THE QUARTILES

While the median splits data into two groups with an equal number of data, the quartiles separate them into four groups with an equal number of data. If a set of data has Q₁ as its first quartile and Q₃ as its third quartile, theoretically we can conclude that 25% of the data have values less than or equal to Q₁ and 75% of the data have values less than or equal to Q₃. Additionally, the median of the data can be viewed as the second quartile, Q₂. Theoretically, half of the data have values less than or equal to Q₂. To elaborate more about the median, click here.

Suppose that we are examining n data at ordinal scale, sorted from the smallest to the largest as follows: X₁, X₂, X₃, …, X_n. (Thus, X₁ ≤ X₂ ≤ X₃ ≤ … ≤ X_n) The quartiles are defined as follows.

The first quartile: Q₁ = X_b where

The third quartile: Q₃ = X_a where

 

Example 1

Fourteen students took a Social Statistics quiz. Their scores are as follows.

47   56   71   65   29  68   78   73   80  75   29  38  65   90

Find the first and the third quartiles.

Answer

Firstly, sort the data from smallest to largest. This results in:

29   29   38   47   56   65   65   68   71   73   75   78   80   90

Let X₁ = X₂ = 29, X₃ = 38, X₄ = 47, X₅ = 56, …, X₁₄ = 90. To find the first quartile, compute .

Thus, Q₁ = X_b = X_3.75. Now, the problem is how to find X_3.75. In the real numbers line, 3.75 is located between 3 and 4. Let the difference between 3.75 and 3 be p and the difference between 4 and 3.75 be q. Thus, p : q = (3.75 – 3) : (4 – 3.75) = 0.75 : 0.25 = 3 : 1. To find X_3.75 it is assumed that the ratio of the difference between Q₁ and X₃ to the difference between X₄ and Q₁ is also p : q. The quartile is calculated by finding the weighted arithmetic mean of X₃ and X₄. The weights are p and q., i.e. 3 and 1. Since Q₁ is closer to X₄ than to X₃, we put more weight to X₄. Consequently, Q₁ is determined as follows.

So, in this example

 

To determine the third quartile, compute

Thus, Q₃ = X_a = X_11.25. To find X_11.25, first note that in the real numbers line, 11.25 is located between 11 and 12. Let the difference between 11.25 and 11 be p and the difference between 12 and 11.25 be q. Thus, p : q = 1 : 3. To find X_11.25 it is assumed that the ratio of the difference between Q₃ and X₁₁ to the difference between X₁₂ and Q₃ is also p : q. The quartile is calculated by finding the weighted arithmetic mean of X₁₁ and X₁₂. The weights are p and q., that is 3 and 1. Since Q₃ is closer to X₁₁ than to X₁₂, we put more weight to X₁₁. As a consequence, Q₃ is determined as follows.

So, in this example:

 

If the data are presented in a frequency distribution table, Q₁ is determined by the formula below.

……………………………………………………………………………………………………………………………………………………………….. (*)

where

L₁ = the lower class boundary (LCB) of the first quartile class

n = the sum of all frequencies (= the number of data)

(Ʃf)₁ = the number of data whose values are less than L₁

f_b = the frequency of the first quartile class

c = the first quartile class’ width

(Regarding LCB and class width, please refer to The Anatomy of Frequency Distribution Tables.)

Note: The first quartile class is the class in the frequency distribution table that contains the k^th data, where k = n/4.

 

The third quartile, Q₃, is determined by the following formula.

………………………………………………………………………………………………………………………………………………………. (**)

where

L₃ = the lower class boundary (LCB) of the third quartile class

n = the sum of all frequencies (= the number of data)

(Ʃf)₃ = the number of data whose values are less than L₃

f_a = the frequency of the third quartile class

c = the third quartile class’ width

Note: The third quartile class is the class in the frequency distribution table that contains the k^th data, where k = 3n/4.

 

Example 2

The frequency distribution table below shows the employees’ monthly expenditure on mobile phone telecommunication. Find the quartiles.

Answer

The first step is to insert an additional column to the right of the the column indicating frequencies, which is the one with the column heading “Number of Employees”. Name the new column “Data Numbers”. Since the first class contains 5 data, the first class’ data numbers are from 1 to 5. The second class contains 13 data, so its data numbers are 6 to 18. Continuing this way, we have the following table.

(We have renamed the third column “Frequency”.)

 

In this example, the sum of all frequencies is n = 50, thus n/4 = 50/4 = 12.5. From the entries of Data Numbers column, we know that the 12.5^th data is in the 2^nd class with the class interval 141-175. This class is called the first quartile class. Then, we calculate the first quartile class’ LCB, i.e. L₁ = 141-0.5 = 140.5. The number of data whose values are less than L₁ is (Ʃf)₁ = 5. The frequency of the first quartile class is f_b = 13. The first quartile class’ width is c = 175.5-140.5 = 35. Substituting these values ​​into (*), we get::

So, the first quartile of the employees’ monthly expenditure on mobile phone telecommunication is IDR 160,692.

 

To calculate the third quartile, we first compute 3n/4 = 3⋅50/4 = 37.5. From the entries of Data Numbers column, we know that the 37.5^th data is in the 3^rd class with the class interval 176-210. This class is called the third quartile class. Then, we calculate the third quartile class’ LCB, i.e. L₃ = 176-0.5 = 175.5. The number of data whose values are less than L₃ is (Ʃf)₃ = 5 + 13 = 18. The frequency of the third quartile class is f_a = 20. The third quartile class’ width is c = 210.5-175.5 = 35. Substituting these values ​​into (**), we get::

So, the third quartile of the employees’ monthly expenditure on mobile phone telecommunication is IDR 209,625.

 

References

Shukla, M. C., S. S., Gulshan, Elements of Statistics for Commerce Students, S. Chand&Co.(Pvt) Ltd., 1971

Spiegel, M. R., Theory and Problems of Statistics, McGraw-Hill Inc., 1981