How would I work out a distribution free 95% confidence interval of the median if given a large sample?

To get the 95% CI I use \mbox{estimate} \pm 1.96 \times \mbox{standard error}. I can work out the estimate of the median, which is just the sample median.

However, to find the standard error do I use var(m) \approx \frac{1}{4nf(m)^2} and then square root that or would that not be distribution free??

Would I use the normal approximation to the binomial? Bi(n,0.5) \approx^d N(0.5n,0.25n)?

To get the 95% CI I use \mbox{estimate} \pm 1.96 \times \mbox{standard error}. I can work out the estimate of the median, which is just the sample median.

However, to find the standard error do I use var(m) \approx \frac{1}{4nf(m)^2} and then square root that or would that not be distribution free??

Would I use the normal approximation to the binomial? Bi(n,0.5) \approx^d N(0.5n,0.25n)?

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