Bounding "an average"? I'm doing a practice problem that is about processing orders.
The assumptions are:
"On average 40 orders are received"
"50 orders per day can be processed"
The task is to formulate notions about the probability of the shop being able to fulfill the processing of orders.
But what does "on average" mean here? It cannot mean that "every day 40 orders are processed" since then the probability to succeed in managing those would always be 1.
Perhaps there's some "standard" way of finding out what the "mean deviation" can be? Or the variance?
Specifically I need to find an "upper bound".
 A: Such a phrase ("on average 40 orders are received") can either mean that the sample average over some sample is 40 or it means that the long run (i.e. population) average is 40. In this case it means the second one.
It certainly doesn't mean "every day 40 orders are processed" (that's not to say that this isn't the case in this instance; it's a possible situation that's consistent with the information -- but so are many other possibilities).
It's telling you about a population quantity (the mean of the distribution of the number of orders received) for the process.
It could be that there's little variation in orders received (that they're nearly always close to 40) or it could be that there's a great deal of variation in the orders received (i.e. that the actual number in a day is generally not very close to 40)
Your last sentence isn't clear. From the information given there's no way to infer any information about variation about the mean, such as mean deviation - it could be as low as zero or it could be arbitrarily large.

Edit: it now appears that you're asking about how to bound the proportion above 50.
That can be done. Clearly the minimum proportion above 50 is 0.
So what about the maximum? You might like to consider 
a. the effect on the mean of an element of probability above the mean but close to it
b. the effect on the mean as you shift that element of probability further above the mean
c. that the same effect obtains below the mean
d. that therefore you can "balance" out some probability at some distance above the mean with some other probability at a distance below the mean in such a way that the mean of both contributions is 40.
e. that therefore if you take some distribution with mean 40 that doesn't achieve the maximum, you can move some proportion up and some other proportion down without changing the mean
f. that there's a limit to how far you can move some of the probability down.
without completely giving it away, the result follows from such considerations. 
Of course, you can apply a theorem -- there's an obvious inequality that applies -- but there's simply no need; the result is obvious from basic considerations.
(Once you work out the answer, the general inequality itself becomes more or less obvious.)
[The OP figured out that the Markov inequality was the relevant one.]

Since it appears that it's not obvious to you (you appear to want to jump straight to applying Markov without consideration of just how elementary this is), consider step d. Slightly informally, we have some small amount of probability, $\epsilon$ say, at $40+x$, and some other amount, $\delta$ say, at $40-y$. The remaining $1-\epsilon-\delta$ of probability is distributed (in some fashion) so that its conditional mean is 40. In order for the whole thing to average 40, how would we choose $\epsilon, \delta, x$ and $y$ to get $\epsilon$ of the total probability above 50?
So the contribution to the mean of the $\epsilon$ of probability at $40+x$ is that it adds $x\epsilon$ to the original mean of 40, and the contribution to the mean of the $\delta$ of probability at $40-y$ is that it subtracts $y\delta$ from the mean, and so the total contribution to the mean of those parts is 
$x\epsilon-y\delta$
Note that the constraint that the mean remain 40 imposes one linear restriction on the four quantities. 
Note also that 


*

*we want $\delta$ to be as small as possible (why? so we can move more probability in similar steps).

*we want $y$ to be as large as possible (so we can make $\delta$ "count" more in its contribution)

*we want $\epsilon$ to be as big as possible (that's the exercise, after all)

*we therefore want $x$ to be as small as possible given the constraints in the question (to minimize the contribution of the $\epsilon$ to the mean).
So how large can $y$ be? How small can $x$ be? If we fix $\epsilon$, what  must $\delta$ be? Can we make $\epsilon$ larger?
