Let the random variable \(X\) represent the number of people who pass the test. \(X\) follows a binomial distribution with parameters \(n = 75\) and \(p = 0.3\).

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

\(\mu = np = 75 \times 0.3 = 22.5\)

\(\sigma^2 = np(1-p) = 75 \times 0.3 \times 0.7 = 15.75\)

The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{15.75} \approx 3.9686\)

We use a continuity correction for the normal approximation:

\(P(X > 20) \approx P\left(Z > \frac{20.5 - 22.5}{\sqrt{15.75}}\right)\)

\(= P\left(Z > \frac{-2}{3.9686}\right)\)

\(= P(Z > -0.504)\)

Using standard normal distribution tables or a calculator, we find:

\(P(Z > -0.504) = \Phi(0.504) \approx 0.693\)

Thus, the probability that more than 20 people pass is approximately \(0.6925 < p < 0.693\).