First, the ordinate is converted from frequency/interval to frequency (interval = 4).

From bottom to top: 0. 1, 0. 15, 0.2, 0.25, 0.3.

According to the problem, the frequency is regarded as probability, represented by the midpoint of each water consumption group (for example, 2 tons is taken for the 0-4 ton group).

Then the average water consumption = 0.15 * 2+0.25 * 6+0.3 *10+0.2 *14+0.1*18 = 9.4 tons.

Variance = 0.15 * (2-9.4) 2+0.25 * (6-9.4) 2+...+0.1* (18-9.4) 2 =

The probability that the monthly water consumption is not less than 12 ton is 0.2+0. 1=0.3.

P(X=0)=0.7*0.7*0.7

P(X= 1)=0.3*0.7*0.7

P(X=2)=0.3*0.3*0.7

P(X=3)=0.3*0.3*0.3

e(X)= 0 * P(X = 0)+ 1 * P(X = 1)+2 * P(X = 2)+3 * P(X = 3)= 1