1. The problem asks for the probability that in a group of 7 randomly selected U.S. residents, exactly 4 or 5 are in their twenties, given that 19% of U.S. residents are in their twenties. 2. Let $p = 0.19$ be the probability a single person is in their twenties, and $n = 7$ be the number of people selected. 3. We use the binomial probability formula: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $X$ is the number of people in their twenties. 4. Calculate $P(X=4)$: $$P(X=4) = \binom{7}{4} (0.19)^4 (0.81)^3$$ Calculate the binomial coefficient: $$\binom{7}{4} = \frac{7!}{4!3!} = 35$$ 5. Calculate $P(X=5)$: $$P(X=5) = \binom{7}{5} (0.19)^5 (0.81)^2$$ Calculate the binomial coefficient: $$\binom{7}{5} = \frac{7!}{5!2!} = 21$$ 6. Compute each probability: $$P(X=4) = 35 \times (0.19)^4 \times (0.81)^3 \approx 35 \times 0.001303 \times 0.531441 = 0.0242$$ $$P(X=5) = 21 \times (0.19)^5 \times (0.81)^2 \approx 21 \times 0.000247 \times 0.6561 = 0.0034$$ 7. Add the probabilities for 4 and 5: $$P(4 \text{ or } 5) = P(X=4) + P(X=5) = 0.0242 + 0.0034 = 0.0276$$ 8. Rounded to four decimal places, the probability is: $$\boxed{0.0276}$$