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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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) ? $\endgroup$
    – Daniel Kim
    Dec 19, 2018 at 7:05
  • $\begingroup$ @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 $e^{-8/3}$. $\endgroup$
    – Godfather
    Dec 19, 2018 at 7:13
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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.

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