Suppose that a 1-day 97.5% VaR is estimated as $13 million from 2.000 observations. The observations on the 1-day changes are approximately normal with mean 0 and standard deviation $6 million. Estimate a 99% confidence interval for the VaR estimate.

Answer is the following from a previous tutor:

The interval will be 13 2.576 *6/2000^.5 = 12.654394 , 13.3456066

Please walk me through step by step of where all the numbers came from and why?

Also, i believe this formula should be used for standard error- 1/f(x) * square root of (1-q)q / n q- is the q-percentile and n-stands for number of observations

when i do this formula i get the answer of .505649. I know that then i would need to use VAR of the 97.5% which is 13M plus or minus (?) x standard error or (.505649)