Compare a student exam score to the community one based on success rate of questions Definition
An exam is composed of x questions that are taken in a pool of questions (so a question can be in several exams). 
Each question gives an amount of points and you can success (gain all the points) or fail (0 points) the question. 
I want to compare a student score on a specific exam to the whole community based on question success rate (the whole community hasn't passed this specific exam but can have passed some questions on another exams).
Example
Exam A:


*

*Question 1 (1pt) : success rate 75% of students who have passed this question (no matter the examen) have succeed to this question 

*Q2 (10pts): 25% success rate

*Q3 (5pts): 50% success rate

*...


The student passes the exam A and has the following result:


*

*Q1: success (gain 1pt)

*Q2: fail (0pts)

*Q3: success (gain 5pts)

*...


Problem
Is there a way (or an approximate one) to say: "this student is better than XX% of the community on this exam" ?
What I've tried
I tried the following approach but I'm not a mathematician and I don't know if it's legal or not:


*

*compute a score for the exam by balancing the question point depending on the success rate:


*

*Q1: 75% (success rate) of 1pt = 0.75pts

*Q2: 25% of 10pts = 2.5pts

*Q3: 50% of 5pts = 2.5pts

*...

*= (0.75 + 2.5 + 2.5) / (1 + 10 + 5) = 36%


*compute the score of the student (by cumulating the points of good answers): (1 + 5) / (1 + 5 + 10) = 37,5%

*try to find the student score on a distribution which is centered on the exam score (36%) and which height is calculated using the standard deviation of the question success rates of the exam

 A: If we want to distinguish students who solve hard problems, the score of a student may be $\sum_j \big( s_j f(q_j)- (1-s_j)(1-f(q_j))$, where $q_j$ is the ratio of correct answers to the $j$-th question, $s_j=1$ if the student has solved the $j$-th question and $s_j=0$ otherwise, $f$ is some decreasing function such that $f(q)\in [0,1]$. 
For example if a student is given questions Q1,Q2,Q3 with 40%,10%,50% of correct answers and he solved only Q1,Q2 then his score will be $f(0.4)+f(0.1)-(1-f(0.5))$.
For instance, one may put $f(q)=1-q$ or $f(q)={\rm logistic}(c(0.5-q))$ with some constant $c$. 
The sense is that the score must increase significantly when the student solves  a hard question and decrease slightly when the student fails to solve a hard question. Symmetrically, the score must decrease significantly when the student fails to solve an easy question and increase slightly when the student does solve an easy question.
A: The simplest approach may be not to consider the total success rate for the questions. One may calculate the percentage of points gained by each student as
$P/T$, where $P$ is the total number of points gained by a student, and $T$ is the total number of points of the questions the student tried to solve. Then, for a given student, we may just calculate the percentage of students with lower $P/T$ rating.
For example, let the questions Q1,Q2,Q3,Q4 be of $10,30,5,15$ points, respectively. Student A is given Q1 and Q2, he solves Q1 and fails to solve Q2, so A's score is $10/(10+30)=0.25$. Student B is given Q2,Q3,Q4, he solves Q2 and fails to solve Q3 and Q4, so B's score is $30/(30+5+15)=0.6$. Thus student B is thought to be better than student A.
