The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 120\)

- Probability of success, \(p = 0.7\)

We approximate the binomial distribution with a normal distribution using the Central Limit Theorem.

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 120 \times 0.7 = 84\)

\(\sigma^2 = np(1-p) = 120 \times 0.7 \times 0.3 = 25.2\)

Using a normal approximation, we find the probability that more than 75 adults own a car. Apply continuity correction:

\(P(X > 75) \approx P\left(Z > \frac{75.5 - 84}{\sqrt{25.2}}\right)\)

Calculate the standard score (z-score):

\(Z = \frac{75.5 - 84}{\sqrt{25.2}} = \frac{-8.5}{5.02} \approx -1.693\)

Using the standard normal distribution table, find:

\(P(Z > -1.693) = 0.955\)

Thus, the probability that more than 75 of them own a car is approximately 0.955.