Cumulative Frequency Diagrams & Box Plots

Cumulative Frequency Time t minutes 0≤t<55≤t<1010≤t<1515≤t<2020≤t<25 Number of students 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

Time (seconds) Frequency 0 < t ≤ < t ≤ < t ≤ < t ≤ < t ≤ 252 Cumulative Frequency Add a 3 rd column and do a running total of the frequency column Turn the table round so that it is written vertically = = = = 80

Time (seconds) Frequency 0 < t ≤ < t ≤ < t ≤ < t ≤ < t ≤ 252 Cumulative Frequency 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

t mins Cumulative freq 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 = ½ = 7½ mins

Box Plot t mins 0 Lowest value Upper Quartile Highest value Lower Quartile Median

t mins Cum freq 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 minutes 0–56–1011–1516–2021–25 Number of students 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

Time t minutes 0–56–1011–1516–2021–25 Number of students 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 minutes 0–56–1011–1516–2021–25 Number of students A group of 80 students were timed on an exam question. The results are shown in the table. 0–5.55.5– – – –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 – – – – – Cumulative Frequency Add a 3 rd column and do a running total of the frequency column Turn the table round so that it is written vertically = = = = 80 So this time the points are plotted at 5.5, 10.5, 15.5, 20.5 and 25.5

5½10½ 15½ 20½25½t mins Cumulative freq 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

5½10½ 15½ 20½25½t mins Cumulative freq x x x x x The graph can also be used to read off values. How many students took less than 12 minutes? So 34 students took less than 12 minutes

5½10½ 15½ 20½25½t mins Cumulative freq 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