Independent solution

How to solve this Exponential Distribution question

Setup

Setup

The recorded time is the larger of the actual lifetime and two. It can be written as a guaranteed two years plus any lifetime excess beyond two.

X=max⁡(T,2)=2+(T−2)+X=\max(T,2)=2+(T-2)_+

Model

Model

The expected excess above two is the survival integral from two to infinity.

E[(T−2)+]=∫2∞Pr⁡(T>t)dt=∫2∞e−t/3dtE[(T-2)_+]=\int_2^\infty\Pr(T>t)dt=\int_2^\infty e^{-t/3}dt

Compute

Compute

Evaluating the exponential survival integral and adding the guaranteed two years gives approximately 3.540251.

E[X]=2+3e−2/3=3.540251E[X]=2+3e^{-2/3}=3.540251

Answer

Answer

The mean recorded time is approximately 3.5403, selecting choice D.

2+3e−2/3(D)\boxed{2+3e^{-2/3}\quad\text{(D)}}