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Quartile Deviation
Statistics - Quartile Deviation
It depends on the lower quartile ${Q_1}$ and the upper quartile ${Q_3}$. The difference ${Q_3 - Q_1}$ is called the inter quartile range. The difference ${Q_3 - Q_1}$ spanided by 2 is called semi-inter quartile range or the quartile deviation.
Formula
${Q.D. = frac{Q_3 - Q_1}{2}}$
Coefficient of Quartile Deviation
A relative measure of dispersion based on the quartile deviation is known as the coefficient of quartile deviation. It is characterized as
${Coefficient of Quartile Deviation = frac{Q_3 - Q_1}{Q_3 + Q_1}}$
Example
Problem Statement:
Calculate the quartile deviation and coefficient of quartile deviation from the data given below:
| Maximum Load (short-tons) | Number of Cables |
|---|---|
| 9.3-9.7 | 22 |
| 9.8-10.2 | 55 |
| 10.3-10.7 | 12 |
| 10.8-11.2 | 17 |
| 11.3-11.7 | 14 |
| 11.8-12.2 | 66 |
| 12.3-12.7 | 33 |
| 12.8-13.2 | 11 |
Solution:
| Maximum Load (short-tons) | Number of Cables (f) | Class Bounderies | Cumulative Frequencies |
|---|---|---|---|
| 9.3-9.7 | 2 | 9.25-9.75 | 2 |
| 9.8-10.2 | 5 | 9.75-10.25 | 2 + 5 = 7 |
| 10.3-10.7 | 12 | 10.25-10.75 | 7 + 12 = 19 |
| 10.8-11.2 | 17 | 10.75-11.25 | 19 + 17 = 36 |
| 11.3-11.7 | 14 | 11.25-11.75 | 36 + 14 = 50 |
| 11.8-12.2 | 6 | 11.75-12.25 | 50 + 6 = 56 |
| 12.3-12.7 | 3 | 12.25-12.75 | 56 + 3 = 59 |
| 12.8-13.2 | 1 | 12.75-13.25 | 59 + 1 = 60 |
${Q_1}$
Value of ${frac{n}{4}^{th}}$ item =Value of ${frac{60}{4}^{th}}$ thing = ${15^{th}}$ item. Thus ${Q_1}$ pes in class 10.25-10.75.
$ {Q_1 = 1+ frac{h}{f}(frac{n}{4} - c) \[7pt] ,Where l=10.25, h=0.5, f=12, frac{n}{4}=15 and c=7 , \[7pt] , = 10.25+frac{0.5}{12} (15-7) , \[7pt] , = 10.25+0.33 , \[7pt] , = 10.58 }$${Q_3}$
Value of ${frac{3n}{4}^{th}}$ item =Value of ${frac{3 imes 60}{4}^{th}}$ thing = ${45^{th}}$ item. Thus ${Q_3}$ pes in class 11.25-11.75.
$ {Q_3 = 1+ frac{h}{f}(frac{3n}{4} - c) \[7pt] ,Where l=11.25, h=0.5, f=14, frac{3n}{4}=45 and c=36 , \[7pt] , = 11.25+frac{0.5}{14} (45-36) , \[7pt] , = 11.25+0.32 , \[7pt] , = 11.57 }$