Independent solution

How to solve this Exponential Distribution question

Setup

Setup

For an exponential lifetime with mean θ, variance is θ squared and survival at time t is exp(-t/θ).

θA2=5.60\theta_A^2=5.60
Pr⁡(TA>t)=e−t/θA,Pr⁡(TB>t)=e−t/θB\Pr(T_A>t)=e^{-t/\theta_A},\qquad \Pr(T_B>t)=e^{-t/\theta_B}

Model

Model

Relate the two survival values without first solving for the common time.

0.49=(0.70)20.49=(0.70)^2
e−t/θA=(e−t/θB)2=e−2t/θBe^{-t/\theta_A}=\left(e^{-t/\theta_B}\right)^2=e^{-2t/\theta_B}

Compute

Compute

Equal exponents imply the second mean is twice the first; square that ratio for variances.

θB=2θA\theta_B=2\theta_A
Var⁡(TB)=θB2=4θA2=4(5.60)=22.40\operatorname{Var}(T_B)=\theta_B^2=4\theta_A^2=4(5.60)=22.40

Answer

Answer

The second lifetime variance is 22.40.

22.40(E)\boxed{22.40\quad\text{(E)}}