(i) To find \(\mu\), we use the standard normal distribution. Given \(\text{P}(X > 20) = 0.04\), we find the corresponding \(z\)-value: \(z = \pm 1.751\).

Using the standardization formula: \(\frac{20 - \mu}{\mu/4} = 1.751\).

Solving for \(\mu\), we get \(\mu = 13.9\).

(ii) To find \(\text{P}(10 < X < 20)\), we first find \(\text{P}(X < 10)\).

Standardizing: \(\text{P}(z < \frac{10 - 13.91}{13.91/4}) = \text{P}(z < -1.124)\).

\(\text{P}(z < -1.124) = 1 - 0.8694 = 0.131\).

Thus, \(\text{P}(10 < X < 20) = 0.96 - 0.131 = 0.829 \text{ or } 0.830\).

(iii) For 250 observations, \(\mu = 250 \times 0.96 = 240\) and \(\sigma^2 = 250 \times 0.96 \times 0.04 = 9.6\).

We need \(\text{P}(\geq 235) = 1 - \Phi\left(\frac{234.5 - 240}{\sqrt{9.6}}\right)\).

\(\Phi(1.775) = 0.962\).