What else can be deduced from the following class grade summary information? Lets say for example a class of students and their grades are the data set.
Lets say there are around 35 students.
Here is what you know:
Your mark
The class average
The median mark
The standard deviation

Are there any other conclusions about this data that can be made given this information?
Thanks
 A: You can probably also compute upper bounds on the number of students who did better or worse than you did. The population version would be via Chebyshev's inequality.
For example, if $X_{me}$ is 'my' score, $s^2$ is the sample variance, $\bar{X}$ is the sample mean, and $X_i$ are the class scores, including 'mine', we have
$$s^2 = \frac{1}{n-1}\sum_{i} (X_i - \bar{X})^2$$
Now partition the sum into parts which are more extreme than my score and those which are less extreme:
$$(n-1) s^2 = \sum_{i \in I} (X_i - \bar{X})^2 + \sum_{i \not\in I}(X_i - \bar{X})^2$$,
where $I$ is the set of indices $i$ such that $|X_i - \bar{X}| \ge |X_{me} - \bar{X}|$. Now bound the summands in the right sum by zero from below, and bound those in the left sum by the condition defining $I$:
$$(n-1) s^2 \ge \sum_{i \in I} (X_{me} - \bar{X})^2 + \sum_{i \not\in I}0 = |I| (X_{me} - \bar{X})^2$$
And thus you have
$$|I| \le (n-1)s^2 / (X_{me} - \bar{X})^2$$
Now $I$ includes all those in the class that did better than I did, if I beat the mean, and thus you have an upper bound on that number in that case.
Using the sample median and a sample version of Gauss' Inequality, you may be able to prove a tighter bound, but with some more assumptions on the distribution, and the proof is not as clear. 
(n.b. This is, I believe, essentially how Chebyshev's inequality is proved.)
A: Since the data set is grades, you probably also know the minimum and maximum possible scores, e.g. 0 and 100.  Given the median and first two moments, you can fit a variety of models to the sample statistics, say a scaled beta distribution.  This would give you tighter estimates than Chebyshev's inequality, although you'd be making some serious regularity and smoothness assumptions.  (The nifty thing with a beta distribution is that the mean and variance completely define it, so you could use the median as a validation statistic.)  Sounds like fun!
