(i) To find the probability that a building is classified as tall, we need to calculate the probability that the height is greater than 70 metres. Using the standard normal distribution, we standardize the height:

\(z = \frac{70 - 50}{16} = 1.25\)

The probability \(P(z > 1.25)\) is found using the standard normal distribution table:

\(P(z > 1.25) = 1 - P(z \leq 1.25) = 1 - 0.8944 = 0.106\)

(ii) Let \(P(\text{short})\) be the probability that a building is classified as short. Since there are twice as many medium buildings as short ones, \(P(\text{medium}) = 2 \times P(\text{short})\). The total probability for medium and short buildings is:

\(P(\text{medium}) + P(\text{short}) = 1 - P(\text{tall}) = 1 - 0.106 = 0.894\)

Substituting \(P(\text{medium}) = 2 \times P(\text{short})\), we have:

\(2P(\text{short}) + P(\text{short}) = 0.894\)

\(3P(\text{short}) = 0.894\)

\(P(\text{short}) = \frac{0.894}{3} = 0.298\)

To find the height below which buildings are classified as short, we find the z-score corresponding to \(P(z < z_0) = 0.298\). From the standard normal distribution table, \(z_0 \approx -0.53\).

Standardizing, we have:

\(-0.53 = \frac{x - 50}{16}\)

Solving for \(x\):

\(x = 50 + (-0.53 \times 16) = 41.5\) metres