Mathematics for Construction

8 Mathematics for Construction

TASK 2
Scenario 1
Revenue Number of customers
(£1000)
January July
Less than 5 27 22
5 and less than 10 38 39
10 and less than 15 40 69
15 and less than 20 22 41
20 and less than 30 13 20
30 and less than 40 4 5
Solution
a) The given table can be organised in following manner -
Revenue Number of customers
(£1000)
January July
0 to 5 27 22
5 to 10 38 39
10 to 15 40 69
15 to 20 22 41
20 to 30 13 20
30 to 40 4 5
1

Histogram of each distribution can be calculated after converting the unequal class interval
into equal class interval –
Revenue
Number of
customers
(£1000)
January
0 to 5 27
5 to 10 38
10 to 15 40
15 to 20 22
20 to 30 13
30 to 40 4
Equal class interval -
Revenue
Number of
customers
(£1000)
January
0 to 10 27 + 38 = 65
10 to 20 40 + 22 = 62
20 to 30 13
2

30 to 40 4
From this histogram, it has been analysed that group 0 to 10 has highest frequency, so, taking
this class to find mode of data of grouped frequency in following way –
Mode (z) = l + f1 – f0 x h
2 f1 – f0 - f2
here, f1 is the highest frequency = 65
f0 is the above frequency = 0 and,
f2 is the below frequency = 62
h is the class range = 10 and,
l is the lower interval of mode class = 0,
so, mode can be calculated as -
Mode (z) = 0 + 65 – 0 x 10
2x65 – 0 – 62
= 0 + 65 x 10
130 – 62
= 0 + 650 / 68
3
0 to 10 10 to 20 20 to 30 30 to 40
0
10
20
30
40
50
60
70 65 62
13
4
Number of customers (£1000)
January

= 9.55
For July month -
Revenue
Number of
customers
(£1000)
July
0 to 5 22
5 to 10 39
10 to 15 69
15 to 20 41
20 to 30 20
30 to 40 5
Revenue
Number of
customers
(£1000)
July
0 to 10 22 + 39 = 61
10 to 20 69 + 41 = 110
20 to 30 20
30 to 40 5
4