Answer:
Option D
Step-by-step explanation:
Given that the BLS reported in February 2012 that the labor force participation rate in the United States was 63.7%
Let P = 63.7% = 0.637
Sample size n=120
Sample proportion p will be normal with mean = 0.637 and
std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{{0.637(1-0.637)}{120} } \\=0.0439[/tex]
required probability
= the probability that the proportion of people who are in the labor force is greater than 0.65
=[tex]P(p>0.65)\\=P(Z>\frac{0.65-0.637}{0.0439} \\=P(Z>+0.296)\\= 0.5-0.1179\\=0.3821[/tex]
APproximately option D is right.