f=n/50 where n is the number of defective brackets in a box. So f ranges between 1/50=0.02 and 12/50=0.24.
We don't know how many boxes out of the 200 contain 1 defective; or how many contain 2; etc. It would appear that every box has at least 1 defective bracket and no box has more than 12. We could assume that the numbers of defectives are related to the bare coefficients of the expansion of the binomial (p+q)12. This would imply that if there were 4095 boxes:
1 box would contain 12 defectives;
12 boxes would contain 11 defectives;
66 ................................ 10
220 .............................. 9
495 .............................. 8
792 .............................. 7
924 .............................. 6
792 .............................. 5
495 .............................. 4
220 .............................. 3
66 ................................ 2
12 ................................ 1
Total: 4095 boxes. If these are scaled down to 200 boxes then we get:
| Number of defects |
Number of boxes |
| 12 |
0.05 (0) |
| 11 |
0.59 (1) |
| 10 |
3.22 (3) |
| 9 |
10.74 (11) |
| 8 |
24.18 (24) |
| 7 |
38.68 (39) |
| 6 |
45.13 (45) |
| 5 |
38.68 (39) |
| 4 |
24.18 (24) |
| 3 |
10.74 (11) |
| 2 |
3.22 (3) |
| 1 |
0.59 (1) |
The numbers in parentheses are rounded to the nearest integer, and the sum total of these integers is 201, while the sum total of the actual numbers is 200.
Therefore the sum total of the f values (using the integers) is:
1×0.22+3×0.2+11×0.18+24×0.16+39×0.14+45×0.12+39×0.1+24×0.08+11×0.06+3×0.04+1×0.02=
0.22+0.6+1.98+3.84+5.46+5.4+3.9+1.92+0.66+0.12+0.02=24.12. The sum of the squares is 97.3928.
These figures are highly likely to be meaningless, of course, because they're based on assumptions which are almost certainly invalid. What's missing in the givens seems to be the probability that a bracket is defective. The probability of one defective in a box should be much higher than the probability of 12 defectives. For example, if the probability of 1 defective in a box is 0.5 (1 out of two defective), then the probability of 12 defectives in a box is very small, around 0.00024. The likelihood of one defective bracket in a box could be as low as 1 in 200 (p=0.005), which makes it almost inconceivable that one box in 200 could contain 12 defectives.
For there to be 12 defectives in one box, p would be about 0.64, which is a very high failure rate. The question would indicate that the count of defectives in a box could reach as high as 12 (0.6412=0.005 approximately, so in 200 boxes there could be 200×0.005=1 box containing 12 defectives).