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Interquartile Range

Interquartile range  is the smallest of all the measures of dispersion. It is defined as the difference between the two extreme observation of the distribution. In other words, range is the difference between the greatest ( maximum ) and the smallest (minimum) observation of the distribution.

Range = Xmax - Xmin

Where Xmax is the greatest observation and Xmin is the smallest observation of the variable values.

In case of the grouped frequency distribution or the continuous frequency distribution, range is defined as the difference between the upper limit of the higher class and the lower limit of the smallest class.

Uses of Range

  • Range is used in industry for the statistical quality control of the manufactured product by the construction of R-chart, i.e., the control chart for range.
  • Range is by far the most widely used measure of variability in our day-to-day life.
  • Range is also used as a very convenient measure by meteorological department for weather forecasts since the public is primarily interested to know the limits within which the temperature is likely to vary on a particular day.
The values, which divide the whole series into a number of equal parts, is called Partitional values. They help us to know the position of the items in the series. So, they are also known as positional averages.

Suppose there is a continuous data with uncountable number of items. Then, we can find the middle number or the position of a particular number using partitional values

There are mainly three types of Partitional values. They are :
1. Quartiles
2. Deciles
3. Percentiles

 

Interquartile Range Definition

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Inter quartile range is defined as the difference between the third and the first quartile. The formula is given by $Q_{3}-Q_{1}$
where $Q_{1}$ is the first quartile and $Q_{3}$ is the third quartile of the series.

Quartiles definition
Quartiles are the partitional values which divide the whole series into 4 equal parts. So, there are 3 quartiles. First Quartile is denoted by $Q_{1}$, second Quartile denoted by $Q_{2}$ and third Quartile denoted by $Q_{3}$. First quartile is also known as Lower quartile and third quartile is also known as Upper quartile. Second Quartile is nothing but Median. Since it denotes the position of the item in the series, it is a positional average.

 When we find quartiles we should arrange the data in the ascending order.

Interquartile Deviation or Interquartile Range

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Q1- Lower Quartile Part

Q2 – Median

Q3 - Upper Quartile Part

  • It is a measure of dispersion based on the upper quartile Q3 and the lower quartile Q1.

Inter-quartile Range = Q3 - Q1

  • Quartile deviation is obtained from inter-quartile range on dividing by 2, hence also known as semi interquartile range

Semi inter-quartile Range = (Q3 - Q1) / 2

Procedure to calculate the interquartile range in mathematical statistics:

  • Arrange the given set of numbers into increasing or decreasing order.

  • Then count the given values. If it is odd then center value is median otherwise obtain the mean value for two center values. This is known as Q2 value.

  • Median equally cuts the given values into 2 equal parts. They are described as Q1 and Q3 part.

  • From Q1 values we have to find one median value.

  • From Q3 values we have to find one median value.

  • Finally we can subtract the median values of Q1 and Q3.

  • This gives the result as interquartile range.

Example: Find the interquartile range value for the first ten prime numbers.

Solution:

The first ten prime numbers are 2,3,5,7,11,13,17,19,23,29

This is already in the decreasing order.

Here number of values = 10

10 is a even number, therefore, median is mean of 11 and 13

That is Q2 = 12.

Now we have to get 2 parts. That is known as Q1 and Q3.

Q1 part : 2, 3, 5,7,11

Here number of values = 5

5 is an odd number, therefore, center value is 5, that is Q1= 5

Q3 part : 13, 17,19,23,29

Here number of values = 5

5 is an odd number, therefore, center value is 19, that is Q3= 19

The subtraction of Q1 and Q3 value is 19-5 = 11

Therefore, 11 is the interquartile range value.

Merits and Demerits of Semi Interquartile Range

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Merits:
Quartile deviation is quite easy to understand and calculate. It has a number of obvious advantages over range as a measure of dispersion. For example:

  • As against range which was based on two observations only, Quartile deviation makes use of 50% of the data, so, it is obviously a better measure than range.
  • Since Quartile Deviation ignores 25% of the data from the beginning of the distribution and another 25% of the data from the top end, it is not affected at all by extreme observations.
  • Quartile Deviation can be computed from the frequency distribution with open end classes. In fact, Quartile Deviation is the only measure of dispersion which can be obtained while dealing with a distribution having open end classes.
Demerits:
  • Quartile Deviation is not based on all the observations since it ignores 25% of the data at the lower end and 25% of the data at the upper end of the distribution. Hence, it cannot be regarded as a reliable measure of variability.
  • Quartile Deviation is affected considerably by fluctuations of sampling.
  • Quartile Deviation is not suitable for further mathematical treatment.

Thus, Quartile Deviation is not a reliable measure of variability, particularly for distributions in which the variation is considerable.

Interquartile Range Examples

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You can see the example on interquartile range given below -

Example 1: Find Interquartile Range, Quartile Deviation for the following distribution:

  Class Interval  
 0-15 
  15-30 
  30-45 
  45-60 
  60-75 
  75-90 
   90-105 
          f   8     26    30    45    20    17      4

Solution:


  Class Interval  
    f  c.f  
        0-15    8   8
       15-30   26   34
       30-45   30   64
       45-60   45  109
       60-75   20  129
       75-90   17  146
      90-105    4  150   
       Total  N = 150   

Here N = 150 , $\frac{N}{4}$ = 3705. The cummulative frequency just greater than 37.5 is 64.

Hence Q1 lies in the corresponding class 30-45

Q1 = l + $(\frac{h}{f})$ ( $(\frac{N}{4})$ - C )

= 30 + $(\frac{15}{30})$ ( 37.5 - 34)

= 31.75

$\frac{3N}{4}$ = 112.5 The c.f just greater than 112.5 is 129.

Hence Q3 lies in the corresponding class 60-75

Q3 = l + $(\frac{h}{f})$ ( $(\frac{3N}{4})$ - C )

     = 1 + $(\frac{15}{20})$ ( 112.5-109)

     = 60 + 2.625 = 62.625

  • Inter-quartile Range = Q3 - Q1

     = 62.625 - 31.750
     = 30.875

  • Quartile Deviation = $\frac{( Q3 - Q1 )}{ 2}$

= $\frac{30.875}{ 2}
= 15.44

Example 2: - Find the Inter quartile range of the following data: 4, 6, 7, 8, 10, 23, 34.

Solution : Here the numbers are arranged in the increasing order. n = 7

First quartile, $Q_{1}$ = $\frac {n+1^{2k}}{4}$ item = $\frac {(7+1)}{4th}$ item = $2^{nd}$ item = 6

Third quartile, $Q_{3}$ = $\frac {3(n+1)^{2k}}{4}$ item = $\frac {3(7+1)}{4th}$ item = 6^{th}$item = 23

Inter quartile range (IQR) = $Q_{3}-Q_{1}$ = 23 - 6 = 17

Answer: Inter quartile range = 17

Example 2: - Find the Inter quartile range of the following marks: 22,13,37,16,26,35,26,35.

Solution : First we have to arrange the numbers in the ascending order

13,16,22,26,26,35,35,37

n = 8

Lower Quartile, $Q_{1}$ = $ \frac {n+1^{th}}{4}$ item= $\frac {(8+1)}{4^{th}}$ item = $2.25^{th}$ item

= $2^{nd}item+0.25(3^{rd}item-2^{nd}item)$

= 15+0.25(21-15) = 16+0.25(6) = 17.5

Upper Quartile,$Q_{3}$ = $3\frac{n+1^{th}}{4}$ item = $ \frac{3(8+1)}{4^{th}}$ item = $6.75^{th}$ item

= $6^{th}item+0.75(7^{th}item-6^{th}item)$

= 35+0.75(35-35) = 35+0.75(0) = 35

Inter quartile range (IQR) = $Q_{3}-Q_{1}$  = 35-17.5 = 17.5

Answer: Inter quartile range = 17.5

While finding Inter quartile range in discrete and continuous case, we use the corresponding formulas for quartiles and then use the formula for Inter quartile range.

Examples calculating Inter quartile range in discrete series

Example 1:  Find the Inter quartile range of the following data

  X       12  
  14 
  16  
  18  
  20  
  F     6    3
   4
   1
   5

Solution :

   X    
  F    
   Cumulative
   Frequency      

  12   6          6
  14   3       6+3=9
  16   4     9+4=13
  18   1    13+1=14
  20   5    14+5=19

First quartile $Q_{1}$ = Size of $\frac{N+1^{th}}{4}$ item = Size of $\frac{19}{4}^{th}$ item = Size of

$4.75^{th}$ item = 12

Third Quartile, $Q_{3}$ = Size of 3 $\frac{N+1^{th}}{4}$ item = Size of $\frac{3(19)}{4^{th}}$ item

= Size of $14.25^{th}$ item = 20

Inter quartile range

(IQR) = $Q_{3}-Q_{1}$ = 20-12 = 8

Answer: Inter quartile range = 8

Example 2:

Find the Inter quartile range of the following data

 X 
  2  
  8    16    25   38 
 F  12   4   6   8  10

Solution:  N = 40

First quartile $Q_{1}$ = Size of $\frac{N+1^{th}}{4}item$ = Size of $\frac{41}{4^{th}}item$ = Size of $10.25^{th}$item = 2

Third Quartile, Q3= Size of $3\frac{N+1^{th}}{4}$ item = Size of $\frac{3(41)}{4^{th}}$ item

= Size of $30.75^{th}$ item = 38

Inter quartile range (IQR) = $Q_{3}-Q_{1}$ = 38 – 2 = 36

Answer: Inter quartile range = 36

Calculate Interquartile Range

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Example of Inter quartile range in continuous series :

Example 1: 
Find the Inter quartile range of the following data:-

 Class 
 10-20  
  20-30  
  30-40 
  40-50  
  50-60  
    F
    4
      2      3
     1
      5

Solution :

   Class     F  
  Cumulative Frequency  
  10-20   4              4
  20-30   2       4 + 2 = 6
  30-40   3       6 + 3 = 9
  40-50   1       9 + 1 = 10
  50-60   5     10 + 1 = 15

N = 15
$\frac {N}{4}$ = $\frac{15}{4}$ = 3.75 which lies in 10-20

$Q_{1}$ class = 10-20
l = 10
cf = 0 c = 10
f = 4

Therefore Lower Quartile = l + $\frac{\left ( \frac{N}{4}-cf \right )c}{f}$ =  10 + $\frac{(3.75-0)10}{4}$ = 10+9.38 = 19.38

$\frac{3N}{4}$ = $\frac{3(15)}{4}$ = 11.25 which lies in 50-60

Q3 class = 50-60
l = 50
cf = 10 c = 10
f = 5

Therefore Upper Quartile $Q_{3}$ = l + $\frac{\left ( \frac{3N}{4}-cf \right )c}{f}$

= 50 + $\frac{(11.25-10)10}{5}$ = 50+2.5 = 52.5

Inter quartile range (IQR) = $Q_{3}-Q_{1}$ =52.5-19.38=33.12

Answer: Inter quartile range = 33.12

Example 2: Find the Inter quartile range of the following marks obtained by 30 students

          Marks  10-15 
 15-20 
 20-25 
 25-30 
 30-35 
 35-40  
 Number Of Students       5     4     6     6     4     5

Solution :

   Class  
 
  Cumulative
  Frequency
  
   10-15   5  
     5
   15-20   4
   5 + 4 = 9
   20-25   6
  9 + 6 = 15
   25-30   6
  15 + 6 = 21
   30-35   4   21 + 4 =25
   35-40   5 
  25 + 5 = 30

N = 30
$\frac{N}{4}$ = $\frac{30}{4}$= 7.5 which lies in 15-20
Q3 class = 15-20
l = 15
cf = 5 c = 5
f = 4

Therefore Lower Quartile $Q_{1}$ = $l+\frac{\left ( \frac{N}{4}-cf \right )c}{f}$= 15+$\frac{(7.5-5)5}{4}$ = 15 + 3.125 = 18.125

$\frac{3N}{4}$ = $\frac{3(30)}{4}$ = 22.5 which lies in 30-35

$Q_{3}$ class = 30-35
l = 30
cf = 21 c = 5
f = 4

Therefore Upper Quartile $Q_{3}$ = l + $\frac{\left ( \frac{3N}{4}-cf \right )c}{f}$ = 30 + $\frac{(22.5-21)5}{4}$

= 30+1.875

= 31.875

Inter quartile range (IQR) = $Q_{3}-Q_{1}$ = 31.875 – 18.125 = 13.75

Answer: Inter quartile range = 13.75

Half of the inter quartile range is called quartile deviation. Quartile deviation is one of the important measure of variation.


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