So10× (0.005+0.01.02+A+0.025+0.01) =1... (1min).
The answer is a = 0.03...(2 points)
(2) Solution: According to the histogram of frequency distribution, the frequency with a score of not less than 60 points is1-10× (0.005+0.01) = 0.85 ... (3 points).
Because there are 640 students in Grade One of our school, using the idea of sample estimation, it can be estimated that there are about 640×0.85=544 students whose math scores are not less than 60 in Grade One of our school. ... (5 points)
(3) Solution: The number of people with scores less than [40,50] is 40×0.05=2, and they are respectively marked as A and B(6 points).
The number of people whose scores are less than [90, 100] is 40×0. 1=4, which are marked as C, D, E, F...(7 points).
If two students are randomly selected from the students whose math scores are within [40,50] and [90, 100], then all the basic events are: (a, b), (a, c), (a, d), (a, e), (a, f), (.
If both students' math scores are within [40,50] or [90, 100], the absolute value of the difference between the two students' math scores shall not be greater than 10. If one score is within [40,50], the other score is within [90,65438+.
Note that "the absolute value of the difference between two students' math scores does not exceed 10" as event M, then the basic events contained in event M are: (a, b), (c, d), (c, e), (c, f), (d, e), (d, f), (e, f).
So the probability is p (m) = 7 15...( 12 points).