| No. of matches | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
| No. of boxes | 6 | 10 | 18 | 25 | 21 | 12 | 8 |
(i) Calculate, correct to one decimal place, the mean number of matches per box.
(ii) Determine how many extra matches would have to be added to the total contents of the 100 boxes to; bring the mean up to exactly 39 matches. (1997)
Solution:
(i)
| No. of matches (x) | Number of boxes (f) | fx |
| 35 | 6 | 35×6 = 210 |
| 36 | 10 | 36×10 = 360 |
| 37 | 18 | 37×18 = 666 |
| 38 | 25 | 38×25 = 950 |
| 39 | 21 | 39×21 = 819 |
| 40 | 12 | 40×12 = 480 |
| 41 | 8 | 41×8 = 328 |
| Total | Ʃf = 100 | Ʃfx = 3813 |
Mean = Ʃfx/Ʃf
= 3813/100
= 38.13
= 38.1
Hence the mean is 38.1.
(ii)New mean = 39
Ʃfx = 39×100 = 3900
So number of extra matches to be added = 3900-3813 = 87
Hence the number of extra matches to be added is 87.