(i) Let \(X\) be the number of cloudy days in November. \(X\) follows a binomial distribution with parameters \(n = 30\) and \(p = 0.8\).

We approximate \(X\) using a normal distribution with mean \(\mu = np = 30 \times 0.8 = 24\) and variance \(\sigma^2 = npq = 30 \times 0.8 \times 0.2 = 4.8\).

Standard deviation \(\sigma = \sqrt{4.8}\).

To find the probability of fewer than 25 cloudy days, we use a continuity correction: \(P(X < 25) \approx P(Y < 24.5)\) where \(Y\) is normally distributed.

Calculate the z-score: \(z = \frac{24.5 - 24}{\sqrt{4.8}} = 0.228\).

Using standard normal distribution tables, \(P(Z < 0.228) = 0.590\).

(ii) The normal approximation is justified because both \(np = 24\) and \(nq = 6\) are greater than 5.