$P(X \ge 450)$ 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.

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
$$P(X_1+X_2+\cdots+X_{60} \ge 450) = 1-\Phi\left( \frac{450-480}{\sqrt{480}}\right) \approx 0.915$$
