Cambridge International Mathematics

286 Analysis of discrete data (Chapter 13)

THE RANGE

Therangeis the difference between themaximum(largest) data value

and theminimum(smallest) data value.

range=maximum data value¡minimum data value

Example 3 Self Tutor

Find the range of the data set: 5 , 3 , 8 , 4 , 9 , 7 , 5 , 6 , 2 , 3 , 6 , 8 , 4.

range=maximum value¡minimum value=9¡2=7

THE QUARTILES AND THE INTERQUARTILE RANGE

Themediandivides an ordered data set into halves, and these halves are divided in half again by the

quartiles.

The middle value of the lower half is called thelower quartile. One quarter, or25%, of the data have

values less than or equal to the lower quartile. 75%of the data have values greater than or equal to the

lower quartile.

The middle value of the upper half is called theupper quartile. One quarter, or25%, of the data have

values greater than or equal to the upper quartile. 75%of the data have values less than or equal to the

upper quartile.

Theinterquartile rangeis the range of the middle half (50%) of the data.

interquartile range=upper quartile¡lower quartile

The data is thus divided into quarters by the lower quartile Q 1 , the median Q 2 , and the upper quartile Q 3.

So, the interquartile range, IQR=Q 3 ¡Q 1.

Example 4 Self Tutor

For the data set: 7 , 3 , 4 , 2 , 5 , 6 , 7 , 5 , 5 , 9 , 3 , 8 , 3 , 5 , 6 find the:

a median b lower and upper quartiles c interquartile range.

The ordered data set is: 233345555667789 ( 15 of them)

a As n=15,

n+1

2

=8

b As the median is a data value, we now ignore it and split the remaining data into two:

lower

z }| {

2333455

upper

z }| {

5667789

Q 1 =median of lower half=3

Q 3 =median of upper half=7

c IQR=Q 3 ¡Q 1 =7¡3=4

) the median=8th score=5

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Y:\HAESE\IGCSE01\IG01_13\286IGCSE01_13.CDR Friday, 3 October 2008 4:18:11 PM PETER