Probability — Question 8

Question 8

A lifetime satisfies $P(X>t)=e^{-(t/5)^2}$, $t\ge0$. Find its density and $P(3<X<4)$.

Question 8 – Solution

Step 1: Convert the survival function to the CDF. For $t\ge0$, $$P(X>t)=e^{-(t/5)^2}=e^{-t^2/25}.$$ Therefore, $$F(t)=P(X\le t)=1-P(X>t)=1-e^{-t^2/25}.$$

Step 2: Differentiate to obtain the density. $$\begin{align*} f(t)&=F'(t)\\ &=-\frac{d}{dt}\left(e^{-t^2/25}\right)\\ &=\frac{2t}{25}e^{-t^2/25}, \qquad t\ge0. \end{align*}$$ Also, $f(t)=0$ for $t<0$.

Step 3: Compute the interval probability. Using the CDF, $$\begin{align*} P(3<X<4) &=F(4)-F(3)\\ &=\left(1-e^{-16/25}\right) -\left(1-e^{-9/25}\right)\\ &=e^{-9/25}-e^{-16/25}. \end{align*}$$ $$\boxed{P(3<X<4)=e^{-9/25}-e^{-16/25}}$$ The same result follows by integrating $f(t)$ from $3$ to $4$.