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20 14 three-mode mathematics in Jiangsu province
Solution: (1) The frequency of the score in [70,80] is:1-(0.010+0.015+0.015+0.025+0.005) ×/kloc.

Therefore, 0.3 10 = 0.03,

As shown in the figure: (4 points) (Find frequency (2 points) and plot 2 points)

(2) The average score is: x = 45×0. 1+55×0. 15+65×0. 15+75×0.3+85×0.25+95×0.05 = 7 1。

(3) According to the meaning of the question, the number of people in the [60,70] score segment is 0.15× 60 = 9; (7 points)

The number of people in [70,80] section is: 0.3× 60 =18; (8 points)

* Take a sample with a capacity of 6 from [60,80] students,

∴[60, 70] Select two people in the score section, which are marked as M and N respectively; [70, 80] Four people were selected from the score section and marked as A, B, C and D respectively;

Suppose two people are randomly selected from the sample, and the score segment [70,80] is the most 1 human event a,

Then the basic event space includes: (m, n), (m, a), (m, b), (m, c), (m, d), (c, d)*** 15, (10).

Then the basic events contained in event A are: (m, n), (m, a), (m, b), (m, c), (m, d), (n, a), (n, b), (n, c), (n, d) * *.

∴ p (a) = 9 15 = 35。 ( 12)