Independent solution

How to solve this Exponential Distribution question

Answer in brief

For an exponential lifetime with mean theta, the median is theta times log 2. The stated gap determines theta as 12.3838, so the variance is 153.358 and choice D.

Setup

Setup

Parameterize the exponential lifetime by its mean theta. Its variance is theta squared, so first recover theta 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 theta 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)}}