(2) When the "service satisfaction" is 3, the average of the five data is (3+7+8+8+4)÷5=6.

So the standard deviation s = 1.5 [(3-6) 2+(7-6) 2? + (8-6) 2 + (8-6) 2 + (4-6) 2 ] = 4.4 ≈2. 1

(3) The three people with "service satisfaction" of 2 points are marked as A, B and C respectively. Two people whose "service satisfaction" is 1 are marked as m and n respectively.

Remember "X < 3, Y < 3", choose two people to comment. At least one person's "service satisfaction" is 1 as event A, and the basic events are (a, b), (a, c), (a, m), (a, n), (b, c), (.

And event a contains seven basic events (a, m), (a, n), (b, m), (b, n), (c, m), (c, n) * *, so "choose two people from five people with x < 3 and y < 3 to make suggestions.