Solution 8.7.2.1.
Using the given information, apply the Poisson distribution. For one hour \(\mu = 10\) so that for three hours the expected number of customers would be triple with 30 expected customers.
With \(\mu_1 = 10\text{,}\)
\begin{align*}
P(X \lt 10) & = F(9)\\
& = \frac{10^0 e^{-10}}{0!} + \frac{10^1 e^{-10}}{1!} + \frac{10^2 e^{-10}}{2!} + ... + \frac{10^8 e^{-10}}{8!} + \frac{10^9 e^{-10}}{9!}\\
& = e^{-10} \cdot ( 1 + 10 + \frac{100}{2} + ... + \frac{10^8}{8!} + \frac{10^9}{9!} )
\end{align*}