Let the random variable \(X\) represent the number of students who pass the examination. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.8\).

To approximate this binomial distribution, we use a normal distribution because both \(np\) and \(n(1-p)\) are greater than 5.

The mean of the binomial distribution is \(np = 200 \times 0.8 = 160\).

The variance is \(np(1-p) = 200 \times 0.8 \times 0.2 = 32\).

Thus, the standard deviation is \(\sqrt{32}\).

We approximate \(X\) by a normal distribution \(N(160, 32)\).

We need to find \(P(X > 166)\). Using continuity correction, this is approximately \(P(X > 166.5)\).

Standardizing, we have:

\(z = \frac{166.5 - 160}{\sqrt{32}} \approx 1.149\)

Thus, \(P(X > 166.5) = P(Z > 1.149) = 1 - P(Z \leq 1.149)\).

Using standard normal distribution tables, \(P(Z \leq 1.149) \approx 0.8747\).

Therefore, \(P(Z > 1.149) = 1 - 0.8747 = 0.125\).

For part (iii), the justification for using the normal approximation is that both \(np = 160\) and \(nq = 40\) are greater than 5, which satisfies the conditions for the normal approximation to the binomial distribution.