Independent solution

How to solve this Exponential Distribution question

Answer in brief

Express five-year survival as q. The supplied interval probability forces q(1-q)=0.25 and hence q=0.5; the requested interval probability is then q^2-q^4=0.1875, corresponding to choice C.

Setup

Setup

Represent survival over one five-year block by a single quantity.

q=e−5λ=Pr⁡(X>5),0<q<1q=e^{-5\lambda}=\Pr(X>5),\qquad 0<q<1

Model

Model

Write the probability of falling in the first stated interval as the difference of two survival probabilities.

Pr⁡(5<X<10)=e−5λ−e−10λ=q−q2\Pr(5<X<10)=e^{-5\lambda}-e^{-10\lambda}=q-q^2
q(1−q)=0.25q(1-q)=0.25

Compute

Compute

The quadratic has a repeated admissible root. Use it to evaluate the later interval by survival differences.

q2−q+0.25=(q−0.5)2=0⟹q=0.5q^2-q+0.25=(q-0.5)^2=0\quad\Longrightarrow\quad q=0.5
Pr⁡(10<X<20)=e−10λ−e−20λ=q2−q4\Pr(10<X<20)=e^{-10\lambda}-e^{-20\lambda}=q^2-q^4
q2−q4=0.25−0.0625=0.1875q^2-q^4=0.25-0.0625=0.1875

Answer

Answer

The interval probability matches choice C.

Pr⁡(10<X<20)=0.1875(C)\boxed{\Pr(10<X<20)=0.1875\quad\text{(C)}}