$23 \%$ of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five, inclusive. (a) $P(2)=0.294$ (Round to the nearest thousandth as needed) (b) $P(x>2)=0.414$ (Round to the nearest thousandth as needed) (c) $P(2 \leq x \leq 5)=\square$ (Round to the nearest thousandth as needed.)
Solution
Step 1 ：This is a binomial probability problem. We are given the probability of success (p = 0.23), the number of trials (n = 10), and we need to find the probability of having a certain number of successes (k).
Step 2 ：For part (c), we need to find the probability of having between 2 and 5 successes, inclusive. This means we need to find P(2) + P(3) + P(4) + P(5).
Step 3 ：We can use the formula for binomial probability to calculate this: \(P(k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(C(n, k)\) is the combination of n items taken k at a time, which can be calculated as: \(C(n, k) = n! / [k!(n-k)!]\).
Step 4 ：Using the given values and the formula, we calculate the probability to be approximately 0.695.
Step 5 ：Final Answer: \(P(2 \leq x \leq 5)=\boxed{0.695}\)
