At airports, the service time (checking passport and ticket) of a passenger at the gate has an exponential distribution with mean $10$ seconds. Assume that service times of different passengers are independent random variables.
Approximate the probability that the total service time of $100$ passengers is less than $15$ minutes.

From the Expectation (mean) I can compute that $E[X_1,...,X_{100}]=100*E[X]=100*10=1000$ so $\lambda=0.001$, so I know that $P(x<k)=1-e^{-\lambda k}$ where $k=15$ and and $\lambda=0..01$, so my result is 0.014 but it should be 0.1587, where I'm wrong?

  • 1
    $\begingroup$ The basic error is that the sum of exponential variables is not another exponential variable. $\endgroup$ – David K Jan 22 at 15:57
  • $\begingroup$ What do you mean? $\endgroup$ – Mark Jacon Jan 22 at 15:59
  • $\begingroup$ What @DavidK means is that if $X$ and $Y$ are two random variables with exponential distribution, you can't just conclude that $X+Y$ has an exponential distribution too. You're looking for the total service time. That's $X_1 + \cdots + X_{100}$ which is a sum of random variables with exponential distribution. $\endgroup$ – stressed out Jan 22 at 16:05
  • $\begingroup$ You computed the correct expected value, which is $1000$ seconds, but then you computed $P(x<k)$ as if you were dealing with an exponential distribution with mean $1000.$ It's like saying I can predict the total service time by measuring the time for the first passenger and then multiplying by $100.$ That would be an incorrect method, and you can see right away that it's incorrect if you look at the variance. $\endgroup$ – David K Jan 22 at 16:07
  • $\begingroup$ @DavidK oh yes, you are right, but then which method should I use? $\endgroup$ – Mark Jacon Jan 22 at 16:08


Since your random variables are i.i.d, use the Central Limit Theorem.

Edit: You want to approximate the probability of the total service time of $100$ passengers. Assuming that $X_i$ is the service time for the $i$-th passenger, you want to approximate $S= X_1+\cdots+X_{100}$. As David mentioned in the comments, the distribution of $S$ is not known to us because the sum of two random variables whose distributions are exponential does not necessarily have an exponential distribution. But since $100$ is a relatively large number (any number bigger than $10$ is usually good enough for approximation by the CLT), we can say that $S$ has approximately the same distribution as the normal distribution $\mathcal{N}(n\mu,n\sigma^2)$ where $\mu$ is the mean and $\sigma^2$ is the variance of our original distribution and $n=100$ in our case.

For an exponential distribution with the rate $\lambda$, i.e. $P(X \leq t) = \lambda \exp(-\lambda t)$, the mean is given by $\frac{1}{\lambda}$ and the variance is given by $\frac{1}{\lambda^2}$. So, we have to determine what $\lambda$ is. The question says that the mean is $10$ seconds. So, $1/\lambda=10 \implies \lambda = 0.1$. This tells us that the variance is $100$ seconds. So, now you know that $S \sim \mathcal{N}(100\times 10, 100\times 100)$ approximately. To get the final answer, we have to convert $15$ minutes to seconds because we are measuring time in seconds. Therefore, we have:

$$P(S \leq 15\times 60) = P(\frac{S-1000}{100} \leq \frac{900-1000}{100})=P(Z < -1)=0.1587$$

where $Z=\frac{S-1000}{100}$ and $Z \sim \mathcal{N}(0,1)$.

  • $\begingroup$ $P(Z<15)=P(Z<\frac{15/100-0.1}{\frac{\sqrt{0.01}}{\sqrt{100}}})=P(Z<5)$ right? $\endgroup$ – Mark Jacon Jan 22 at 15:55
  • $\begingroup$ Well, have you calculated the mean and the standard deviation of your exponential distribution already? If yes, what are they? $\endgroup$ – stressed out Jan 22 at 15:56
  • $\begingroup$ Should not be $E[X_{100}]=100*E[X]$, $Var(X_{100})=100*Var(X)$? $\endgroup$ – Mark Jacon Jan 22 at 15:59
  • $\begingroup$ Well, since you know that your distribution is exponential, you just need to determine $\lambda$ in $p(x \leq t)=\lambda \exp(-\lambda t)$. You know that the mean is $10$ seconds. So, $1/\lambda = 10$ and $\lambda = 0.1$. And the variance is equal to $0.01$ because for an exponential distribution, the variance is $1/\lambda^2$. Do you agree? Now, what can you conclude from the CLT? $\endgroup$ – stressed out Jan 22 at 16:02
  • 1
    $\begingroup$ Now it's clear, thanks a lot! $\endgroup$ – Mark Jacon Jan 22 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.