1. The problem is to understand the values of $n=5$ and $p=0.55$ in a binomial distribution context. 2. Here, $n$ represents the number of trials, which is 5. 3. The parameter $p$ is the probability of success on each trial, which is 0.55. 4. This setup is typical for a binomial random variable $X \sim \text{Binomial}(n=5, p=0.55)$. 5. You can calculate probabilities such as $P(X=k) = \binom{5}{k} (0.55)^k (0.45)^{5-k}$ for $k=0,1,2,3,4,5$. 6. For example, the probability of exactly 3 successes is: $$ P(X=3) = \binom{5}{3} (0.55)^3 (0.45)^2 = 10 \times 0.166375 \times 0.2025 = 0.336. $$ 7. This means there is approximately a 33.6% chance of getting exactly 3 successes in 5 trials with success probability 0.55. Final answer: $n=5$, $p=0.55$ define a binomial distribution with these parameters.