10(0.006+0.006+0.0 1+0.054+x+0.006)= 1，
x = 0.0 18；
(2) 0.0 18× 10×50=9 out of 80 ~ 90 people with a score of not less than 80 in science.
There are 0.006× 10×50=3 people over 90 years old.
Two people were randomly selected from this 12 people, and the number of people with 90 points or above was recorded as ξ.
ξ=0、 1,2;
∴p(ξ=0)=c29c2 12=6 1 1，
P(ξ= 1)=C 19？ C 13C2 12=922，
p(ξ= 2)= c23c 2 12 = 122；
The distribution list of ∴ξprobability is:
The mathematical expectation of ξ 012p611922122 ξ is e ξ = 0× 61+1× 922+.
(Text) There are 0.0 18× 10×50=9 students among 80 ~ 90 students whose grades are not less than 80.
There are 0.006× 10×50=3 people over 90 years old.
Three people were randomly selected from 12, and the basic number of events was12×1×103× 2×1= 220.
The basic number of events for two of these three people is 12×3=36, and the basic number of events for all three people is 1.
What is the probability that at least two of these three people scored more than 90 points (including 90 points)?
P=36220+ 1220=37220。