Consider the time to wait for your bus is modeled by $X\text{~}Exp(\lambda)$.
What is the $90/100th$ quantile for $\lambda=1$?
Given $90/100th$ quantile is 20 min ($\lambda$ is unknown). Give an upper bound on the probability that you wait more than 30 min. (The upper bound must be less than $0.5$).
My attempt:
For the first part, I can just use the formula that $\mathbb{P}(X\geq x_p)=1-p$ where $p=90/100$, and the formula for Exponential distribution, which is $f(x)=\lambda e^{-\lambda x}$. $$1e^{-1x}=1-90/100$$ $$x=-\ln(10/100)=2.3025$$
However, I don't understand what is meant by find the upper bound for the probability. Which formulas are involved and how should I approach this problem? Thank you.
$X \sim Exp(\lambda)$
The PDF $f(x) = \lambda.e^{-\lambda x}$
The CDF $F(x) = P(X \leq x) = 1 - e^{-\lambda x}$
The quantile function $F^{-1}(p) = \frac{1}{\lambda}\ln \frac{1}{1-p}$
For $\lambda=1$ the $\frac{90}{100}$ quantile is $\ln \frac{1}{1-\frac{9}{10}} = \ln 10 = 2.302585$ mins
Given, $F^{-1}(90/100) = \frac{1}{\lambda}\ln \frac{1}{1-90/100} = \frac{\ln 10}{\lambda} = 20 \implies \lambda = \frac{\ln 10}{20}$
Note that it implies $P(X \leq 20) = F(20) = \frac{90}{100}$ and $P(X > 20) = \frac{10}{100}$
$\therefore P(X > 30)$
$= 1 - P(X \leq 30) = 1 - F(30) = e^{-\lambda. 30} = e^{-\frac{\ln 10}{20}. 30}$
$= (e^{-\ln 10})^\frac{3}{2} = \left(\frac{1}{10}\right)^\frac{3}{2} = 0.03162278$