Quartiles, Median, Deciles, and Percentiles are all statistical measures used to analyze and understand data distributions. These concepts divide data into different parts to help identify patterns, trends, and outliers.
Median is the middle value in a dataset. It represents the point dividing the data into two equal halves when ordered from least to greatest.
Quartiles break the data into four equal parts. The first quartile (Q1), second quartile (Q2, which is the median), and third quartile (Q3) help to understand the spread. They also explain the central tendency of the data.
Deciles divide the data into ten equal parts, giving more granular insight into the distribution.
Percentiles split the data into 100 equal parts, offering even finer detail for analyzing specific data points.
Together, these tools are invaluable for understanding the spread of a dataset. They help identify central tendencies, making them essential for data analysis.
Quartiles divides a set of data into four equal parts. A median divides a set of data into two parts each with equal number of items.
The first Quartiles, mostly referred to as the lower Quartiles contains 25% of the total data items. Lower Quartiles can be described as the median of the bottom half.
Second Quartiles is actually the median of the whole data(50%).
The third Quartiles is usually referred to as upper Quartiles and contains 75% of total data items. It can be described as the median of the upper half the data set.
Formula for the getting the first Quartiles Q1
The formulae for general expression of a quartile of a data is expressed as below:
Where
- L is the lower class boundary of the quartile class.
- n is the total frequency
- c is the cumulative frequency above the quartile class
- i is the class interval
- f is the frequency of the lower quartile class
Formula for the getting the second quartile Q2
Second quartile Q2 is actually the median of the data
it is calculated from:
where
- L is the lower class boundary of the median class.
- n is the total frequency
- c is the cumulative frequency above the median class
- i is the class interval
- f is the frequency of the median class
Formula for the getting the third quartile Q3
The general expression of the third quartile would be:
where
- L is the lower class boundary of the upper quartile class.
- n is the total frequency
- c is the cumulative frequency above the third quartile class
- i is the class interval of the upper quartile class
- f is the frequency of the upper quartile class
Deciles
Deciles divides a set of data into ten equal parts.
First decile is when n is divided by 10. that is;
where n is the total frequency for the data
Percentiles
Percentiles divides a set of data into hundred equal parts.
one percentile is given as :
In quartiles, deciles and percentiles, data is arranged in ascending order
Example 5
The table below shows the distribution of heights to the nearest cm of students in a school.
Find (a) the median
(b)(i) lower quartile (ii) upper quartile (iii) 80th percentile.
Solution
(a) The new frequency table for the data is shown here
There are 130 students . Therefore, the median height is the 65th student. that is;
The 65th student falls in the 150-159 class. This class is called the median class.
Using the formula for the median:
Lower class boundary of Q2 class = 149.5, hence:
hence the median = 155.056
(b) (i)
The general expression of the lower quartile(first Quartile is):
The total frequency is 130 students: hence;
(130/4)th student is the one at 32.5th position.
This position falls 140-149 class
The lower class boundary that has the first quartile is 139.5; hence
(ii)
The general expression of the third quartile(Upper quartile) is:
hence:
The 97.5th student falls into 160-169 class which has a lower class boundary 159.5
hence:
(C)
The 80th percentile of the data is given by 80/100)*130=104th value.
The 104th student falls in the 160-169 class
80th percentile= L+(80/100n-C)i/f
The complete solution is as below:
80th percentile student is a student at 104th position according to the expression:
This value falls in (160 – 169) class
The lower class boundary of this class is 159.5, hence
hence Q3 = 169.500
Example
Determine the lower quartile and upper quartile for the following set of data
15, 20, 16, 15, 18, 17, 13, 9, 17, 18, 11
solution
arranging in ascending order
9, 11, 13, 15, 15, 16, 17, 17, 18, 18, 20
The median number is 16. On left of 16 there are 5 values and on the right 5 values.
16 is at the center of the data list
9, 11, 13, 15, 15 | 16 | 17, 17, 18, 18, 20
The first half contains: 9, 11, 13, 15, 15
The central value in that lower half is 13 and it is the first quartile of the data
The upper half includes: 17, 17, 18, 18, 20
The central value is 18 and is hence the upper quartile for the data list
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