Quartile deviation is one of the measures of dispersion. Before getting into a deeper understanding, let’s recall quartiles and how we can define them. Quartiles are the values that divide a list of numerical data into three-quarters, such as Q_{1}, Q_{2} and Q_{3}. The middle part of the three quarters measures the central point of distribution and shows the data values near the midpoint (or the central value; this is referred to as the median). The lower part of the quarters indicates just half the information set, which comes under the median, and the upper part shows the remaining half, which falls above the median. Thus, the quartiles represent the distribution or dispersion of the given data set.
Read more: |
Quartile Deviation in Statistics can be defined as the statistic that measures the dispersion. Here, the Dispersion is the state of getting dispersed or spread. Statistical dispersion means the extent to which numerical data is likely to vary about an average value. In other words, dispersion helps to understand the distribution of the data.
Quartile Deviation Definition
The Quartile Deviation can be defined mathematically as half of the difference between the upper and lower quartile. Here, quartile deviation can be represented as QD; Q_{3} denotes the upper quartile and Q_{1} indicates the lower quartile.
Quartile Deviation is also known as the Semi Interquartile range.
Quartile Deviation Formula
Suppose Q_{1} is the lower quartile, Q_{2} is the median, and Q_{3} is the upper quartile for the given data set, then its quartile deviation can be calculated using the following formula.
QD = (Q_{3} – Q_{1})/_{2}
In the next section, you will learn how to calculate these quartiles for both ungrouped and grouped data separately.
Quartile Deviation for Ungrouped Data
For an ungrouped data, quartiles can be obtained using the following formulas,
Q_{1} = [(n+1)/4]th item
Q_{2} = [(n+1)/2]th item
Q_{3} = [3(n+1)/4]th item
Where n represents the total number of observations in the given data set.
Also, Q_{2} is the median of the given data set, Q_{1} is the median of the lower half of the data set and Q_{3} is the median of the upper half of the data set.
Before, estimating the quartiles, we have to arrange the given data values in ascending order. If the value of n is even, we can follow the similar procedure of finding the median.
Click here to get the quartiles calculator for ungrouped data.
Quartile Deviation for Grouped Data
For a grouped data, we can find the quartiles using the formula,
Here,
Q_{r} = the 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 = the cumulative frequency of the class preceding the quartile class
N = Number of observations in the given data set
Quartile Deviation Example
Let’s understand the quartile deviation of ungrouped and grouped data with the help of examples given below.
Example 1:
Find the quartiles and quartile deviation of the following data:
17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28
Solution:
Given data:
17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28
Ascending order of the given data is:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Number of data values = n = 16
Q_{2} = Median of the given data set
n is even, median = (1/2) [(n/2)th observation and (n/2 + 1)th observation]
= (1/2)[8th observation + 9th observation]
= (10 + 14)/2
= 24/2
= 12
Q_{2} = 12
Now, lower half of the data is:
2, 5, 7, 7, 8, 8, 10, 10 (even number of observations)
Q_{1} = Median of lower half of the data
= (1/2)[4th observation + 5th observation]
= (7 + 8)/2
= 15/2
= 7.5
Also, the upper half of the data is:
14, 15, 17, 18, 24, 27, 28, 48 (even number of observations)
Q_{3}= Median of upper half of the data
= (1/2)[4th observation + 5th observation]
= (18 + 24)/2
= 42/2
= 21
Quartile deviation = (Q_{3} – Q_{1})/2
= (21 – 7.5)/2
= 13.5/2
= 6.75
Therefore, the quartile deviation for the given data set is 6.75.
Example 2:
Calculate the quartile deviation for the following distribution.
Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
80-90 |
90-100 |
Frequency |
5 |
3 |
4 |
3 |
3 |
4 |
7 |
9 |
7 |
8 |
Solution:
Let us calculate the cumulative frequency for the given distribution of data.
Class |
Frequency |
Cumulative Frequency |
0 – 10 |
5 |
5 |
10 – 20 |
3 |
5 + 3 = 8 |
20 – 30 |
4 |
8 + 4 = 12 |
30 – 40 |
3 |
12 + 3 = 15 |
40 – 50 |
3 |
15 + 3 = 18 |
50 – 60 |
4 |
18 + 4 = 22 |
60 – 70 |
7 |
22 + 7 = 29 |
70 – 80 |
9 |
29 + 9 = 38 |
80 – 90 |
7 |
38 + 7 = 45 |
90 – 100 |
8 |
45 + 8 = 53 |
Here, N = 53
We know that,
Finding Q_{1}:
r = 1
N/4 = 53/4 = 13.25
Thus, Q1 lies in the interval 30 – 40.
In this case, quartile class = 30 – 40
l_{1} = the lower limit of the quartile class = 30
l_{2} = the upper limit of the quartile class = 40
f = the frequency of the quartile class = 3
c = the cumulative frequency of the class preceding the quartile class = 12
Now, by substituting these values in the formula we get:
Q_{1} = 30 + [(13.25 – 12)/3] × (40 – 30)
= 30 + (1.25/3) × 10
= 30 + (12.5/3)
= 30 + 4.167
= 34.167
Finding Q_{3}:
r = 3
3N/4 = 3 × 13.25 = 39.75
Thus, Q_{3} lies in the interval 80 – 90.
In this case, quartile class = 80 – 90
l_{1} = the lower limit of the quartile class = 80
l_{2} = the upper limit of the quartile class = 90
f = the frequency of the quartile class = 7
c = the cumulative frequency of the class preceding the quartile class = 38
Now, by substituting these values in the formula we get:
Q_{3} = 80 + [(39.75 – 38)/7] × (90 – 80)
= 80 + (1.75/7) × 10
= 80 + (17.5/7)
= 80 + 2.5
= 82.5
Finally, the quartile deviation = (Q_{3} – Q_{1})/2
QD = (82.5 – 34.167)/2
= 48.333/2
= 24.1665
Hence, the quartile deviation of the given distribution is 24.167 (approximately).