P(ξ=0)=0.38
P(ξ=2)=0.5
P(ξ=4)=0. 12
So the distribution list of ξ is
0 2 4
0.38 0.5 0. 12
eξ= 0×0.38+2×0.5+4×0. 12 = 1.48
(2) Solution: Since the sequence an=n2-6/5ξn+ 1(n∈N*) is strictly monotonous, the sequence 3/5 < ξ < 3/2,
Namely ξ < 5/2
P(A)= P(ξ< 5/2)= P(ξ= 0)+P(ξ= 2)= 0.88