How can I get the CI of a couple of numbers? Is it possible to get the Confidence Interval of a couple of numbers (or more) without knowing the distribution or anything like that ? 
Thanks.
 A: I assume that you observed a (big enough) number $n$ of packets, amoung which $x$ packets were lost.  You have an estimation of the proportion $p$ of packet loss, $\hat p = {x\over n}$.
The usual Confidence Interval procedure gives a 95% CI
$$ \left[ \hat p - 1.96 \sqrt{\hat p (1 -\hat p) \over n} ; \hat p + 1.96 \sqrt{\hat p (1 -\hat p) \over n}\right],$$ 
which is usually considered as valid if $n p > 5$ and $n(1-p)>5$ on the whole interval.
As I assume that the proportion $p$ you estimate is small, I give you this Confidence Interval procedure which is robust for small values of $p$.
Let $\Phi(p) = \arcsin(\sqrt p)$ for $p\in [0,1]$.
A 95% CI on $\Phi(p)$ is given by
$$\left[ \Phi\left({ x - 0,5 \over n}\right) - 1,96 { 1 \over 2 \sqrt n} ; 
  \Phi\left({x + 0,5 \over n}\right) + 1,96 {1 \over 2 \sqrt n} \right]$$
To get a CI on $p$, use the inverse transformation $\Phi^{-1}(y)=\sin(y)^2$ on the bounds of this interval.
A: Generally a confidence interval is about a parameter of a population/distribution, not the observed values.  As such there needs to be some assumptions (even saying that your are confident that the mean lies between minus infinity and infinity assumes that it is a real number).
Here is an article that derived a formula for a confidence interval for the mean with a sample of size 1 (they do make some assumptions): 

An Effective Confidence Interval for the Mean With Samples of Size One
  and Two  Melanie M Wall, James Boen, Richard Tweedie. The American
  Statistician. May 1, 2001, 55(2): 102-105.
  doi:10.1198/000313001750358400.

If you are looking at proportions or counts then you can use binomial or poisson distributions that can be estimated using 1 or 2 data points.  If you are not happy with those assumptions then you will need to make some others.
A: Converting my comment into an answer in view of the OP's response, 
given any two numbers $x_1$ and $x_2$, their average value (or sample mean) is 
$$\bar{x} = \frac{x_1+x_2}{2}.$$
The sample standard deviation is
$$
\sigma 
= \sqrt{\left.\left.\frac{1}{2-1}\right[(x_1-\bar{x})^2+(x_1-\bar{x})^2\right]}
= \frac{|x_2-x_1|}{\sqrt{2}} \approx 0.707|x_2-x_1|
$$
Assume without loss of generality that $x_1 \leq x_2$.  Obviously,
$[x_1, x_2]$ is a $100\%$ confidence interval for the observations
which does not even require any calculations of $\bar{x}$ or $\sigma$,
but even greater confidence can be generated among the non-cognoscenti
by saying that $\left[\bar{x}-\sigma/\sqrt{2},\bar{x}+\sigma/\sqrt{2}\right]$ is
a $100\%$ confidence interval for the observations.
A: Assuming that you have enough observations, you can then use the central limit theorem. This will mean that you have a standard normal distribution to work with.
From then on, it depends on what percentage your significant level needs to be to compute the confidence interval. Calculate the sample variance (s^2) and your confidence interval will be:
Xbar ± s/sqrt(N) where N is the number of observations.
