Answer
Yes, I would be surprised because it is an unusual event.
Work Step by Step
$μ_X=np=300\times0.267=80.1$
$σ_X=\sqrt {np(1-p)}=\sqrt {300\times0.267(1-0.267)}=7.66$
$P(X\gt100)=P(X\geq101)$
Let's find the z-score for 100.5:
$z=\frac{X-μ}{σ}=\frac{100.5-80.1}{7.66}=2.66$
According to Table V, the area of the standard normal curve to the left of z-score equal to 2.66 is 0.9961
But, we want the area of the standard normal curve to the right (more than 100) of z-score equal to 2.66:
$1-0.9961=0.0039$
$P(X\gt100)=0.0039\lt0.05$. It is an unusual event.
