Let the random variable \(X\) represent the number of faulty components in a sample of 200. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.15\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 200 \times 0.15 = 30\)

\(\sigma^2 = np(1-p) = 200 \times 0.15 \times 0.85 = 25.5\)

Thus, \(X\) is approximately \(N(30, 25.5)\).

We need to find \(P(X > 40)\). Using a continuity correction, this is equivalent to \(P(X > 40.5)\).

Standardizing, we have:

\(P(X > 40.5) = P\left(Z > \frac{40.5 - 30}{\sqrt{25.5}}\right)\)

\(= P\left(Z > \frac{10.5}{5.04975}\right)\)

\(= P(Z > 2.079)\)

Using standard normal distribution tables or a calculator, \(P(Z > 2.079) = 1 - \Phi(2.079) \approx 1 - 0.9812 = 0.0188\).