In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 Americans 18 or older more than 30 are single?

Given that p = 0.22, q = 1 - p = 0.78

n = 200

mean = miu = np = 200 x 0.22 = 44

standard deviation = √ (npq) = √ (200 x 0.22 x 0.78) = 5.8583

P (X>30) = P (X - mean/SD > 30 - 44/5.8583)

= P (z > - 2.3897)

= 1 - P (z<-2.3897) = 1 - 0.0084

= 0.9915

Hence P (X>30) = 0.9915

Hence the probability that in a random sample of 200 Americans 18 or older more than 30 are single is 0.9915