Test analysis: (1) Use the median of each group of data to estimate the average score of the sampled students, which is similar to the weighted average algorithm. Multiply the median of each segment by the frequency corresponding to that segment to get the average value, and use the average value of the sample to estimate the overall average value;

(2) According to the probability formula of equal possible events, the probability that two numbers are exactly the math scores of two students is obtained. The possible values of random variable ξ are 0, 1, 2, 3, and the variable conforms to binomial distribution. According to binomial distribution, the distribution table and expectation can be written, and it can also be solved by the general method of finding expectation.

Analysis of test questions: (1) estimate the average score of sample students with median;

45×0.05+55×0. 15+65×0.2+75×0.3+85×0.25+95×0.05=72.？ (3 points)

The estimated model value is 75 points? (5 points)

So it is estimated that the average score of this exam is 72. (6 points)

(Note: The modes and averages here are estimates. If the words such as estimation or approximation are omitted, 1 minute will be deducted. )

(2) All possible basic results of extracting two numbers from 95, 96, 97, 98, 99, 100 are,

There are 15 results, and the number of people with [90, 100] is 0.005× 10×80=4 (people).

These two figures are just the basic results of two students' math scores.

Two numbers are just the probability of two students' math scores (8 points)

The possible values of random variables are 0, 1, 2, 3, and then there are.

∴

The distribution list of variable ∴ is:

1

2

three

P

(10)

(12)

Solution 2. Random variables satisfy the independent repetition test, so they are binomial distribution, that is, (10 point).

(12).