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I came across an old exam question as follows:

If the life of one computer component (in years) has Gamma distribution with mean $6$ and variance $18$, how can we find the probability that this component has a lifetime of at least $9$ years?

What is the method of solving such question?

The last answer is $4e^{-3}$. I think my book is wrong, any hint or idea?

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  • $\begingroup$ Please add the self-study tag, read its tag wiki and alter your question accordingly; in particular you will need to explain what you tried, and how you decided the answer was wrong. $\endgroup$ – Glen_b -Reinstate Monica Apr 18 '15 at 15:03
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Your translation looks correct.

As you know, the Gamma distribution has two parameters and may be parameterized in two different ways. I believe the most prevalent one is the scale parameterization and indeed, this gives the required answer for your question.

Now you are given the values of the mean and the variance. These values are functions of the parameters and two equations is all we need to recover them. It is an easy exercise to show that for the scale parameterization $\mu=k \theta$ and $\sigma^2=k\theta^2$.

Solving simultaneously, one gets $k=2$ and $\theta=3$. And now you have you all you need to compute the required probability. Inserting these values into the density

$$f_X(x)=\begin{cases} \frac{x^{k-1}}{\Gamma(k) \theta^{k}} e^{-\frac{x}{\theta}} & 0<x<\infty \\ 0 & \text{otherwise} \end{cases} $$

the required probability turns out to be

$$P\left(X\geq 9 \right)=\int_{9}^{\infty} f_X (x) \mathrm{dx}=\int_{9}^{\infty} \frac{x}{9} e^{-\frac{x}{3}}\mathrm{dx}=\frac{1}{9} \int_{9}^{\infty} x e^{-\frac{x}{3}} \mathrm{dx} $$

We will use integration by parts to evaluate this integral. If we let $u=x$ and $dv=e^{-\frac{x}{3}}$ and for now leaving out the $\frac{1}{9}$ factor what we get is $$\int_{9}^{\infty} x e^{-\frac{x}{3}} \mathrm{dx}=-3 xe^{-\frac{x}{3}} \Big|^{\infty}_{9}+3\int_{9}^{\infty} e^{-\frac{x}{3}}\mathrm{dx}=27e^{-3}-3e^{-\frac{x}{3}} \Big|_{9}^{\infty}=36 e^{-3}$$

And multiplying by $\frac{1}{9}$ we get $4e^{-3}$ or $0.199$, as required.

R can help you with the calculations

1-pgamma(9,shape=2,scale=3)
[1] 0.1991483

but be sure to specify the scale argument, otherwise it defaults to rate and the answer is misleading.

Hope this helps!

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  • 1
    $\begingroup$ my main problem is via integration :) is it possible show how solve it? +1 $\endgroup$ – Dr. Hoshang Apr 18 '15 at 14:41
  • $\begingroup$ @Dr.Hoshang Can you explain where your problem with integration lies? For routine bookwork questions, including "old exam questions" we are not supposed to give complete solutions, but guidance and hints -- in return for attempted solutions and a clear explanation of the specific help needed. JohnK did give a hint (use integration by parts). Did you try that? Can you show your attempt? (in an edit to your question) $\endgroup$ – Glen_b -Reinstate Monica Apr 18 '15 at 14:57
  • $\begingroup$ @Dr.Hoshang I have included some additional details but I hope you can fill in-and understand- the intermediate steps. $\endgroup$ – JohnK Apr 18 '15 at 15:03

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