The distribution of the difference (sample.mean-population.mean) depends on the population standard deviation and the sample size (in particular, the standard deviation of the difference is related to both -- it's $\sigma/\sqrt{n}$).
This is the (true) standard error of the sample mean.
The distribution of percentage difference will depend on where the population mean is.
Consider that percentage is basically dividing the raw difference by the mean and multiplying by 100. (I'll leave aside the x100 for now.)
So instead of $\sigma/\sqrt{n}$ for the standard deviation of the difference, you're dealing with $\frac{\sigma}{\mu\sqrt{n}}$ for the standard deviation of the relative difference.
Consider I have a mean of 100, and and a standard error of 5. I compute a relative standard error of 0.05 (5%). Now if I subtract 90 from every observation, my standard error is unchanged (it doesn't involve the mean), but my relative standard error has jumped to 50. So I can't make general comments about the size of the relative standard error without reference to where the mean is. It depends on that mean, quite directly.
This is a warning -- when you do simulation, you can't just make conclusions based on one set of parameter values unless you really understand how it will generalize to other values. If you don't, you have to let the simulation tell you how things change as you play with mean and standard deviation independently.
Now $\frac{\sigma}{\mu}$ is called the coefficient of variation. It's often useful in situations where spread tends to increase proportionally when the mean increases (generally the same time that percentage changes make the most sense, basically).
I'm assuming infinite population size.
The best estimate of the population mean is the sample mean, $\bar{x}$.
The best estimate of the population standard deviation is the sample standard deviation, $s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2}$
Since the sample size is less than 30 (10 in this case) and the population standard deviation is unknown, I would prefer to use the T-distribution to develop an interval.
$\bar{x} \pm t(\frac{s}{\sqrt{n}})$
where $t$ is the critical value from the $t_{n-1}$ distribution. In your case with n=10 and a desired confidence level of 95%, $t_{9}=1.833$.
More details can be found in this Example.
Best Answer
The distributions of the sample mean and variance of a normal distribution are well-known (normal for the mean, Chi square for the variance). As whuber says, you can't find the pdfs of the sample mean $\overline{x}$ and, especially, the variance $s^2$ except in special situations. Given only the population mean $\mu$ and variance $\sigma^2$ and nothing else, all you can find exactly are sample mean and variance of $\overline{x}$ and the mean of $s^2$ (but not $s$):
Let the sample size be $n$. Then every introductory text on statistical theory demonstrates that:
$$ E(\overline{x})= \mu $$
$$ Var(\overline{x}) = \frac{\sigma^2}{n} $$ and $$ E(s^2) = \sigma^2 $$
If you know, in addition to $\mu$ and $\sigma^2$, the population fourth central moment $ \mu_4 = E[(X =\mu)^4]$, you can also compute the exact variance of $s^2$
$$Var(s^2) = \frac{(n-1)^2}{n^3}\left(\mu_4 - \frac{n-3}{n-1}\sigma^2\right) $$
Reference:
CR Rao (1965) Linear Statistical inference and its applications, Wiley, New York, p.368.