$P(X \ge 450)$ in Possion distribution The number of pedestrians that cross the street in one minute has Poisson distribution $\def\Pois{\operatorname{Pois}}\Pois(8)$. Find the probability that at least $450$ pedestrians will cross the street in $1$ hour.
The number of pedestrians that cross the street in one hour has $\Pois(480)$ distribution. So, $$\begin{align}
P(X \ge 450) & = 1 - P(X < 450) \\
& = 1 - \sum_{i = 0}^{449}\frac{450^i}{i!}e^{-450} \\
& = 1 - e^{-450}\sum_{i = 0}^{449}\frac{450^i}{i!},\end{align}$$ and I am stuck here.
 A: Presumably with appropriate software we can evaluate that unpleasant sum. But I would not care to attack it by hand.
However, note that the number crossing in an hour is a sum of $60$ independent identically distributed Poisson random variables.
So the sum should  have nearly normal distribution, Recall that the variance of a Poisson is equal to the parameter, as is the mean.  We conclude that the number crossing in an hour has a close to normal distribution,  with mean $480$ and variance $480$. 
Now you can use a probably familiar calculation to find the probability that this normal is $\ge 450$. Depending on what is expected in your course, you may need to make a continuity correction. To me it would not make much sense, since the Poisson traffic model gives at best a modestly good fit. 
A: What you really want is the number in 60 minutes, which is
$$X_1+X_2+\cdots+X_{60}$$
Where $X_k$ is the number of pedestrians in the $k$th minute.  As each $X_k$ is iid, we may approximate the distribution of the sum using a normal distribution, with $\mu = 60 \cdot 8 = 480$ and variance $\sigma^2 = 60 \cdot 8 = 480$.  You then want the cumulative
$$P(X_1+X_2+\cdots+X_{60} \ge 450) = 1-\Phi\left( \frac{450-480}{\sqrt{480}}\right) \approx 0.915$$
