Exam P Practice Problem 3

Problem 3a
Suppose that the lifetime (in years) of an electronic device has the following probability density function:

\displaystyle f(x)=0.0625 \ x \ e^{-0.25 x} \text{where } x>0

The device is sold with a one-year warranty. The manufacturer is considering offering an extended warranty for an additional one year. What proportion of all devices that are found to be working at the expiration of the regular warranty will be working at the end of the extended warranty?

Problem 3b
Losses under the two policies (policy 1 and policy 2) are independent and each of the losses follows an exponential distribution with mean 2. if the total loss is greater than 3, what is the probability that the loss from policy 1 is greater than 3?

Solution is found below.

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Solution to Problem 3a
The problem is to find the probability that a device will survive two years given that it has already survived one year. Let X be the lifetime. Then we need to find P(X>2 \lvert X>1). Note that the pdf given in the problem is a Gamma distribution with parameters 0.25 and 2. We need to find the following right tail of this distribution:

\displaystyle \int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx

The integral can be evaluated by the method of integration by parts. We can also evaluate using a Poisson distribution. The integral is equivalent to P(N \le 1) where N has the Poisson distribution with parameter 0.25t (see Evaluating the Gamma right tail).

\displaystyle \begin{aligned}\int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx&=e^{-0.25t}+e^{-0.25t} \ (0.25t) \\&=e^{-0.25t} \ (1+0.25t)  \end{aligned}

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\displaystyle \begin{aligned}P(X>2 \lvert X>1)&=\frac{e^{-0.5} \ 1.5}{e^{-0.25} \ 1.25} \\&\text{ } \\&=\frac{1.5}{1.25} \ e^{-0.25} \\&\text{ }  \\&=0.93456  \end{aligned}

Answer to Problem 3b


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