Answer
The sampling distribution is approximately normal with mean = 0.28 and $σ_{p̂} = 0.02$.
Work Step by Step
np(1-p) = 500 x 0.28 x 0.72 = `100.8 $\geq 10$
$μ_{p̂} = p = 0.28$
$σ_{p̂} = \sqrt \frac{p(1-p)}{n}$
$ = \sqrt \frac{0.28 * 0.72}{500}$
= 0.02
