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 $P(x<k)=1-e^{-\lambda k}$ where $k=15$ and and $\lambda=0..01$, so my result is 0.014 but it should be 0.1587, where I'm wrong?
 A: 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 $Z \sim \mathcal{N}(0,1)$.
