Approximation of Poisson distribution problem 
Question: A certain type of particle is emitted at a rate of 900 per
  hour. What is the probability that more than 950 particles will be
  emitted in a given hour if the counts form a Poisson process?

I tried to solve this by myself and then I got stuck. I don't know how the book got its answer either :^( If anyone can help I'd really appreciate it.

 A: The idea behind the solution given in the textbook is that when the rate parameter $\lambda$ is very large, then the Poisson distribution is approximately normal, with mean $\mu = \lambda$ and variance $\sigma^2 = \lambda$.  So, by standardizing the Poisson random variable $X$ that counts the random number of events per hour, and using an approximation, we get $$\Pr[X > 950] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{950 - 900}{\sqrt{900}} \right] \approx \Pr[Z > 5/3],$$ where $Z \sim \operatorname{Normal}(0,1)$ is a standard normal random variable.  Then you simply look up the resulting probability in a normal distribution table:  if $\Phi(z) = \Pr[Z \le z]$ is the CDF of the standard normal distribution, then the desired probability is simply $$\Pr[X > 950] \approx 1 - \Phi(5/3),$$ as the book claims.  Again, the takeaway here is that when $\lambda$ is big, $X$ is approximately normally distributed with mean and variance equal to the Poisson rate, therefore, $(X - \mu)/\sigma = (X - \lambda)/\sqrt{\lambda}$ is also approximately standard normally distributed.
If you want, you could also employ continuity correction, which would give you $$1 - \Phi(101/60) \approx 0.0461553,$$ instead of the book's answer.  This is because the Poisson count $X$ is a discrete variable, whereas the normal distribution is continuous; thus you need a correction factor to account for this.  But this is likely to be beyond the scope of your text.
If you have access to a computer algebra system, you can compute the exact probability using the Poisson distribution directly:  $$\Pr[X > 950] = \sum_{k=951}^\infty e^{-900} \frac{900^k}{k!} = 0.047119018020199065215 \ldots,$$ and you can see that this exact value falls between the approximations with and without continuity correction.  So here, we see that continuity correction doesn't give us much benefit.

To give you a sense of how well the approximation works, here is a rather crude animation which displays the PMF of a $\operatorname{Poisson}(\lambda)$ distribution (in blue) along with the PDF of a $\operatorname{Normal}(\mu = \lambda, \sigma^2 = \lambda)$ distribution (in orange), for increasing values of $\lambda$ from $1$ to $100$.  At the beginning, they are different; but as we increase $\lambda$, the approximation improves considerably.

It seems too good to be true.  The curves seem to match so well, right?  Actually, though, if we plot the function $$\rho(x) = \frac{\Pr[X = x]}{f_{Z \sqrt{\lambda} + \lambda}(x)},$$ that is to say, the ratio of the Poisson PMF to the density of its normal approximation, we find that even for $\lambda = 100$, we have the following picture:

This suggests that the Poisson distribution "tail" is quite a bit heavier than its normal approximation, and that the relative quality of the approximation is poor when we are looking more than about $4$ standard deviations away from the mean.
