to get right, we suppose that each is marked correctly by this peer with probability 0.72. This gives a binomial distribution of 6 marks with a probability of 0.72. On the 4 questions that they got wrong, we suppose also that they are marked correctly by this peer (as false) with probability 0.72, and incorrectly (as true) with probability 1-0.72. This gives again a binomial distribution of 4 marks with a probability of 1-0.72. Hence, the final mark given to the agent by their peer is the sum of these two binomial distributions: $bin(6,0.72)+bin(4,1-0.72)$ where $bin(m,p)$ is a binomial distribution of $m$ trials with probability $p$.

We tried other marking models including normally distributed peer grades with a standard deviation that is inversely proportional to the grade of the marking agent, and uniform distributed peer grades with a range that is also inversely proportional to the grade of the marking agent. As we obtained similar results with these other marking models, we focus here on our simple sum of binomials model.

Mark distributions
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We begin with a simple binomial distribution of marks. We let the actual mark of the agents be a binomial distribution of 100 trials with a 