Then n×(0.02+0.06)=4 and the solution is n=50.

(2) Let the frequencies of group I be π and xi, respectively, and it can be seen from the figure that p 1=0.02, p2=0.06, p3=0.3, p4=0.4, p5=0. 12, p6=0.08, and p7=0.02.

Then from Xi = 50×π, we can get x 1= 1, x2=3, x3= 15, x4=20, x5=6, x6=4, x7= 1.

The average daily discretionary time of senior one students is:

. t = 5x 1+ 15 x2+25 x3+35 x4+45 X5+55x 6+65x 750 = 33.6 < 45

Then the school needs to reduce the amount of homework.

(3) The frequencies of the third and fourth groups are 15 and 20 respectively. If 7 people are sampled by layers, the third group should be 7× 15 15+20 = 3 (people), and the fourth group should be 7× 20 15+20 = 4 (people).

According to the meaning of the question, X=0, 1 2, while P (X = 0) = C24C27 = 27, p (x =1) = c14c13c27 = 47, p (x =

Then the distribution list of x is

X 0 1 2P 27 47 17 then e (x) = 0× 27+1× 47+2×17 = 67.