一道数学建模面试题
在测试(有 7 个问题)中对班级(50 名学生)进行(相对)评分的最佳方式是什么?
他们不想要传统的百分位区间答案,而是更科学的答案。
这是一个非常开放式的问题,他们要求假设以下框架:
- [m_1,...,m_50],其中每个 m_i 是一个 7 向量,表示 50 名学生中每人 7 个问题的得分。
- [c_1,...,c_7],其中每个 c_i 是每个问题测试的“概念”向量。 c_i 不必是不相交的。我们可以假设 union(c_i) 的元素之间具有重要性排序。
What's the best way to (relative) grade a class (of 50 students) on a test (with 7 questions)?
They did not want the traditional percentile-intervals answer, but a more CS-ey one.
It's a pretty open ended question, they asked to assume the following framework:
- [m_1,...,m_50], where each m_i is a 7-vector for marks scored in the 7 questions for each of the 50 students.
- [c_1,...,c_7], where each c_i is a vector of 'concepts' tested by each question. c_i's need not be disjoint. We can assume to have an importance ordering amongst elements of union(c_i).
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简单化方法:假设所有概念都具有相同的价值,我会将其全部总结起来。到处每个概念一分。
整体方法:概念较多的问题可能比概念较少的问题要难得多(并且价值大于概念的总和)。概念彼此“相互作用”。为了缓解这个问题,我会给每个问题赋予一个值(N over C),其中 N 是概念向量的大小,C 是概念总数。然后我会把这一切总结一下。
真正的整体方法:如果概念在不同的问题中重复,那么我们应该“减弱”它们的影响。但是我不确定如何实现这一点。也许我们应该将每个(N 除以 C)值除以所涉及的每个概念的重复次数。
我忽略了概念的重要性排序,因为我不知道如何对其进行评估。
Simplistic approach: Assuming that all concepts have the same value I would just sum it all up. One point for each concept everywhere.
Holistic approach: It could be that the question with more concepts is significantly harder than the question with fewer (and worth more than the sum of concepts). Concepts "interact" with each other. To alleviate this I would put a value of (N over C) to each question, where N is the size of the vector of concepts, and C is total number of concepts. And then I would sum it all up.
True holistic approach: If concepts are repeated in different questions then we should "tone down" their influence. However I'm not sure how to accomplish this. Maybe we should divide each (N over C) value with the number of repetitions of each concept involved.
I ignored the importance ordering of concepts, because I don't know how to put a value on that.