Determining p-th Quantile From Probability Density Function I'm not sure how to derive the $p$-th quantile. I know that it is point which divides the distribution of $X$ into two parts, but I'm not sure what I'm supposed to do here.

If the random variable $X$ has probability density function 
  $f(x) =  \lambda e^{-\lambda x}$ for $x > 0$. ($\lambda > 0$), 
  determine the $p$-th quantile $x_p$ in terms of $\lambda$ and $p$. 

 A: To be technical, a $p$-th quantile is any point $x_p$ such that $P(X \le x_p)=p$.
There may be more than one such $x_p$, but in our case there is only one. 
So for our random variable with exponential distribution, we want to find $x_p$ such that
$$\int_0^{x_p} \lambda e^{-\lambda x}\,dx=p.$$
Integrate. We get $1-e^{-\lambda x_p}$. Set this equal to $p$, and solve for $x_p$.  The equation $1-e^{-\lambda x_p}=p$ can be rewritten as
$$e^{-\lambda x_p}=1-p.$$
Take the natural logarithm of both sides, and solve for $x_p$.
We get
$$x_p=\frac{-\ln(1-p)}{\lambda}.$$
For a more general random variable $X$, we want to solve the equation $F_X(x_p)=p$, where $F_X$ is the cumulative distribution function of $X$.  It may not be possible to solve this equation by "exact" methods, but usually we can at least get a good numerical approximation.
A: Calculate the cumulative distribution function
$$F_X(\alpha) = \int_0^{\alpha} \lambda e^{-\lambda x} \mathrm dx$$
for $\alpha > 0$.  The $p$-th quantile is $x_p$ where $x_p$
is the solution to
$$F_X(x_p) = \frac{p}{100}$$
if $p$ is stated as a percentage.
