Calculate an upper limit of a probability without given variance I was wondering how you can calculate the upper limit of a probability using only a mean. Without using a sample size or variance.
Example:

The average score for a test is 60 out of 100.
Calculate the upper limit for the probability that a student will score more than 80 out of a 100.

The answer is apparently $\frac{3}{4}$ but I cannot find any explanation how this is calculated.
Every explanation that I can find online always uses a mean, variance and some sort of distribution. In this example you have only a mean and yet somehow it is still possible to find an answer.
Am I missing something super obvious or is the writer of this exercise breaking a fundamental law?
 A: Use Markov's Inequality (proof similar to Chebyshev, see Wikipedia on 'Markov Inequality') states: For a random variable $X$ with $P(X > 0) = 1$ and
$E(X) = \mu,$
$$P(X \ge a) \le \mu/a.$$
Use $\mu = 60,\, a = 80$ to get the stated answer.
Because this inequality holds for a large variety of distributions, you
can't expect the bound to be very good in general. For example, if
$Y$ is normal with $\mu = 60,\,\sigma = 10,$ we would have $P(Y \ge 80)  \approx  0.023 < .75.$ [Technically, to apply Markov's Inequality here, the normal
distribution would have to be truncated to ignore the tiny probability
$P(Y < 0).]$
Addendum: Sketch of proof for a continuous density. Proof for a
discrete distribution is similar, but using sums instead of integrals:
$$\mu = E(X) = \int_0^\infty xf_x(x)\,dx \ge \int_a^\infty xf_X(x)\,dx
 \ge \int_0^a af_X(x)\,dx
= aP(X \ge a).$$
Notice that the first inequality uses the assumption that $P(X > 0) = 1.$
A: The key point is that test scores are bounded below by $0$.  Let $p_i$ be the probability of getting a score of $i$.  Then $$60=\sum ip_i≥80p+\sum_{i< 80} 0p_i=80p\implies p≤\frac 34$$
Note:  this is the best possible bound.  If we imagine that there are only two possible scores, $0,80$, and we set $p=\frac 34$ the bound is realized.
