1. **State the problem:** We have a probability $p=0.008$ that a new car has faulty brakes. A sample of $n=520$ cars is taken, and $X$ denotes the number of cars with faulty brakes. 2. **Distribution of $X$:** Since $X$ counts the number of successes (faulty brakes) in $n$ independent trials, each with probability $p$, $X$ follows a Binomial distribution: $$X \sim \text{Binomial}(n=520, p=0.008).$$ 3. **Suitable approximating distribution:** Because $n$ is large and $p$ is small, the Binomial distribution can be approximated using a Poisson distribution with parameter: $$\lambda = np = 520 \times 0.008 = 4.16.$$ The Poisson approximation is suitable here because $n$ is large and $p$ is small (rule of thumb: $np < 10$ and $p$ close to 0). 4. **Calculate $P(X > 3)$ using the Poisson approximation:** We want: $$P(X > 3) = 1 - P(X \leq 3) = 1 - \sum_{k=0}^{3} P(X = k).$$ For Poisson, $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ Calculate each term: $$P(X=0) = e^{-4.16} \frac{4.16^0}{0!} = e^{-4.16},$$ $$P(X=1) = e^{-4.16} \frac{4.16^1}{1!} = 4.16 e^{-4.16},$$ $$P(X=2) = e^{-4.16} \frac{4.16^2}{2!} = \frac{4.16^2}{2} e^{-4.16},$$ $$P(X=3) = e^{-4.16} \frac{4.16^3}{3!} = \frac{4.16^3}{6} e^{-4.16}.$$ Sum these: $$\sum_{k=0}^3 P(X=k) = e^{-4.16} (1 + 4.16 + \frac{4.16^2}{2} + \frac{4.16^3}{6}).$$ Calculate numeric values: $4.16^2 = 17.3056$ $4.16^3 = 71.9569$ Sum inside parentheses: $1 + 4.16 + \frac{17.3056}{2} + \frac{71.9569}{6} = 1 + 4.16 + 8.6528 + 11.9928 = 25.8056$ Calculate $e^{-4.16} \approx 0.0156$ Therefore: $$\sum_{k=0}^3 P(X=k) \approx 0.0156 \times 25.8056 = 0.4029.$$ Hence, $$P(X > 3) = 1 - 0.4029 = 0.5971.$$ **Final answer: $P(X > 3) \approx 0.597$ (using the Poisson approximation).**