Finding the probability that a student is a random guesser I want to find the probability that my student is a random guesser. On a 360-item multiple choice test with four choices for each question, he got 28.5% or 103 of the questions correctly.  
Here is what I have so far. As everyone knows, the expected score is 25% or 90 items. Assuming that he is indeed a random guesser, I used the binomial distribution to get the variance np(1-p) = 360(.25)(.75) = 67.5; hence, a standard deviation of 8.22. Further assuming that random guessers are normally distributed, his z-score is (103-90)/8.22 = 1.58, making him an outlier. This places him in the top 6% of random guessers. This suggest that either
(1) he is a very good guesser, 
(2) he is a very lucky guesser, or
(3) he is NOT a random guesser at all. 
Now I don't know what other concepts to use to find the probability that he is a random guesser. I don't even know if there is enough information; nor do I know whether all my computations and assumptions make any sense. I hope you can help. Cheers! 
PS: I only had a 3-unit statistics course way back in college. "Dummifying" your explanations would surely be appreciated. Cheers! :-)

Edit: Thanks for all your help. So I guess it's really not that easy to get a good approximation on the said probability.
Having said that, is there a relatively simple way to get even a very crude approximation of the answer? For instance, even before posting the question here, I actually considered the Bayesian probability mentioned above. To make things simple, I assumed that P(getting 103|guesser) is simply ${{360}\choose{103}}*.25^{103}*.75^{360-103}.$ And just to have a starting point, let's just say that 1 out of 5 students are random guessers, so P(guesser) is 0.2. What would be a reasonable initial estimate, albeit inaccurate, for P(getting 103)? 
Then maybe we can play around with the assumed values later to get a spectrum of possibilities.
 A: Once you know the distribution (in this case binomial as you correctly stated), you can use Bayesian probability to ask what the chances are the student is randomly guessing:
$$P(\text{guesser}\mid\text{got }103) = {P(\text{getting }103\mid \text{guesser})P(\text{guesser})  \over P(\text{getting }103)}$$
Note that this depends on your prior belief as to how many students you think don't know anything..
A: Your computations make sense. However, one can never be absolutely sure. But you know that if 100 students relied totally on guessing then about 6 are expected to have a score of 103 or better. It all depends on the confidense you want.
Also, it is possible that the student knows some answers correctly and guesses the rest. While a complete guesser scores $90\pm 8.22$, a student actually knowing 20 answers and guessing the rest scores $20+85\pm 7.98$, that is the student with a score of 103 might be a slightly unlucky guesser with little knowledge. In fact a score of $k$ out of $n$ suggests by some kind of maximum-likelihood argument that the student actually knew $\frac{4k-n}3$ answers and guessed the rest.
This does not even take into account "educated guessing". That is: a student with little knowledge mightsee that one of the 4 answers is obviously wrong but he does not know how to find out about the other three options. Each such question would add $\frac1{12}$ to the expected score (but also lower the deviation).
