# Confidence interval for the 95th percentile of the normal distribution

Let $$X_1, .., X_n \sim Normal(\mu, \sigma^2)$$.

Let $$\tau$$ be the 95th percentile of this distribution. Thus,

$$P(X_i < \tau) = 0.95$$.

What is the $$1 - \alpha$$ confidence interval for $$\tau$$?

I know how to get the maximum likelihood estimator for $$\tau$$; I would invoke the equivariance principle and plug in the MLEs for $$\mu$$ and $$\sigma$$.

$$\hat{\tau} = \bar{X} + S \Phi^{-1}(0.95)$$.

However, I'm struggling to estimate the standard error for it. It likely involves Fisher's information matrix, but I'm stuck at this point.

For normal distribution, $$\bar X$$ and $$S$$ are independent. So $$\mathrm{Var}(\hat \tau) = \mathrm{Var}(\bar X) +\mathrm{Var}(S \Phi^{-1}(0.95)) = \frac {\sigma^2}n + (\Phi^{-1}(0.95))^2 \mathrm{Var}(S)$$
$$\sqrt {n-1} S/\sigma$$ follows chi distribution with $$n-1$$ degree of freedom. Its variance is $$\frac {2[\Gamma(\frac {n-1}2)[\Gamma(1+ \frac {n-1}2)-[\Gamma(\frac {n}2)]}{\Gamma(\frac {n-1}2) } = V$$. So the variance of $$S$$ is $$\frac {\sigma^2}{n-1}V$$.
So $$\mathrm{Var}(\hat \tau) = \frac {\sigma^2}n + (\Phi^{-1}(0.95))^2\frac {\sigma^2}{n-1}V$$