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Question
The contents of 100 match boxes were checked to determine the number of matches they contained.
| No. of matches | 35 | 36 | 37 | 38 | 39 | 40 | 41 |
| No. of boxes | 6 | 10 | 18 | 25 | 21 | 12 | 8 |
- Calculate, correct to one decimal place, the mean number of matches per box.
- 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.
Solution
| No. of matches (x) |
No. of boxes (f) |
fx |
| 35 | 6 | 210 |
| 36 | 10 | 360 |
| 37 | 18 | 666 |
| 38 | 25 | 950 |
| 39 | 21 | 819 |
| 40 | 12 | 480 |
| 41 | 8 | 328 |
| Total | 100 | 3813 |
i. `barx = (sumfx)/(sumf)`
= `3813/100`
= 38.13
ii. In the second case,
New mean = 39 matches
Total contents = 39 × 100 = 3900
But total number of matches already given = 3813
Number of new matches to be added = 3900 – 3813 = 87
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