The number of faulty toys follows a binomial distribution with parameters \(n = 200\) and \(p = 0.08\).

First, calculate the mean and variance:

\(\text{mean} = np = 200 \times 0.08 = 16\)

\(\text{variance} = np(1-p) = 200 \times 0.08 \times 0.92 = 14.72\)

Since \(n\) is large and \(p\) is not too close to 0 or 1, we can use a normal approximation.

Apply continuity correction for \(P(X \geq 15)\):

\(P(X \geq 15) = P(X > 14.5)\)

Standardize using the normal distribution:

\(P(X > 14.5) = 1 - \Phi \left( \frac{14.5 - 16}{\sqrt{14.72}} \right)\)

Calculate the standardized value:

\(\frac{14.5 - 16}{\sqrt{14.72}} = -0.391\)

Using the standard normal distribution table, find:

\(\Phi(-0.391) = 0.391\)

Thus,

\(P(X \geq 15) = 1 - 0.391 = 0.652\)