Error Bars for Monte Carlo Experiment Suppose we have a random variable $X$, where $\mathbb{E}(X)$ and $\text{Var}(X)$ are known. I have computed $N$ number of MC-type samples from the distribution of $X$. 
Let $\bar{x} = \frac{1}{N}\sum x_i$ where $x_i$ represents a single sample from our distribution. From CLT, I know that the error bars for $\bar{x}$ can be constructed in the following way, assuming 90% CI:
$$ \bar{x} \pm (1.64)\sqrt{\frac{\text{Var}(X)}{N}}$$
But I'm interested in computing the error bars for the relative error $\frac{|\bar{x} - \mathbb{E}(X)|}{|\mathbb{E}(X)|}$. Since the relative error itself is a random variable, I'm having difficulty constructing error bars for it . 
Any suggestions? Thanks in advance! 
I wonder if I can use relative error in Variance and substitute that with $\text{Var}(X)$. 
 A: Writing $\mu=E(X)$ and $\sigma^2=\text{Var}(X)$
Let's assume (as you did) that the distribution of $X$ is one for
which the CLT approximately "kicks in" by sample size $n$. That
is $\bar{X} \,\,:\hspace{-.58em}\sim N(\mu,\sigma^2/n)$.
Note that $Y=\bar{X}-\mu$ will have mean 0 and be approximately normal, so $|Y|$ has a scaled chi distribution.
If $Z = \sqrt{n} Y/\sigma$, then $|Z|$ has a standard chi-distribution:

Since the distribution is skew, we won't have any kind of symmetric interval,
so it doesn't really make sense to write it in $\pm$ form.
The 5th percentile of this distribution is the 52.5 percentile of the corresponding standard normal, which is at 0.0627, which the 95th percentile is at the 97.5 percentile
of the standard normal, 1.96. That is, 90% of the standard chi lies in $(0.0627,1.96)$.
So we have:
$0.0627<|Z|<1.96$
$0.0627<\frac{|Y-\mu|}{\sigma/\sqrt{n}}<1.96$
$0.0627\frac{\sigma}{\mu\sqrt{n}}<\frac{|Y-\mu|}{\mu}<1.96\frac{\sigma}{\mu\sqrt{n}}$
with approximate 90% probability.
Hence $(0.0627\frac{\sigma}{\mu\sqrt{n}},1.96\frac{\sigma}{\mu\sqrt{n}})$ will form an approximate 90% interval for $\frac{|\bar{X}-\mu|}{\mu}$.
The mean of $\frac{|\bar{X}-\mu|}{\mu}$ will be $\sqrt{\frac{2}{\pi}}\frac{\sigma}{\mu\sqrt{n}}\approx 0.7979\frac{\sigma}{\mu\sqrt{n}}$ and the median will be $\approx 0.6745\frac{\sigma}{\mu\sqrt{n}}$

($\sigma/\mu$ is sometimes called the coefficient of variation)
