# Quartile Deviation and its Coefficient

__Quartile Deviation__

Quartile deviation is based 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 $${Q_3} – {Q_1}$$ divided by $$2$$ is called semi-inter-quartile range or the quartile deviation. Thus

\[Q.D = \frac{{{Q_3} – {Q_1}}}{2}\]

The quartile deviation is a slightly better measure of absolute dispersion than the range, but it ignores the observations on the tails. If we take difference samples from a population and calculate their quartile deviations, their values are quite likely to be sufficiently different. This is called sampling fluctuation, and it is not a popular measure of dispersion. The quartile deviation calculated from the sample data does not help us to draw any conclusion (inference) about the quartile deviation in the population.

__Coefficient of Quartile Deviation__

A relative measure of dispersion based on the quartile deviation is called the coefficient of quartile deviation. It is defined as:

Coefficient of Quartile Deviation \[Coefficient\,of\,Quartile\,Deviation = \frac{{\frac{{{Q_3} – {Q_1}}}{2}}}{{\frac{{{Q_3} + {Q_1}}}{2}}} = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}}\]

It is a pure number free of any units of measurement. It can be used for comparing the dispersion of two or more sets of data.

__Example__

The wheat production (in Kg) of 20 acres is given as: 1120, 1240, 1320, 1040, 1080, 1200, 1440, 1360, 1680, 1730, 1785, 1342, 1960, 1880, 1755, 1720, 1600, 1470, 1750, and 1885. Find the quartile deviation and coefficient of quartile deviation.

__Solution__:

After arranging the observations in ascending order, we get

1040, 1080, 1120, 1200, 1240, 1320, 1342, 1360, 1440, 1470, 1600, 1680, 1720, 1730, 1750, 1755, 1785, 1880, 1885, 1960.

$${Q_1} = {\text{Value of }}\left( {\frac{{n + 1}}{4}} \right)th{\text{ item}}$$

$$ = {\text{Value of }}\left( {\frac{{20 + 1}}{4}} \right)th{\text{ item}}$$

$$ = {\text{Value of (5}}{\text{.25)}}th{\text{ item}}$$

$$ = 5th{\text{ item}} + 0.25(6th{\text{ item}} – 5th{\text{ item}}) = 1240 + 0.25(1320 – 1240)$$

$${Q_1} = 1240 + 20 = 1260$$

$${Q_3} = {\text{Value of }}\frac{{{\text{3(n + 1)}}}}{{\text{4}}}th{\text{ item}}$$

$$ = {\text{Value of }}\frac{{{\text{3(20 + 1)}}}}{{\text{4}}}th{\text{ item}}$$

$$ = {\text{Value of (15}}{\text{.75)}}th{\text{ item}}$$

$$ = 15th{\text{ item}} + 0.75(16th{\text{ item}} – 15th{\text{ item}}) = 1750 + 0.75(1755 – 1750)$$

$${Q_3} = 1750 + 3.75 = 1753.75$$

\[Q.D = \frac{{{Q_3} – {Q_1}}}{2} = \frac{{1753.75 – 1260}}{2} = \frac{{492.75}}{2} = 246.875\]

\[Coefficient\,of\,Quartile\,Deviation = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}} = \frac{{1753.75 – 1260}}{{1753.75 + 1260}} = 0.164\]

__Example__:

Calculate the quartile deviation and coefficient of quartile deviation from the data given below:

Maximum Load(short-tons) |
Number of Cables |

$$9.3 – 9.7$$ | $$2$$ |

$$9.8 – 10.2$$ | $$5$$ |

$$10.3 – 10.7$$ | $$12$$ |

$$10.8 – 11.2$$ | $$17$$ |

$$11.3 – 11.7$$ | $$14$$ |

$$11.8 – 12.2$$ | $$6$$ |

$$12.3 – 12.7$$ | $$3$$ |

$$12.8 – 13.2$$ | $$1$$ |

__Solution__:

The necessary calculations are given below:

Maximum Load(short-tons) |
Number of Cables($$f$$) |
ClassBoundaries |
CumulativeFrequencies |

$$9.3 – 9.7$$ | $$2$$ | $$9.25 – 9.75$$ | $$2$$ |

$$9.8 – 10.2$$ | $$5$$ | $$9.75 – 10.25$$ | $$2 + 5 = 7$$ |

$$10.3 – 10.7$$ | $$12$$ | $$10.25 – 10.75$$ | $$7 + 12 = 19$$ |

$$10.8 – 11.2$$ | $$17$$ | $$10.75 – 11.25$$ | $$19 + 17 = 36$$ |

$$11.3 – 11.7$$ | $$14$$ | $$11.25 – 11.75$$ | $$36 + 14 = 50$$ |

$$11.8 – 12.2$$ | $$6$$ | $$11.75 – 12.25$$ | $$50 + 6 = 56$$ |

$$12.3 – 12.7$$ | $$3$$ | $$12.25 – 12.75$$ | $$56 + 3 = 59$$ |

$$12.8 – 13.2$$ | $$1$$ | $$12.75 – 13.25$$ | $$59 + 1 = 60$$ |

$${Q_1} = {\text{Value of }}\left( {\frac{n}{4}} \right)th{\text{ item }} = {\text{Value of }}\left( {\frac{{60}}{4}} \right)th{\text{ item = }}15th{\text{ item}}$$

$${Q_1}$$ lies in the class $$10.25 – 10.75$$

$$\therefore $$ $${Q_1} = l + \frac{h}{f}\left( {\frac{n}{4} – c} \right)$$

Where $$l = 10.25$$, $$h = 0.5$$, $$f = 12$$, $$n/4 = 15$$ and $$c = 7$$

$$\therefore $$ $${Q_1} = 10.25 + \frac{{0.5}}{{12}}\left( {15 – 7} \right) = 10.25 + 0.33 = 10.58$$

$${Q_3} = {\text{Value of }}\left( {\frac{{3n}}{4}} \right)th{\text{ item }} = {\text{Value of }}\left( {\frac{{3 \times 60}}{4}} \right)th{\text{ item = 4}}5th{\text{ item}}$$

$${Q_3}$$ lies in the class $$11.25 – 11.75$$

$$\therefore $$ $${Q_3} = l + \frac{h}{f}\left( {\frac{{3n}}{4} – c} \right)$$

Where $$l = 11.25$$, $$h = 0.5$$, $$f = 14$$, $$3n/4 = 45$$ and $$c = 36$$

$$\therefore $$ $${Q_1} = 11.25 + \frac{{0.5}}{{14}}\left( {45 – 36} \right) = 11.25 + 0.32 = 11.57$$

\[Q.D = \frac{{{Q_3} – {Q_1}}}{2} = \frac{{11.57 – 10.58}}{2} = \frac{{0.99}}{2} = 0.495\]

\[Coefficient\,of\,Quartile\,Deviation = \frac{{{Q_3} – {Q_1}}}{{{Q_3} + {Q_1}}} = \frac{{11.57 – 10.58}}{{11.57 + 10.58}}\]

\[Coefficient\,of\,Quartile\,Deviation = \frac{{0.99}}{{22.15}} = 0.045\]

Priyanshu tomar

August 13@ 6:55 pmHow to calculate the quartile deviation