Independent solution

How to solve this Exponential Distribution question

Setup

Setup

Let U and V denote the earlier and later independent intervals. For an exponential variable, variance is the square of its mean.

E⁡[U]=0.80\operatorname{E}[U]=0.80
Var⁡(U)=(0.80)2\operatorname{Var}(U)=(0.80)^2
Var⁡(V)=μ2\operatorname{Var}(V)=\mu^2

Model

Model

Independence removes the covariance term, so the variance of the combined duration is the sum of the two component variances.

Var⁡(U+V)=Var⁡(U)+Var⁡(V)\operatorname{Var}(U+V)=\operatorname{Var}(U)+\operatorname{Var}(V)
2.65=(0.80)2+μ22.65=(0.80)^2+\mu^2

Compute

Compute

Isolate the later interval's variance and take the positive square root because an exponential mean is positive.

μ2=2.65−0.64=2.01\mu^2=2.65-0.64=2.01
μ=2.01=1.4177446879\mu=\sqrt{2.01}=1.4177446879

Answer

Answer

The requested expected interval is approximately 1.42 years.

1.42(C)\boxed{1.42\quad\text{(C)}}