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I am trying to find the waiting time which will occur with the probability of 0.99 . To do so, I do $P(X=x)=0.99$ and I did $\lambda*\exp(-\lambda*x)=0.99$ where $\lambda=0.5$. I find the $x$ value as $-1,36$ which should not be because the pdf is defined for $x\geq$ 0 . To be able to find $x$ values which are greater than $0$, the $\lambda$ should be greater than $1$. How should I interpret this case? What is wrong here?

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  • $\begingroup$ You are working with the PDF. You need to consider the cumulative distribution function, CDF. $\endgroup$ – Stephan Kolassa Jul 19 '18 at 14:17
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An exponentially distributed random variable is a continuous random variable, which implies that $ P( X = x) = 0 $ for all $ x $.

What you're looking for is the probability that $ X \geq x $, which you can get by using $ 1 - F(x) $ where $ F(x) = P(X \leq x) $.

The cdf of an exponentially distributed random variable is $ 1 - e^{-\lambda x} $.

Here you're confusing the density function with the distribution function.

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From the cumulative distribution function of an exponential distribution with rate $0.5$ (mean $2$):

  • There is a probability of $0.99$ that the waiting time is less than $-2 \log_e(1-0.99)\approx 9.21034$
  • There is a probability of $0.99$ that the waiting time is more than $-2 \log_e(0.99) \approx 0.02010$
  • There is a probability of $0.99$ that the waiting time is between $-2 \log_e(1-0.995)\approx 0.010025$ and $-2 \log_e(0.995)\approx 10.59663$

The density is never $0.99$ - in fact it never exceeds the rate of $0.5$ - but that does not matter if your questions are about probabilities, the integrals of the density

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  • $\begingroup$ , Thank you so much , I totally understand what you say but can you explain me what they compute with "PDF" in this example . They say that they compute the value of probability. courses.lumenlearning.com/introstats1/chapter/… $\endgroup$ – Angelıque Jul 19 '18 at 15:15
  • $\begingroup$ @Angelıque - the first example and try it in your link are about density. The others, with areas (i.e. integrals of densities), are about probabilities $\endgroup$ – Henry Jul 19 '18 at 15:48
  • $\begingroup$ Thank you so much , I am confused for the meanings of "probability " and "density" What do I find with pdf function $\lambda$*exp(-$\lambda$*x)= say 0.2 what do I find with 1-exp(-$\lambda$*x)= say 0.05 . How can I interpret it ? The integral , probability of (X<x) is 0.05 ok , but what is 0.2 ,then? Thank you so much $\endgroup$ – Angelıque Jul 19 '18 at 17:08
  • $\begingroup$ @Angelıque I would suggest you treat the density simply as a mechanism for finding the probability though integration. You can say for an exponential distribution that the highest densities are at or just above $0$, but knowing the density at some value $x\gt 0$ is $0.2$ and is close to $0.2$ when close to $x$ just tells you that the probability of a value in $\left[x-\frac{\delta}{2},x+\frac{\delta}{2}\right]$ is approximately $0.2\,\delta$ when $\delta$ is small $\endgroup$ – Henry Jul 19 '18 at 21:30

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