Probability Analysis of Opposition to Government-Controlled Health Plans

Question

In a May, 2009 Rasmussen Reports survey, 58% of U.S. adults surveyed opposed "any kind of government-controlled health plan if it meant they had to change their own insurance company". We will assume that 58% is the true proportion of all adults who feel this way. If we randomly select 20 adults and ask the same question, what is the probability that we would find fewer than 6 or more than 17 in opposition? (Round answer to three decimal places)

Answer 1

You would use the Binomial Distribution with p = .58 and n = 20

You need to add up all these probabilities:

P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=18) + P(X=19) + P(X=20)

Where P(X=i) = _nC_i pⁱ (1-p)^n-i

Give it a try!