Answer
No, it is not unusual because $P(more~than~4)\gt0.05$
Work Step by Step
$P(x)=\frac{(λt)^x}{x!}e^{-λt}$
λ = 2 (2 times per month)
t = 1 (any given month)
$P(no~more~than~4)=P(X\leq4)=P(0)+P(1)+P(2)+P(3)+P(4)=\frac{(2\times1)^0}{0!}e^{-2\times1}+\frac{(2\times1)^1}{1!}e^{-2\times1}+\frac{(2\times1)^2}{2!}e^{-2\times1}+\frac{(2\times1)^3}{3!}e^{-2\times1}+\frac{(2\times1)^4}{4!}e^{-2\times1}=\frac{1}{1}e^{-2}+\frac{2}{1}e^{-2}+\frac{4}{2}e^{-2}+\frac{8}{6}e^{-2}+\frac{16}{24}e^{-2}=0.9473$
The probability that $x\gt4$ is the complement of the probability that $x\leq4$
Using the Complement Rule (see page 275):
$P(more~than~4)=P(X\gt4)=1-P(X\leq4)=1-0.9473=0.0527\gt0.05$
It is not unusual.
