# An extended warranty pays a benefit of $100$ if failure occurs between time $t = 1.5$ and $t = 8$. Find $P(w<79)$.

The time until failure, $T$, of a product is modeled by a uniform distribution on $[0, 10]$. An extended warranty pays a benefit of $100$ if failure occurs between time $t = 1.5$ and $t = 8$. The present value, $W$, of this benefit is $$w(t)=\begin{cases} {0} & \text{if } 0\leq t<1.50\\ 100e^{-0.04} & \text{if } 1.5\leq t< 8.0,\\ 0 & \text{if } 8\leq t\leq 10,\end{cases}$$

\begin{align*} P(w<79) &= p(100 e^{-.04t}<79)\\ &= p(t>5.89)\\ &= 1-p(t<5.89)\\ &= 1-F(5.89)\\ &= 1-.589\\ &= 0.411 \end{align*}

correct answer is $0.59$

• Please review my edits and use them as a model for your future posts. Some of your notation was not clear. – Em. Aug 27 '16 at 18:55

You must consider the conditional CDF of W under all three conditions on T. First, since W=0 for T in [0,1.5], then:

Pr(W<79| 0

Secondly, since W= 100exp(-.04t) for T in [1.5,8], then:

Pr(W<79|1.5 < T < 8)=.411 as you calculated above

Lastly, since W=0 for T in [8,10],then:

Pr(W<79|8 < T < 10)=Pr(0<79| 0 < T < 10)=1

The CDF of a a mixed distribution like W, which is weighted based on values of T in the form:

Pr(W <= 79)=Pr( W <= 79 | 0 < T < 1.5) Pr(0 < T < 1.5) + Pr(W<= 79| 1.5 < T < 8) Pr( 1.5 < T < 8) + Pr( W <= 79 | 8 < T < 10) Pr(8 < T < 10)

The weights on the conditional CDF distributions, which are based on probabilities in T, are calculated from the PDF of the continuous uniform distribution on [0,10]. For example,

Pr(0 < T < 1.5) =(1.5)/10= 3/20

and so on...

You can then plug these values into the above conditional CDF form to get the correct answer.

Hope this helps!