Answer
No, it would not be unusual.
Work Step by Step
$P(x)=\frac{(λt)^x}{x!}e^{-λt}$
λ = 4.5 -> (4.5 calls per shift)
t = 1 -> (in a shift)
$P(fewer~than~2)=P(x\lt2)=P(0)+P(1)=\frac{(4.5\times1)^0}{0!}e^{-4.5\times1}+\frac{(4.5\times1)^1}{1!}e^{-4.5\times1}=\frac{1}{1}e^{-4.5}+\frac{4.5}{1}e^{-4.5}=0.0611$
$P(fewer~than~2)\gt0.05$
No, it would not be unusual.
