# Central limit theorem for sample quantiles

I am reading Ruppert's Statistics and Data Analysis for Financial Engineering, which contains the following theorem:

Let $Y_1$, $...$ , $Y_n$ be an i.i.d. sample with a CDF $F$. Suppose that $F\;$ has a density $f\;$ that is continuous and positive at $F^{-1}(q)$, 0 < $q$ < 1. Then for large $n\;$, the $q^{th}$ sample quantile is approximately normally distributed with mean equal to the population quantile $F^{-1}(q)$ and variance equal to:

$$\frac{q(1-q)}{n \left[f\;\left(F^{-1}(q)\right)\right]^2}$$

However, consider a Normal density with mean $\mu$ and variance $\sigma^2$. The distribution of the mean of $n\;$ samples will have mean $\mu$ and variance $\sigma^2/n$. If I am understanding it correctly, for q = 0.5 the above formula gives the variance as $$\frac{\pi\sigma^2}{2n}$$

What am I missing here? Any insights are appreciated.

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Although the population mean may equal the population median, the sample mean and sample median are not always the same and do not have the same sampling distributions. –  whuber Aug 27 '11 at 14:53
Actually it's just the opposite: var(median) > var(mean). –  Ringold Aug 27 '11 at 19:31
you're right; I deleted my previous comment –  Andre Holzner Aug 27 '11 at 20:31
n <- 100; n.rep <- 1000