(i) The probability of waiting between 5 and 10 minutes is given by the difference between the probability of waiting less than 10 minutes and the probability of waiting less than 5 minutes:

\(P(5 < X < 10) = P(X < 10) - P(X < 5) = 0.88 - 0.16 = 0.72\)

(ii) Let \(X\) be the number of people who wait between 5 and 10 minutes. \(X\) follows a binomial distribution \(X \sim B(180, 0.72)\).

Using a normal approximation, \(X\) can be approximated by a normal distribution \(X \sim N(np, npq)\) where:

\(np = 180 \times 0.72 = 129.6\)

\(npq = 180 \times 0.72 \times 0.28 = 36.288\)

Thus, \(X \sim N(129.6, 36.288)\).

We need to find \(P(X > 115)\). Using continuity correction, this becomes \(P(X > 115.5)\).

Standardizing, we have:

\(P\left( z > \frac{115.5 - 129.6}{\sqrt{36.288}} \right) = P(z > -2.341)\)

Using standard normal distribution tables, \(P(z > -2.341) = 0.990\).