Probability — Question 2

Question 2 – Solution

Step 1: Use the definition of a median. The median $m$ satisfies $$P(T\le m)=\frac12.$$ Because $T\ge0$, $$\int_0^m\lambda e^{-\lambda t}\,dt=\frac12.$$

Step 2: Evaluate the integral. $$\begin{align*} \int_0^m\lambda e^{-\lambda t}\,dt &=\left[-e^{-\lambda t}\right]_0^m\\ &=1-e^{-\lambda m}. \end{align*}$$ Therefore, $$1-e^{-\lambda m}=\frac12.$$

Step 3: Solve for $m$. $$\begin{align*} e^{-\lambda m}&=\frac12,\\ -\lambda m&=\ln\left(\frac12\right)=-\ln2,\\ m&=\frac{\ln2}{\lambda}. \end{align*}$$ $$\boxed{m=\frac{\ln2}{\lambda}}$$ Since $\lambda>0$, the median is positive as expected.