Gr 10 Data Handling 3 - Maths Excellence

Since there is an even number of data values, the median will not be a data
value in the data set. We will need to find the average of the two middle
numbers, 61 and 73 and then insert this number into the data set:
61 + 73
M = Q 2
= _ = 67
2
20 32 43 54 55 61 67 73 78 89 90 91 98
The lower quartile is the median of the lower half of the data set. The
lower half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 43 and 54 and then insert this
number into the data set:
43 +54
Q 1
= _ = 48,5
2
Q 1
20 32 43 48,5 54 55 61
The upper quartile is the median of the upper half of the data set. The
upper half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 89 and 90 and then insert this
number into the data set:
89 + 90
Q 3
= _ = 89,5
2
Q 3
73 78 89 89,5 90 91 98
In summary:
20 32 43 48,5 54 55 61 67 73 78 89 89,5 90 91 98
(b) Consider the following set of 10 marks obtained by a class on a class
test out of 150 marks. The number of marks is even.
12 60 95 105 120 125 130 135 140 142
Since there is an even number of data values, the median will not be a data
value in the data set. We will need to find the average of the two middle
numbers, 120 and 125 and then insert this number into the data set:
120 +125
M = Q 2
= __ = 122,5
2
12 60 95 105 120 122,5 125 130 135 140 142
The lower quartile is the median of the lower half of the data set. The
lower half contains an odd number of data values. Therefore Q 1
= 95.
12 60 95 105 120
The upper quartile is the median of the upper half of the data set. The
upper half contains an odd number of data values. Therefore Q 2
= 135.
125 130 135 140 142
In summary:
12 60 95 105 120 122,5 125 130 135 140 142
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10 LC G10 MATHS LWB.indb 17 2008/09/09 12:22:44 PM