Measure Of Dispersion (Cont.)
2. The Quartile Deviation
§Quartiles divide a set of ordered data into four groups with equal numbers of values
The three “dividing points” are the first quartile (Q1), median, (sometimes called the second quartile, Q2), and the third quartile (Q3)
§ The interquartile range is Q1 – Q3, which is the range of the middle of the data.
§ The semi-interquartile range is one half of the interquartile range.
§ Both these ranges indicate how closely the data are clustered around the median.
The quartile deviation is called the semi-quartile range IT is defined as the amount of dispersion present in the middle of 50% of the values, hence the equation is given by,
Q.D = Q3 – Q1
2
|
Q.D = the quartile deviation
Q1 = the first quartile
Q2 = the third quartile
Q1 = L + (N/4 – F1) i
F
|
Where:
Q = the first quartile
L = the lower limit of the first quartile class (i.e. the class containing the N/4th item)
Strictly, L is the lower boundaries of the 1st quartile class.
F1 = the cumulative frequency just above the 1st quartile class.
F = the frequency of the 1st quartile class
N = the total frequency
I = the class interval
Class Intervals
|
Frequency
|
M
|
Quartile class
|
Cf>
|
45 – 49
|
3
|
47
|
3
| |
50 – 54
|
4
|
52
|
7
| |
55 – 59
|
6
|
57
|
Q1
|
13
|
60 – 64
|
7
|
62
|
20
| |
65 – 69
|
10
|
67
|
30
| |
70 – 74
|
7
|
72
|
37
| |
75 – 79
|
6
|
77
|
43
| |
80 – 84
|
4
|
82
|
47
| |
85 – 89
|
3
|
87
|
50
|
Solution:
Q1 = L + (N/4 – F1) i
F
N/4 = 50/4 = 12.5
Q1 = 54.5 (12.7 – 7)5
6
Q1 = 59.08
|
3rd Quartile:
Formula :
Q3 = L + (3N/4 – f1) i
F
|
Where :
F3 = the cumulative frequency just above the 3rd quartile class.
Example :
Q3 = L + (3N/4 – f1) i
F
3N/4 = 150/4 = 37.5
Q3 = 74.5 + (37.5 – 37) 5
6
74.5 + (O.O833) 5
74.5 + O.4165
74.9165
=74.92
Q.D = Q3 – Q1/2
= 74.92 – 59.08/2
= 7.92
|
Wow! thanks for posting this about this topic, it is easier to understand rather than other maths blog that i've searchs.
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