Solution: (1) According to the meaning of the question,

There are 3+x+9+15+18+y = 6018+y3+x+9+15 = 23.

X = 9y = 6...(2 points)

∴p=0. 15,q=0. 10.

Complete the frequency distribution histogram as shown in the figure. ... (4 points)

(2) Select 10 people by stratified sampling.

Then there are 10× 25 = 4 "online shoppers".

Have 10× 35 = 6 "non-online shoppers" ... (6 points)

Therefore, the possible values of ξ are 0, 1, 2,3; P(ζ= 0)= c04c 3 10 = 16，P(ζ= 1)= c 14c 26 c 3 10 = 12，P(ζ)

So the distribution list of ξ is:

ξ0 1 23p 16 12 3 10 130∴eξ= 0× 16+ 1× 12+2×365438。