A class was divided into two groups, A and B, and took part in the truant competition. The number of participants in both groups is equal. After the competition, a statistical chart was drawn accordin

A class was divided into two groups, A and B, and took part in the truant competition. The number of participants in both groups is equal. After the competition, a statistical chart was drawn according to the scores of the two groups of students. (1) The average score of students in group A is: (5× 5+6× 2+7× 3+8+9× 4) ÷15 = 6.8 (points).

Because * * * has the number of 15, the median of group A is 7 points;

Because the average value is 6.8 < 7, from the average point of view, group B performs better;

Since the median is 7 > 6, from the median point of view, Group A scored better.

(2) Group B should be selected,

Variance of S2 in group B =115 [(5-7) 2+7× (6-7) 2+5× (8-7) 2+2× (9-7) 2] =1.6 < 2.56]

Then the results of group b are more neat;

(3) There are 4+2=6 students with 9 points.

From the students who scored 9 points in this competition, 4 people were selected to participate in the competition. The probability of students with 9 points participating in the competition is 46 = 23.