(1) The frequency of the score distribution in the interval [70,90] is1-10 (0.004+0.006+0.02+0.016) = 0.54,

Let the sample size be n, then 0.54=27n, and the solution is n = 50.

(ii) According to (i), the frequency of the rank in the interval [80, 90] is 0.24,

So, the average score of these N students is

45×0.04+55×0.06+65×0.2+75×0.3+85×0.24+95×0. 16=76.2;

(iii) There are 50×0.04=2 students in the sample, and the score is [40,50], marked as X and Y;

There are 50×0.06=3 students with scores of [50,60], which are marked as A, B and C respectively.

The basic events of choosing two people from the above five people are: {x, y}, {x, a}, {x, b}, {x, c}, {y, a}, {y, b}, {y, c}, {a, b}, {a, c}, {b}

Note that "Choose two of the above two people, at least 1 person in [40,50]" is event A, then the basic events contained in event A are: {x, y}, {x, a}, {x, b}, {x, c}, {y, a}, {y,

So the probability p (a) = 7 10.