It is given that \(P(X \geq 2) = 0.25\) for an exponentially distributed random variable \(X\) with \(E[X] = \frac{1}{\lambda}\), where \(E[X]\) denotes the expectation of \(X\). What is the value of \(\lambda\)?
(\(\ln\) denotes natural logarithm)
Answer
(A)
Explanation
An exponential random variable with mean \(\frac{1}{\lambda}\) has rate \(\lambda\) and tail probability \(P(X \geq x) = e^{-\lambda x}\) for \(x \geq 0\). So
\[e^{-2\lambda} = 0.25 = \frac{1}{4} \;\Rightarrow\; 2\lambda = \ln 4 = 2\ln 2 \;\Rightarrow\; \lambda = \ln 2.\]
Check: \(e^{-2\ln 2} = 2^{-2} = 0.25\). Option B is the value of \(2\lambda\), not \(\lambda\), and option D is negative (\(\ln 0.25 = -\ln 4\)), which is impossible for a rate.
Answer: A