As can be seen from the histogram of frequency distribution:

F7 = 1-(0.004+0.0 1+0.0 1+0.02+0.02+0.0 16+0.008)× 10 = 0. 12，

So the probability of students with scores above 260 is P≈f72+f8=0. 14.

Therefore, among these 2,000 students, about 2,000× 0.14 = 280 people were qualified for the interview. -(4 points)

(2) We set three students as A, B and C, and the score of A is above 270.

Note that events M, N, and R respectively represent events in which A, B, and C obtained Grade B qualification.

Then P(M)= 1- 14=34, p (n) = p (r) =1-kloc-0/8 = 78, -(6 points).

So P(X = 0)= P(m. n. r)= 1256,

P(X = 1)= P(M . n . r+Mn . r+M . NR)= 17256，

P(X=2)=P(MN。 R+。 MNR+M.NR)=9 1256，

P(X=3)=P(MNR)= 147256

So the distribution list of random variable x is:

x 0 1 23p 1 256 1 7256 9 1 256 1 47256∴ex = 0× 1× 1× 1× 17256+2。