Answer
$P(accepts)=0.0113$
Since $P(accepts)\lt0.05$, it would be unusual for the engineer to accept the shipment. But, he will refuse 96% of the nondefective resistors.
Work Step by Step
$p=0.04=4$%
If there are 10 defective resistors in the sample.
$p ̂=\frac{x}{n}=\frac{10}{500}=0.02$
$σ_{p ̂}=\sqrt {\frac{p(1-p)}{n}}$
Let $d$ be the number of defective resistors in the sample.
$P(accepts)=P(d\lt10)=P(p ̂\lt0.02)=P(z\lt\frac{0.02-0.04}{\sqrt {\frac{0.04(1-0.04)}{500}}})=P(z\lt-2.28)=0.0113$
