Question

Hello guys, I have question to ask.

I don't understand how answer got the working out for continuous random variable and the integral for part A and B.


Can someone explain this to me?

Cheers!

Answer 1

As pointed out in comments, part (A) is not possible to do by hand unless we introduce the so-called exponential integral function.

The integrals in part (B) however are easy to compute.

The probability that a guinea pig lives over 6 years is

`\displaystyle P(T>6) = \int_6^\infty L(t) \, dt = 0.0977 \int_6^\infty e^(0.4t) e^{-0.2 e^(0.4t)} \, dt`

Substitute

`u = -0.2 e^(0.4t) \implies du = -0.08 e^(0.4t) \, dt`

Then

`\displaystyle P(T>6) = -(0.0977)/(0.08) \int_(-0.2 e^(2.4))^(-\infty) e^u \, du \\\\ ~~~~~~~~ = (0.0977)/(0.08) \int_(-\infty)^{-0.2 e^(2.4)} e^u \, du \\\\ ~~~~~~~~ = (0.0977)/(0.08) \left(e^{-0.2e^(2.4)} - \lim_(u\to-\infty) e^u\right) \\\\ ~~~~~~~~ = \frac{0.0977}{0.08e^{0.2e^(2.4)}} \approx 0.134693`

On the other hand, the probability that a guinea pig lives for less than 1 year is

`\displaystyle P(T < 1) = 0.0977 \int_0^1 e^(0.4t) e^{-0.2e^(0.4t)} \, dt \\\\ ~~~~~~~~ = -(0.0977)/(0.08) \int_(-0.2)^{-0.2e^(0.4)} e^u \, du \\\\ ~~~~~~~~ = -(0.0977)/(0.08) \left(e^{-0.2e^(0.4)} - e^(-0.2)\right) \approx 0.0936702`