}=[\frac{50(N)}{100}]^{th}~item

P_{50}=[\frac{50(60)}{100}]^{th}~item

P₅₀ = 30^th item

Now, the 30^th item falls under the cumulative frequency 38 and the salary against this cf value lies between 700-800.

P_{50}=l+\frac{\frac{50(N)}{100}-m}{f}\times{c}

P_{50}=700+\frac{\frac{50(60)}{100}-22}{16}\times{100}

P_{50}=700+\frac{30-22}{16}\times{100}

P₅₀ = ₹750

Related Articles

• Range
• Calculation of Range and Coefficient of Range
• Interquartile Range and Quartile Deviation

Practice Questions on Quartiles, Deciles, and Percentiles

Question 1: Given the dataset: 5, 7, 8, 12, 15, 16, 18, 20, 22, 25, find the quartiles Q1, Q2, and Q3.

Question 2: Consider the dataset: 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90. Calculate the deciles D3, D5, and D7.

Question 3: Given the dataset: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, determine the 25th percentile P25 and the 75th percentile P75

Question 4: Using the dataset: 1, 4, 7, 8, 10, 12, 14, 15, 18, 20, 22, find the 40th percentile P40 and the 90th percentile P90.