# Approximate the probability that the total service time of $100$ passengers is less than $15$ minutes.

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 $$P(x where $$k=15$$ and and $$\lambda=0..01$$, so my result is 0.014 but it should be 0.1587, where I'm wrong?

• The basic error is that the sum of exponential variables is not another exponential variable. – David K Jan 22 at 15:57
• What do you mean? – Mark Jacon Jan 22 at 15:59
• 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. – stressed out Jan 22 at 16:05
• 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. – David K Jan 22 at 16:07
• @DavidK oh yes, you are right, but then which method should I use? – Mark Jacon Jan 22 at 16:08

## 1 Answer

Hint:

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 $$Z \sim \mathcal{N}(0,1)$$.

• $P(Z<15)=P(Z<\frac{15/100-0.1}{\frac{\sqrt{0.01}}{\sqrt{100}}})=P(Z<5)$ right? – Mark Jacon Jan 22 at 15:55
• Well, have you calculated the mean and the standard deviation of your exponential distribution already? If yes, what are they? – stressed out Jan 22 at 15:56
• Should not be $E[X_{100}]=100*E[X]$, $Var(X_{100})=100*Var(X)$? – Mark Jacon Jan 22 at 15:59
• 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? – stressed out Jan 22 at 16:02
• Now it's clear, thanks a lot! – Mark Jacon Jan 22 at 17:18