Answer
P(p̂ > 0.5) =0.1977. This implies that about 20 out of 100 samples of n=200 will contain more than 50% Americans who can order in a foreign language.
Work Step by Step
Using data from part a for p̂ > 0.5, we have:
$z = \frac{0.5 - 0.47}{0.035} = 0.86$
P(p̂ > 0.5) = P(z > 0.86) = 0.1977
