Analysis: The number of people who want to pass the exam first requires P(90≤X≤ 150), and to find this probability, we need to transform the problem into the probability form of several special values of normal variables, and then use symmetry to solve it.

Solution: ∫ x-n (1/kloc-0,202), ∴ μ = 1 10, σ = 20.

p( 1 10-20 < X < 1 10+20)= 0.683。

The probability of ∴ x > 130 is × (1-0.683) = 0.158 5.

The probability of X≥90 is 0.683+0.1.585 = 0.841.5.

∴ Qualified number 54× 0.84 1.3 ≈ 45 (person),

The number of people with scores above 130 is 54×0. 158 5≈9 (people).

Green channel: this question is to use the symmetry of normal curve combined with three special probability values to find the probability. We should learn to use this method.