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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.

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    A true random can be on outside the probability range, because we often expect a random sequence to be uniformly distributed, but in real it can be nothing like you would expect. This is a very informal video that explains what I'm saying http://youtu.be/Lf4ZmWc_jmA?t=8s2013-03-22

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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..

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    @Hagen: The marginal probability $P(103)$ in the denominator is just a normalizing factor that depends on what strategies _other_ than random guessing the students might have available, and how likely you find each of them _a priori_. Specifically, $P(103) = \sum_\mathrm{strategy} P(103 | \mathrm{strategy}) P(\mathrm{strategy})$. As for the prior strategy probabilities $P(\mathrm{strategy})$, those really are to some degree subjective and arbitrary.2012-09-09
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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).

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    Yes, I'm a guesser for fraction additions2012-09-12