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. (i) Find the probability that a randomly selected battery lasts more than $5$ years. (ii) Find the mean lifetime $E[T]$.

Step-by-step solution

1. For an exponential distribution, $P(T>t)={\int }_t^{\infty }\lambda e^{-\lambda x}dx=e^{-\lambda t}$.
2. Substitute $t=5$ and $\lambda =0.2$: $P(T>5)=e^{-0.2\times 5}=e^{-1}$.
3. Numeric value: $e^{-1}\approx 0.3679$.
4. The mean (expected value) of an exponential distribution is $E[T]=\frac{1}{\lambda }$.
5. With $\lambda =0.2$, $E[T]=\frac{1}{0.2}=5$ years.

Concept tested: Exponential distribution and its expectation

Published 5 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked