Let the random variable \(X\) represent the number of substandard cameras in a sample of 300. \(X\) follows a binomial distribution with parameters \(n = 300\) and \(p = 0.072\).

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

\(\mu = np = 300 \times 0.072 = 21.6\)

\(\sigma^2 = np(1-p) = 300 \times 0.072 \times 0.928 = 20.0448\)

Using a continuity correction, we find \(P(X < 18)\) by calculating \(P(X < 17.5)\).

Standardizing, we have:

\(P(X < 17.5) = P\left( Z < \frac{17.5 - 21.6}{\sqrt{20.0448}} \right)\)

\(= P(Z < -0.9157)\)

Using the standard normal distribution table, \(P(Z < -0.9157) = 1 - 0.8201 = 0.180\).