I am having a hard time finding the confidence interval of the median of an exponential distribution. I am currently studying for an upcoming test. 

I found the mle of $\lambda = \frac{5}{61}$

I solved $S(t>z)=0.5$ and found the median is $z=\frac{\log(2)} {\lambda}$

Now I need to find the 95% confidence interval. I am unsure of how to do this. My first thought is to try something like:

$$
CI = \frac{\log(2)}{\lambda}
 \pm 1.96\sqrt{Var\left(\frac{\log(2)}{\lambda}\right)}
$$
where $\lambda$ is the mle I found above. 

Is that how you correctly solve for the CI for the median? If so, how do I find $Var(\frac{\log(2)}{\lambda})$ ?

EDIT: i solved for 5/61 by maximizing the log-likelihood function. I am dealing with discrete data that is subject to right censoring

I would greatly appreciate someones help, thanks!