The number of packets arriving at a multiplexer in any one second interval is a Poisson random variable with mean $15$. Assume that the number of arrivals in nonoverlapping one second intervals is independent.
$(a)$ Use the Central Limit Theorem to estimate the probability that more than $950$ packets arrive in one minute.
$(b)$ Compute the exact probability for the event described in $(a)$.
For the first part, I thought to let $S_{60} = \sum_{i=1}^{60} X_i$ where $X_i$ is a Poisson random variable representing the number of packets arriving in the $i$th second. Then we want $P(S_{60} > 950)$ and can use the Central Limit Theorem with mean $60×15=900$ and variance $60×15 = 900$ where I used the fact that the mean and variance of a Poisson random variable is the same as well as sum of Poisson random variables is Poisson. Is this correct??
So you get $1 - \Phi(1.66666)$ for part $(a)$.
Now I don't really know how to do part b. Can someone please help me?
https://en.wikipedia.org/wiki/Poisson_distribution
Why don't you just apply the formula of the PMF of the poisson distribution?