The table below shows the weekly profit in naira from a mini-market.
Weekly profit (N)
1-10
11-20
21-30
31-40
41-50
51-60
Freq
6
6
12
11
10
5
(a) Draw the cumulative frequency curve of the data;
(b) From your graph, estimate the: (i) median; (ii) 80th percentile
(c) What is the modal weekly profit?
(a) Cumulative frequency table and curve (ogive)
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
Weekly profit (N)
Upper class boundary
Frequency (f)
Cumulative frequency
1 - 10
10.5
6
6
11 - 20
20.5
6
12
21 - 30
30.5
12
24
31 - 40
40.5
11
35
41 - 50
50.5
10
45
51 - 60
60.5
5
50
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
Ogive: cumulative frequency plotted against the upper class boundaries. Median (cf = 25) reads about N31; 80th percentile (cf = 40) reads about N45.5.
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
(ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]
where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
Weekly profit (N)
Upper class boundary
Frequency (f)
Cumulative frequency
1 - 10
10.5
6
6
11 - 20
20.5
6
12
21 - 30
30.5
12
24
31 - 40
40.5
11
35
41 - 50
50.5
10
45
51 - 60
60.5
5
50
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
Ogive: cumulative frequency plotted against the upper class boundaries. Median (cf = 25) reads about N31; 80th percentile (cf = 40) reads about N45.5.
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
(ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]
where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):