(i) Let the frequencies of the fourth group and the fifth group be x and y respectively, then 2y=x+0.005× 10①.

x+y = 1-(0.005+0.0 15+0.02+0.035)× 10②

X=0. 15 and y=0. 10 from ① ② solution.

So as to obtain a histogram (as shown in the figure)

m = 95×0.2+ 105×0. 15+ 15×0.35+ 125×0. 15+ 135×0. 1+ 145×0.05 = 1 14.5

(2) According to the meaning of the question, the number of finalists is 40.05× (0.15+0.10+0.05) = 24, and then fill in the contingency table as follows:

[120, 140] [140, 150] 5 3 8 participants in training 15 16 participants in training.

And by k2 = 24 (5× 1? 15× 3) 220× 4×16× 8 = 3.75 < 6.635, so there is no 99% certainty that "the students who enter the finals become seed players are related to expert training.