The lifetime $T$ (in years) of a certain type of battery follows an exponential distribution with pdf $f(t)=\lambda e^{-\lambda t}$ for $t\ge 0$, where $\lambda =0.2$ per year. What is the probability that a randomly selected battery lasts more than its mean lifetime?

Step-by-step solution

1. For an exponential distribution, the mean (expected value) is $E[T]=1/\lambda$.
2. Given $\lambda =0.2$, the mean lifetime is $1/0.2=5$ years.
3. The required probability is $P(T>5)$.
4. Use the survival function of the exponential: $P(T>t)=e^{-\lambda t}$.
5. Substitute $t=5$: $P(T>5)=e^{-0.2\times 5}=e^{-1}$.

Concept tested: Exponential distribution mean and tail probability

Published 16 September 2026 · How this solution was prepared and checked