1. Finding averages from the frequency table

2. In this screencast Mean from frequency table Mean from frequency table with intervals Mode from frequency table Median from frequency table

3. The Mean

4. Find the total of the data and divide by the number of data items

5. 12, 12, 12, 15, 17, 17, 19, 19, 19

6. Total is 142 Mean is 15.8

7. 12 x 3 + 15 x 1+17 x 2 + 19 x 3

9. Mean from a frequency table

10. Days absent (x) No of staff (f) 0126354 12141913974

11. Days absent (x) No of staff (f) 0126354 12141913974 This is the variable

12. Days absent (x) No of staff (f) 0126354 12141913974 This is the frequency This is the variable

13. Days absent (x) No of staff (f) 0126354 12141913974 Sub-total X x F 0143839363524

14. Days absent (x) No of staff (f) 0126354 12141913974 Sub-total X x F 0143839363524 78 186 T

15. Days absent (x) No of staff (f) 0126354 12141913974 Sub-total X x F 0143839363524 78 186 T Mean = 186 / 78 = 2.38

16. Intervals

17. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3

18. Height (mm) Frequency F 10  X  1515  X  20 712 Mid-point X 12.517.5 20  X  25 15 22.5 25  X  30 3 27.5

19. Height (mm) Frequency F 10  X  1515  X  20 712 Mid-point X 12.517.5 Sub-total X x F 87.5210 20  X  25 15 22.5 337.5 25  X  30 3 27.5 82.5

20. Height (mm) Frequency F 10  X  1515  X  20 712 Mid-point X 12.517.5 37 T Sub-total X x F 87.5210 20  X  25 15 22.5 337.5 25  X  30 3 27.5 82.5

21. Height (mm) Frequency F 10  X  1515  X  20 712 Mid-point X 12.517.5 37 T Sub-total X x F 87.5210717.5 20  X  25 15 22.5 337.5 25  X  30 3 27.5 82.5

22. Height (mm) Frequency F 10  X  1515  X  20 712 Mid-point X 12.517.5 37 T Sub-total X x F 87.5210717.5 20  X  25 15 22.5 337.5 25  X  30 3 27.5 82.5 Mean = 717.5 / 37 = 19.4 mm

23. The mode

24. The mode is the most common value in a data set

25. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3

26. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3 15 is largest frequency so 20  X  25 is the modal class

27. Days absent (x) No of staff (f) 0126354 12141913974

28. Days absent (x) No of staff (f) 0126354 12141913974 19 is largest frequency so 2 days is the most common absence period

29. The median

30. Median is value of ‘middle’ data item

31. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3

32. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3 Half of 37 is 18.5… 37 T

33. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3 Cumulative frequency 719 3437

34. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3 Cumulative frequency 719 3437 First value that is above 18.5

35. Height (mm) Frequency F 10  X  1515  X  20 712 20  X  25 15 25  X  30 3 Cumulative frequency 719 3437 First value that is above 18.5 Median is in this interval

36. Days absent (x) No of staff (f) 0126354 12141913974

37. Days absent (x) No of staff (f) 0126354 1214191397478 T

38. Days absent (x) No of staff (f) 0126354 1214191397478 T Half of 78 is 39…

39. Days absent (x) No of staff (f) 0126354 1214191397478 T Cumulative Frequency 12264558677478

40. Days absent (x) No of staff (f) 0126354 1214191397478 T Cumulative Frequency 12264558677478 First value that is above 39

41. Days absent (x) No of staff (f) 0126354 1214191397478 T Cumulative Frequency 12264558677478 First value that is above 39 Median number of days absent is 2

42. Summary

43. Remember Mean: multiply midpoints by frequencies and add the sub-totals. Divide by the total of the frequencies Mode: find the largest frequency – the corresponding value is the modal value or modal class Median: calculate a running total of the frequencies – the first interval that is above half the total contains the median

44. Your turn…