Independent solution

How to solve this Exponential Distribution question

Setup

Setup

Parameterize the exponential lifetime by its mean θ. Its variance is θ squared, so first recover θ from the mean-median gap.

E⁡[T]=θ,Var⁡(T)=θ2\operatorname{E}[T]=\theta,\qquad \operatorname{Var}(T)=\theta^2

Model

Model

The median solves a one-half survival probability and therefore equals θ times the natural logarithm of two.

e−m/θ=12e^{-m/\theta}=\frac12
m=θln⁡2m=\theta\ln 2

Compute

Compute

Use the stated difference to solve for the mean and then square that mean.

θ−θln⁡2=3.80\theta-\theta\ln2=3.80
θ=3.801−ln⁡2=12.38378714…\theta=\frac{3.80}{1-\ln2}=12.38378714\ldots
Var⁡(T)=θ2=153.35818399…\operatorname{Var}(T)=\theta^2=153.35818399\ldots

Answer

Answer

The variance rounds to 153 at the precision of the choices.

Var⁡(T)≈153(D)\boxed{\operatorname{Var}(T)\approx153\quad\text{(D)}}