1. Cumulative Frequency Diagrams & Box Plots

2. Cumulative Frequency Time t minutes 0≤t<55≤t<1010≤t<1515≤t<2020≤t<25 Number of students 101630222 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 5 This 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

3. Time (seconds) Frequency 0 < t ≤ 510 5 < t ≤ 1016 10 < t ≤ 1530 15 < t ≤ 2022 20 < t ≤ 252 Cumulative Frequency 10 26 56 78 80 Add a 3 rd 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

4. Time (seconds) Frequency 0 < t ≤ 510 5 < t ≤ 1016 10 < t ≤ 1530 15 < t ≤ 2022 20 < t ≤ 252 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 axis X axis

5. 510 152025t mins Cumulative freq 20 40 60 80 x x x x x Median=½(n+1) Middle Value QUARTILES Lower Quartile=¼(n+1) way from bottom Upper Quartile=¼ (n+1) way from top Interquartile Range 8½ 1612½ = 16 - 8½ = 7½ mins

6. Box Plot 510 152025t mins 0 Lowest value Upper Quartile Highest value Lower Quartile Median

7. 510 152025t 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

8. Time t minutes 0–56–1011–1516–2021–25 Number of students 101630222 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 5 This means any time from 6 up to 10 0–5 6–10 21–25 This means any time from 21 up to 25

9. Time t minutes 0–56–1011–1516–2021–25 Number of students 101630222 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

10. Time t minutes 0–56–1011–1516–2021–25 Number of students 101630222 A group of 80 students were timed on an exam question. The results are shown in the table. 0–5.55.5–10.5 10.5–15.515.5–20.5 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

11. Time (seconds) Frequency 0 – 5.510 5.5 – 10.516 10.5 – 15.530 15.5 – 20.522 20.5 – 25.52 Cumulative Frequency 10 26 56 78 80 Add a 3 rd 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

12. 5½10½ 15½ 20½25½t 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

13. 5½10½ 15½ 20½25½t 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

14. 5½10½ 15½ 20½25½t mins Cumulative freq 20 40 60 80 x x x x x What time did the first 50 students complete it in? 14½ 50 So 50 students took less than 14½ minutes