Answer
$P(p̂ \leq 0.4) = 0.0239$. This implies that about 2 out of 100 samples of n=200 will contain 40% or less Americans who can order in a foreign language. This result is unusual as p < 0.05.
Work Step by Step
80/200 = 0.4. Using data from part a for $p̂ \leq 0.4$, we have:
$z = \frac{0.4 - 0.47}{0.035} = -2$
$P(p̂ \leq 0.4) = P(z\leq -2) = 0.0239$
