Answer
It is not an unusual event.
Work Step by Step
$p̂=\frac{x}{n}=\frac{10}{1500}=0.006667$
The z-score for 0.006667:
$z=\frac{p̂−μ_{p̂}}{\sqrt {\frac{p(1-p)}{n}}}=\frac{0.006667−0.01}{\sqrt {\frac{0.01(1-0.01)}{1500}}}=-1.30$
$P(p̂\lt0.006667)=P(z\lt-1.30)=0.0968\gt0.05$. It is not an unusual event.
