# life expectancy of new computer with exponential distribution

I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.

Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?

what I have: f(x) = $$\lambda \cdot e^{-\lambda \cdot x}$$ given: $$\lambda = 3, x = 8$$ so simply $$= 3 \cdot e^{-3 \cdot 8} = 3\cdot e^{-24}$$?

sorry if this is too low level question. But I'm a bit confused.

If $$X$$ is exponentially distributed with parameter $$\lambda$$, the probability $$P(X>x)=e^{-\lambda x}$$. You can substitute $$\lambda = 3$$ and $$x=8$$ to get the probability that the computer is still working after 8 years.

By the way, some clarity is required on what $$\lambda$$ represents. Typically, $$\lambda$$ is the parameter of the distribution and $$1/\lambda$$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $$\lambda = 1/3$$.

You substituted $$x=8$$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $$P(X=x)$$. However, you can compute the probability $$P(X>x)$$ or $$P(X \leq x)$$ using the CDF (which is obtained by integrating the PDF).

Hope this helps.

• Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ? Dec 19, 2018 at 7:05
• @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $\lambda = 1/3$ (since the mean of an exponential distribution is $1/\lambda$). So your answer is $e^{-8/3}$. Dec 19, 2018 at 7:13

As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $$\frac{1}{\lambda} = 3$$ and so $$\lambda = \frac{1}{3}$$ simply by definition. Let $$X \sim exp(-\frac{1}{3})$$

Since the exponential distribution is continuous, the $$P(X = 8) = 0$$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $$P(X>8)$$. we have that \begin{align*} P(X > 8) &= 1 - P(X \leq 8) \\ & = 1 - F_X(8) \\ & = 1 - (1-e^{-\frac{1}{3}x}) \\ & = e^{-\frac{8}{3}} \end{align*}

This says that there is approximately 7% chance that the computer is still working after 8 years.