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What is the formula (if it exists) for the sample variance / confidence interval of a quantile / percentile of the normal distribution?

For example, the 5th percentile for a standard normal population distribution is -1.64, but what is the 95% one-sided confidence interval if I have a sample of n=1,000? I.e. on average the 5th percentile of a standard normal sample will be -1.64 and 95% of the time the sample 5th percentile of a sample of n=1,000 will be below -1.54 (approximate from simulations).

There seems to be some kind of formula as it's an option in SAS (CIQUANTNORMAL) and I think it may be related to the Probit but haven't found an explanation.

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When you are trying to estimate a quantile from data then you can turn the problem into a binomial problem. For 1,000 data points you want to know what value will have 5% of values below it (population values, the sample is just the estimate) and 95% above it. So you can test each possible value using the binomial (how many of your 1,000 values are below the value you are testing). All the values that would not be rejected given your data constitute the confidence interval.

There is some more detail on Wikipedia or by Googling "quantile confidence interval".

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Suppose that $X\sim N(\mu,\sigma)$, where $\mu$ and $\sigma$ are unknown. Since $\bar{x}$ and $s$ are independent, it is pretty easy to calculate confidence bounds for any linear combination of $\mu$ and $\sigma$.

Now, all you need to remember is that the 5th percentile of $X$ is, as you note, $\mu-1.64\sigma$.

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