(2) The number of people with a score of 90 or above in liberal arts mathematics in the city is13120;
(3) The probability that two out of seven people are admitted and 1 is a girl is.
Test analysis: (1) The values of m, n, m and n can be easily obtained from the table. See the test analysis for the histogram of frequency distribution.
(2) According to the classical probability, it can be found that the number of people with 90 points or above in liberal arts mathematics in the city is13120;
(3) Let four boys be named A 1, A 2, A 3 and A 4 respectively, and three girls be named B 1, B 2 and B 3 respectively. Among the seven students, 2 1 student is the basic event of enrolling two books, and 12 meets the requirements, so just 60 of the seven students enroll two books.
Problem analysis: (1) as shown in the figure.
, m = 1000, m = 436, n = 0.436, n = 0.220.5.
(2) Let the number of people with 90 points or above in liberal arts mathematics in the city be X, then X = 13 120. 7.
(3) Assuming that four boys are recorded as A 1, A 2, A 3, A 4, and three girls are recorded as B 1, B 2, B 3, then the basic events of recruiting two students from seven students are as follows:
(A 1，A 2)，(A 1，A 3)，(A 1，A 4)，(A 1，B 1)，A 1，B 2)，(A 1，B 3)，
(A 2，A 3)，(A 2，A 4)，(A 2，B 1)，(A 2，B 2)，(A 2，B 3)，(A 3，A 4)，
(A 3，B 1)，(A 3，B 2)，(A 3，B 3)，(A 4，B 1)，(A 4，B 2)，(A 4，B 3)，
(B 1, B 2), (B 1, B 3), (B 2, B 3), ***2 1 species.
We take "there are 1 girls in 2 people" as event A, and there are:
(A 1，B 1)，(A 1，B 2)，(A 1，B 3)，(A 2，B 1)，(A 2，B 2)，(A 2，B 3)，
(A 3，B 1)，(A 3，B 2)，(A 3，B 3)，(A 4，B 1)，(A 4，B 2)，(A 4，B 3)，
*** 12 species,
Then.
So two out of seven people are admitted, and the probability that 1 is a girl is 0.9.