The percentage of cars is calculated as:

\(100 ext{%} - 20 ext{%} - 16 ext{%} = 64 ext{%}\)

Let \(X\) be the number of cars in the sample of 125 vehicles. \(X\) follows a binomial distribution \(B(n, p)\) where \(n = 125\) and \(p = 0.64\).

We approximate this binomial distribution with a normal distribution \(N(\mu, \sigma^2)\) where:

\(\mu = np = 125 \times 0.64 = 80\)

\(\sigma^2 = np(1-p) = 125 \times 0.64 \times 0.36 = 28.8\)

We need to find \(P(X > 73)\). Using continuity correction, this becomes \(P(X > 73.5)\).

Standardizing, we have:

\(P(X > 73.5) = 1 - \Phi\left( \frac{73.5 - 80}{\sqrt{28.8}} \right)\)

Calculating the standard score:

\(\frac{73.5 - 80}{\sqrt{28.8}} = -1.211\)

Thus:

\(P(X > 73.5) = 1 - \Phi(-1.211) = \Phi(1.211)\)

Using standard normal distribution tables, \(\Phi(1.211) \approx 0.887\).