(i) Since \(P(X > 3.6) = 0.5\), the mean \(\mu\) is 3.6 because the probability of being greater than the mean in a normal distribution is 0.5.

For \(P(X > 2.8) = 0.6554\), we standardize: \(\frac{2.8 - \mu}{\sigma} = -0.4\).

Substitute \(\mu = 3.6\): \(\frac{2.8 - 3.6}{\sigma} = -0.4\).

Solve for \(\sigma\): \(\sigma = \frac{-0.8}{-0.4} = 2\).

(ii) Let \(p = 0.6554\) be the probability that an observation is greater than 2.8. We use the binomial distribution with \(n = 4\) and \(p = 0.6554\).

The probability that at least two observations are greater than 2.8 is:

\(P(X \geq 2) = \binom{4}{2} (0.6554)^2 (0.3446)^2 + \binom{4}{3} (0.6554)^3 (0.3446) + \binom{4}{4} (0.6554)^4\).

Calculate each term:

\(\binom{4}{2} (0.6554)^2 (0.3446)^2 = 0.3061\)

\(\binom{4}{3} (0.6554)^3 (0.3446) = 0.3881\)

\(\binom{4}{4} (0.6554)^4 = 0.1845\)

Sum these probabilities: \(0.3061 + 0.3881 + 0.1845 = 0.879\).