Finding percentiles of the $\chi^2$ distribution

Question

If the random variable $Y$ has a $\chi^2$ distribution with $54$ degrees of freedom, then what is the approximate $84^{\text{th}}$ percentile of $Y$?

I don't understand very good question when it is with $30$ or more degrees of freedom and with approximate percentile.

Please give some explanation to your answer as well.

Answer 1

Using R, the answer is

> qchisq(0.84, 54)
[1] 64.26393

Answer 2

A $\chi_k^2$ distribution is the sum of $k$ IIDs each of mean $1$ and variance $2$, so by the central limit theorem it approximates $N(k,\,2k)$ for large $k$. The $84$th percentile of a Normal variable is $z:=0.9945$ standard deviations above its mean, in this case $54+z\sqrt{108}\approx64.33$. But if you have something more accurate at your disposal, as @kludg did, use it.