Professor sent the following data to his students: Full Grade: 25, Max. grade: 22.5, Mean: 19.9, Median: 20

Assuming the number of students is n. Is it possible to calculate the minimum grade from the given data only?

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PS: you may know some of other students' grades. For example you heard that student x1 got 21, another student x2 got 18 and so on .. but you don't know all the students' grades of course –  Osama Gamal Jan 9 '12 at 11:27
What does "full grade" mean? –  Dan Brumleve Jan 9 '12 at 11:29
@Dan Probably "Full Grade" is the maximum grade you can get in the exam. For instance there are 5 questions 5 points each. –  marvinthemartian Jan 9 '12 at 11:36
@marvinthemartian, is that information relevant? –  Dan Brumleve Jan 9 '12 at 11:38
@Dan Frankly, I do not think it is. But I do not know the answer yet, so maybe. –  marvinthemartian Jan 9 '12 at 11:41

The answer is no. Let us look for an easier example:

Assume $n=5$, our students are called "A","B","C","D","E" and the grades are given by:

\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 2&1&1&1&0 \end{array}

Clearly Maximum, Median and Mean are given by {2,1,1}. But the dataset

\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 2&1&1&1/2&1/2 \end{array}

Also has the same Mean, Median and Maximum. You can always do this by using a convex combination of the last two values as long as they stay below the median. They will not affect the median, nor the mean or the maximal element but they will change the minimal grade given.

Edit: Now with this knowledge we can come back to your question and easily construct a counterexample flor $n=5$:

\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 22.5&20&20&20&17 \end{array}

\begin{array}{c|c|c|c|c} A&B&C&D&E\ \newline 22.5&20&20&19&18 \end{array}

Those do both fit to your data. For higher $n$ you can pad with the mean $19.9$ on the left and right side of the median (odd $n$). For even $n$ there is a similar counterexample. For $n \in \{3,4\}$ refer to the answer given by tom.

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I don't believe so, not for general n at least. We can for some isolated cases:

For $n = 2$, the problem simply doesn't work

For $n = 3$, we need one student to get 22.5, another student to get 20, and the other to get $s_3$ so that:

$$\frac{22.5+20+s_3}{3}=19.9$$ Which works out for $s_3=17.2$, the minimum grade

For $n=4$, lets say we have 4 students with 4 grades $s_1 \geq s_2\geq s_3\geq s_4$. We know then that $s_1=22.5$, that: $$\frac{s_2+s_3}{2}=20$$

and $$\frac{22.5+s_2+s_3+s_4}{4}=19.9$$ We can then solve this for $s_4$ using the fact that $s_2+s_3=40$ to get $s_4=17.1$.

However, for $n=5$, we run into difficulties. We know (using the same system of notation) that $s_1=22.5$, and that $s_3=20$, but beyond that, the fact that

$$\frac{22.5+s_2+20+s_4+s_5}{5}=19.9$$

Will never allow us to say any more about $s_5$. Indeed, this is true for any $n\geq5$ (is this clear?).

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Note that you are searching for integer numbers (for simplification since grades can only be x.0 or x.5) and note that s_2 > s_4 > s_5 –  Osama Gamal Jan 9 '12 at 12:17