6.6 Measures of Dispersion
(A) Determine the range of a set of data
1. For an ungrouped data,
Range = largest value – smallest value.
2. For a grouped data,
Range = midpoint of the last class – midpoint of the
first class.
Example 1:
Determine the range of the following data.
(a) 720, 840, 610, 980, 900
(b)
Time
(minutes)

1 – 6

7 – 12

13 – 18

19 – 24

25 – 30

Frequency

3

5

9

4

4

Solution:
(a)
Largest value of the data = 980
Smallest value of data = 610
Range = 980 – 610 = 370
(b)
Midpoint of the last class
= ½ (25 + 30)
minutes
= 27.5 minute
Midpoint of the first class
= ½ (1 + 6) minutes
= 3.5 minute
Range = (27.5 – 3.5) minute = 24 minutes
(B) Medians and Quartiles
1. The first quartile (Q_{1})
is a number such that $\frac{1}{4}$
of the total number of data that has a value
less than the number.
2. The median is the second quartile which is the value that lies at the centre of the
data.
3. The third quartile (Q_{3}) is a number such that $\frac{3}{4}$of the total number of data that has a value
less than the number.
4. The interquartile range is
the difference between the third quartile and the first quartile.
Interquartile
range = third quartile – first quartile

Example 2:
The ogive in the diagram shows the distribution of time (to the nearest second) taken by 100 students in a swimming competition. From the ogive, determine
(a) the median,
(b) the first quartile,
(c) the third quartile
(d) the interquartile range
of the time taken.
Solution:
$\begin{array}{l}\text{(a)}\frac{1}{2}\text{of100students}=\frac{1}{2}\times 100=50\\ \text{Fromtheogive,median,}M=50.5\text{second}\\ \\ \text{(b)}\frac{1}{4}\text{of100students}=\frac{1}{4}\times 100=25\\ \text{Fromtheogive,firstquartile,}{Q}_{1}\text{=}44.5\text{second}\\ \\ \text{(c)}\frac{3}{4}\text{of100students}=\frac{3}{4}\times 100=75\\ \text{Fromtheogive,thirdquartile,}{Q}_{3}\text{=5}4.5\text{second}\end{array}$
(d)
Interquartile range
= Third quartile – First quartile
= 54.5 – 44.5
=
10.0 second