Question

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)

• A. \( \ln 2 \)
• B. \( \ln 4 \)
• C. \( \ln 3 \)
• D. \( \ln 0.25 \)

Hint:

For exponentially distributed random variables, the survival function is \( P(X \geq x) = e^{-\lambda x} \), and you can use natural logarithms to solve for \( \lambda \) when given a probability.

Answer 1

The Correct Option is A

For an exponentially distributed random variable \( X \), the probability that \( X \geq x \) is given by: \[ P(X \geq x) = e^{-\lambda x}. \] We are given that \( P(X \geq 2) = 0.25 \). So, we can write the equation as: \[ e^{-\lambda \cdot 2} = 0.25. \] Taking the natural logarithm (ln) of both sides: \[ -\lambda \cdot 2 = \ln(0.25). \] Since \( \ln(0.25) = \ln \left( \frac{1}{4} \right) = -\ln 4 \), the equation becomes: \[ -\lambda \cdot 2 = -\ln 4. \] Simplifying: \[ \lambda = \frac{\ln 4}{2} = \ln 2. \] Thus, the value of \( \lambda \) is \( \ln 2 \), which corresponds to Option (A).