Median and Quartiles
a) The marks of a group of students in their October tests are listed below:
41
56
68
59
43
37
70
58
61
47
75
66
Calculate:
(i)
the median;
(ii)
the semi-interquartile range.
b) The teacher arranges extra homework classes for the students before the next
test in December.
In this test, the median is 67 and the semi-interquartile range is 7.
Make two appropriate comments comparing the marks in the October and
December tests.
Solution
a) i) Arrange the data in order to find the median (middle number):
37
41
43
47
45
56
58
59
61
58.5
66
68
70
75
67
12 numbers so the middle number must be between the 6th and 7th number.
A nice way to work it out is to do (12 + 1) / 2 = 13 / 2 = 6.5
The median is between 58 and 59, therefore:
Median = 58.5
ii) To work out semi-interquartile range, you need to work out the upper and
lower quartiles.
Lower Quartile: 6 numbers in the lower quartile so middle of these is between
the 3rd and 4th number, therefore:
The lower quartile is between 43 and 47, so:
Lower Quartile = 45
Upper Quartile: 6 numbers in the upper quartile so the middle of these is
between the 9th and 10th number, therefore:
The upper quartile is between 66 and 68, so
Upper Quartile = 67
Semi interquartile range = Upper Quartile – Lower quartile
2
= 67 - 45 = 22 = 11
2
2
b) 1. On average, the students did better in the December test.
2. In the December test, the results are less varied as the semi-interquartile
range is smaller.
2.
Solution
a) Cumulative frequency is a running total of the frequency.
1
 1+2=3
3+3=6
6 + 5 = 11
11 + 5 = 16
16 + 6 = 24
24 + 1 = 25
b) There are 25 numbers altogether so the median is the middle number, which is
the 13th number. Look at the cumulative frequency column, the 13th number must
be in the 5th row (where no. of books = 4), since there is only up to 11 numbers in
the 4th row.
Median = 4
ii) There are 12 numbers in the lower section, so the middle of these numbers
must be between the 6th and 7th number. Look at the cumulative frequency
column. The 3rd row (where no. of books = 2) contains the 6th number and the 4th
row (where no. of books = 3) contains the 7th number. So the lower quartile must
be half way between 2 and 3, so must be 2.5.
Lower Quartile = 2.5
iii) There are 12 numbers in the upper section, so the middle of these numbers
must be between the 19th and 20th number (6.5 on from the median, which is 13.)
The 19th and 20th number both lie in the 6th row of the table (where no. of books =
5) so upper quartile must be 5.
Upper Quartile = 5
c) Semi-Interquartile Range = Upper Quartile – Lower Quartile
2
= 5 – 2.5
2
= 1.25
d) The number of text books is more spread out for girls.