1. **Stating the problem:** We have a random variable $X$ denoting the number of bent rods in a sample of 400 rods. Each rod has a 1.2% chance of being bent. 2. **(i) Distribution of $X$:** $X$ follows a binomial distribution because we have a fixed number of independent trials (rods), each with the same probability of 'success' (being bent). So, $$X \sim \text{Binomial}(n=400, p=0.012).$$ 3. **(ii) Suitable approximate distribution:** Since $n=400$ is large and $p=0.012$ is small, the binomial distribution can be approximated by a Poisson distribution with parameter $$\lambda = np = 400 \times 0.012 = 4.8.$$ This is suitable because $n$ is large and $p$ is small making $np$ moderate. So, $$X \approx \text{Poisson}(\lambda=4.8).$$ 4. **(iii) Probability that more than 2 rods are bent:** We want $$P(X > 2) = 1 - P(X \leq 2) = 1 - [\!P(X=0) + P(X=1) + P(X=2)].$$ In the Poisson distribution, $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}.$$ Calculate: - $$P(X=0) = e^{-4.8} \frac{4.8^0}{0!} = e^{-4.8}.$$ - $$P(X=1) = e^{-4.8} \frac{4.8^1}{1!} = 4.8 e^{-4.8}.$$ - $$P(X=2) = e^{-4.8} \frac{4.8^2}{2!} = \frac{4.8^2}{2} e^{-4.8} = 11.52 e^{-4.8}.$$ Sum: $$P(X \leq 2) = e^{-4.8} (1 + 4.8 + 11.52) = e^{-4.8} \times 17.32.$$ Using $e^{-4.8} \approx 0.00823$, $$P(X \leq 2) \approx 0.00823 \times 17.32 = 0.1425.$$ Therefore, $$P(X > 2) = 1 - 0.1425 = 0.8575.$$ **Final answer:** The probability that more than 2 rods are bent is approximately **0.8575** or **85.75%**.