Answer
$P(p̂ \leq 0.75) = 0.0344$. This implies that about 3 out of 100 samples of n=100 will contain 75% or less Americans who are satisfied with the way things are going in their life. This result is unusual as p < 0.05
Work Step by Step
Using data from part a for $p̂ \leq 0.75$, we have:
$z = \frac{0.75 - 0.82}{0.038} = -1.82$
$P(p̂ \leq 0.75) = P(z < -1.82) = 0.0344$
