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Cumulative Frequency Diagrams & Box Plots
Cumulative Frequency
Time
t minutes0≤t<5 5≤t<10 10≤t<15 15≤t<20 20≤t<25
Number of students
10 16 30 22 2
A group of 80 students were timed on an exam question. The results are shown in the table.
This means any time from 0 up to but not including 5This means any time from 5 up to but not including 10
0≤t<5 5≤t<10 20≤t<25
This means any time from 20 up to but not including 25
Time (seconds)
Frequency
0 < t ≤ 5 10
5 < t ≤ 10 16
10 < t ≤ 15 30
15 < t ≤ 20 22
20 < t ≤ 25 2
Cumulative Frequency
10
26
56
78
80
Add a 3rd column and do a running total of the frequency column
Turn the table round so that it is written vertically
10+16 = 26
26+30 = 56
56+22 = 78
78+2 = 80
Time (seconds)
Frequency
0 < t ≤ 5 10
5 < t ≤ 10 16
10 < t ≤ 15 30
15 < t ≤ 20 22
20 < t ≤ 25 2
Cumulative Frequency
10
26
56
78
80- as there are 80 students. Final value should always be equal to the number of pieces of data in the question.
Scale for x axis, on your graph, is the end points of the class intervals
Scale for y axis, should be suitable for your data
We now need to show this information in a graph
Y axisX axis
5 10 15 20 25 t mins
Cumulative freq
20
40
60
80
x
x
x
x x
Median=½(n+1) Middle Value
QUARTILES•Lower Quartile=¼(n+1) wayfrom bottom•Upper Quartile=¼ (n+1) wayfrom top
Interquartile Range
8½ 1612½
= 16 - 8½ = 7½ mins
Box Plot
5 10 15 20 25 t mins0
Lowest value
Upper Quartile
Highest value
Lower Quartile
Median
5 10 15 20 25 t mins
Cum freq
20
40
60
80
x
x
x
x x
Median =1/2(n+1) Middle Value
QUARTILES•Lower Quartile = ¼(n+1) way from top•Upper Quartile = ¼(n+1) way from bottom
8½ 1612½
It is easiest to draw the box plot straight under the cumulative frequency graph
Time
t minutes0–5 6–10 11–15 16–20 21–25
Number of students
10 16 30 22 2
A group of 80 students were timed on an exam question. The results are shown in the table.
This means any time from 0 up to 5This means any time from 6 up to 10
0–5 6–10 21–25
This means any time from 21 up to 25
Time
t minutes0–5 6–10 11–15 16–20 21–25
Number of students
10 16 30 22 2
A group of 80 students were timed on an exam question. The results are shown in the table.
So what happens if the time is 5.3minutes?
We have to change the boundaries of each group.
This is done by finding ½ way between the end and start of each successive group
Time
t minutes0–5 6–10 11–15 16–20 21–25
Number of students
10 16 30 22 2
A group of 80 students were timed on an exam question. The results are shown in the table.
0–5.5 5.5–10.5 10.5–15.5 15.5–20.5 20.5–25.5
Group 1 ends at 5 and group 2 starts at 6 so ½ way is 5.5
Group 2 ends at 10 and group 3 starts at 11 so ½ way is 10.5
Group 3 ends at 15 and group 4 starts at 16 so ½ way is 15.5
Time (seconds)
Frequency
0 – 5.5 10
5.5 – 10.5 16
10.5 – 15.5 30
15.5 – 20.5 22
20.5 – 25.5 2
Cumulative Frequency
10
26
56
78
80
Add a 3rd column and do a running total of the frequency column
Turn the table round so that it is written vertically
10+16 = 26
26+30 = 56
56+22 = 78
78+2 = 80
So this time the points are plotted at 5.5, 10.5, 15.5, 20.5 and 25.5
5½ 10½ 15½ 20½ 25½ t mins
Cumulative freq
20
40
60
80
x
x
x
x x
Median = Middle Value
QUARTILES•Lower Quartile = ¼ way•Upper Quartile = ¾ way
Interquartile Range
8¾ 16½ 12¾
= 16½ - 8¾ = 7¾ mins
5½ 10½ 15½ 20½ 25½ t mins
Cumulative freq
20
40
60
80
x
x
x
x x
The graph can also be used to read off values.
How many students took less than 12 minutes?
12
34
So 34 students took less than 12 minutes
5½ 10½ 15½ 20½ 25½ t mins
Cumulative freq
20
40
60
80
x
x
x
x x
What time did the first 50 students complete it in?
14½
50So 50 students took less than 14½ minutes