Statistical Measures of Variation Formulas | List of Statistical Measures of Variation Formulas You Should Know - BYJUS

# Statistical Measures of Variation Formulas

Measures of variation tell us how spread out or dispersed the data is. They represent how far the points of data are from the center of the data set. It can be used to analyze, summarize and describe trends in the data. The measures of variability we will focus here are range, interquartile range and outliers....Read MoreRead Less

### Defining Measures of Variation

Measures of variation can be used when similar quantities are measured. By using the measures of variation, we can understand the reliability of such a dataset. If the dataset has high variability, then extreme values would be frequent, consequently, a low variable dataset would have values that are more uniform.

### Formulas for Measures of Variation

Consider the set of data arranged in ascending order: $$a_1$$  <  $$a_2$$  <   $$a_3$$  <  $$a_4$$  <  $$a_5$$

The range is the difference between the highest and lowest value in the dataset.

Range $$a_5$$ – $$a_1$$

The median is the middle value.

Median = $$a_3$$

The values that lie below the median are the lower half and the values that lie above the median are the upper half.

Quartiles in a data set divide it into 4 equal groups.

The first quartile or the lower quartile is the median of the lower half of the data set. Here, the lower half consists of $$a_1$$ and $$a_2$$.

So, the first quartile $$Q_1$$ = $$\frac{a_1~+~a_2}{2}$$

Similarly, the third quartile or the upper quartile is the median of the upper half of the data set. Here, the upper half consists of $$a_4$$ and $$a_5$$.

So, the third quartile $$Q_3$$ = $$\frac{a_4~+~a_5}{2}$$

The “Interquartile Range” or IQR is the difference of the third quartile and the first quartile. It represents the range of the middle half of the data.

So, IQR  =  $$\frac{a_4~+~a_5}{2}$$ – $$\frac{a_1~+~a_2}{2}$$

Note that extremely high or low values in a dataset are outliers

Outlier boundaries are $$Q_1$$ –  1.5 (IQR) and $$Q_3$$ + 1.5 (IQR)

Hence, outliers are data points that are less than $$Q_1$$ – 1.5 (IQR) or greater than $$Q_3$$ + 1.5 (IQR).

### Solved Examples : Measures of Variation formulas

Example 1: Find the range of the data : 3, 7, 5, 9, 10, 6

Solution:

First arrange the data in increasing order:

3 < 5 < 6 < 7 < 9 < 10

Range  = 10 – 3 = 7

Hence, the range of the data is 7.

Example 2: Five children were measuring the height of a tree . The measurements are given as : 50 cm, 53 cm, 49 cm, 58 cm, 59 cm and 46 cm. Find the range and median of the data.

Solution:

First arrange the data in increasing order:

46 cm < 49 cm < 50 cm < 53 cm < 58 cm < 59 cm

Range = 59 – 46 = 13 cm

Median = $$\frac{50~+~53}{2}$$ = 51.5 cm

Hence the range is 13 cm and the median is 51.5 cm.

Example 3: The marks scored by Janine, Joey, Jeremy, Jade and Jessica in a quiz are given below. Use the interquartile range or IQR to check whether Jeremy’s score is an outlier.

Solution:

Arrange the data in increasing order:

60 < 70 < 78 < 82 < 85

Median = 78

The median divides the data set into two halves.

Lower half : 60, 70

Upper half: 82, 85

First quartile $$(Q_1) ~=~\frac{60~+~70}{2}$$ = 65

Third quartile $$(Q_3) ~=~\frac{82~+~85}{2}$$  = 83.5

IQR = 83.5 – 65

=18.5

Outlier boundaries:

$$(Q_1)$$  –  1.5 (IQR) = 65 – 1.5 x 18.5 = 37.25

$$(Q_2)$$ + 1.5 (IQR) = 83.5 + 1.5 x 18.5 = 111.25

Hence, data values less than 37.25 or data values greater than 111.25 are outliers.

Since Jeremy’s score is greater than 37.25, it is not an outlier.

Example 4:

The heights of four men are given below. Find the interquartile range of the data.

Solution:

Arrange the data in increasing order:

175 < 180 < 185 < 188

Median = $$\frac{180~+~185}{2}$$ = 182.5

Lower half : 175, 180

First quartile $$(Q_1)~=~\frac{175~+~180}{2}$$ = 177.5

Upper half : 185, 188

Third quartile $$(Q_3)~=~\frac{185~+~188}{2}$$ = 186.5

The interquartile range = $$~Q_3~-~Q_1$$

= 186.5 – 177.5

= 9

So the interquartile range of the given data is 9 cm.