Exponential Distribution answer check.

Question

so I missed a lecture on the exponential distribution due to illness, so I'm a bit confused as to how to works properly.

I have this is part of a homework problem so I'm not looking for the answer, just some non-specific insight as I'm not sure if I have the right understanding or not.

So the question is along the lines of this;

A patient is treated with a drug for a chronic illness. The doctor estimates there is a 50% chance that the patient will remain symptom free for 30 days.

Find the rate parameter.

My attempt is this.

We are given $P(0\leq X\leq 30) = \frac{1}{2}$. We also know that $E(X)=\frac{1}{\lambda}$. As $P(0\leq X\leq 30) = E(X)$, we have $E(X)=\frac{1}{2}$.

Hence the rate of the distribution is $\lambda=2$??

Thanks.

Answer 1

The cumulative distributive function (CDF) of an exponential variable $X$ is $$F_X(x) = P(X \le x) = \int_0^x \lambda e^{-\lambda t} \,dt = e^{-\lambda t} \rvert_0^\infty = 1 - e^{-\lambda x}.$$

It's given that (note that $X \ge 0$)

$$P(0 \le X \le 30) = F_X(30) = 1-e^{-30\lambda},$$

so we have

\begin{align} 1-e^{-30\lambda} &= \frac12 \\ e^{-30\lambda} &= \frac12 \\ e^{30\lambda} &= 2 \\ \lambda &= \frac{\ln2}{30} \end{align}