Finding the maximum number of students who made a certain score In a class of 100 students, the mean score of the exam is 75, with a standard deviation of 6. What is the maximum number of students who could have made 100 on the exams?
What I tried
$P(X = 100) = P(X>99)\leq (1/99)E(X) = 75$ students. Apparently this is not correct. 
Thanks,
 A: I'll assume that the possible scores are integers $0$ to $100$ (fractional scores are not allowed).
Suppose $x_i$ students get score $i$, $i = 0 \ldots 100$.  We have
$$ \eqalign{\sum_{i=0}^{100} x_i &= 100\cr
           \sum_{i=0}^{100} i x_i &= 7500\cr
           \sum_{i=0}^{100} i^2 x_i &=  100 (6^2 + 75^2) = 566100\cr
\text{all}\ x_i \ge 0} $$
and you want to maximize $x_{100}$.
As a linear programming problem, this has optimal solution
$x_{73} = 1100/27 \approx 40.74$, $x_{74} = 700/13 \approx 53.85$, 
$x_{100} = 1900/351 \approx 5.41$.  Of course we want integer
solutions, but this does tell us it's impossible to do better than $x_{100} = 5$.  It turns out there are integer solutions with $x_{100} = 5$, e.g.:
$$ \eqalign{x_{67} &= 3\cr
            x_{71} &= 2\cr
            x_{72} &= 1\cr
            x_{73} &= 17\cr
            x_{74} &= 69\cr
            x_{76} &= 1\cr
            x_{77} &= 1\cr
            x_{85} &= 1\cr
            x_{100} &= 5\cr
\text{all other}\ x_i &= 0\cr} $$
(I found this using Cplex)
Thus the maximum possible number of students getting $100$ is $5$.
