There are 100 randomly selected cars. There are five types of colors in the sample of cars such as white, black, gray, silver, red.

Requirements for normal approximation of binomial distribution:

The sample is a simple random sample with size n from a population with the proportion of success p or probability of success p.\(np \ge 5,nq \ge 5.\)

Let X be the random variable for the number of black cars in the sample of 100.

The proportion of cars with black color is 0.18(p).

It is probability of success is obtained using percentage of each colour car.

The sample size and probability of success are;

\(\begin{aligned}{l}n = 100\\p = 0.18\end{aligned}\)

The requirement for the normal approximation to binomial distribution, that is, \(np \ge 5,nq \ge 5.\)

Thus,

\(\begin{aligned}{c}np = 100 \times 0.18\\ = 18\\ = 18 \ge 5\end{aligned}\)

\(\begin{aligned}{c}nq = n \times \left( {1 - p} \right)\\ = 100 \times \left( {1 - 0.18} \right)\\ = 82\\ \ge 5\end{aligned}\)

Thus, the approximation to normal distribution is verified.

To approximate binomial distribution into normal, it is important to find the mean and standard deviation.

Hence, the formula for mean and standard deviation is given below,

\(\begin{aligned}{l}\mu = n \times p\\\sigma = \sqrt {n \times p \times q} \end{aligned}\)

Now, substitute the all values in the formula,

\(\begin{aligned}{c}\mu = n \times p\\ = 100 \times 0.18\\ = 18\end{aligned}\)

\(\begin{aligned}{c}\sigma = \sqrt {n \times p \times q} \\ = \sqrt {100 \times 0.18 \times 0.82} \\ = \sqrt {14.76} \\ = 3.8419\end{aligned}\)

Now, the mean of the normal distribution is \(\mu = 18\) and the standard deviation of normal distribution is \(\sigma = 3.8419\).

The probability that the number of cars is at least 25 is expressed as,

\(P\left( {X \ge 25} \right) = P\left( {X > 24.5} \right)\)

The formula for z-score is,

\(z = \frac{{x - \mu }}{\sigma }\)

The z-score associated with 24.5,

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{24.5 - 18}}{{3.8419}}\\ = 1.6919\end{aligned}\)

The z-score is 1.6919.

From the standard normal table, the cumulative probability for the value 1.69 us 0.9545.

Thus, probability that the number of cars is at least 25 is,

\(\begin{aligned}{c}P\left( {X > 24.5} \right) = P\left( {Z > 1.6919} \right)\\ = 1 - P\left( {Z < 1.69} \right)\\ = 1 - 0.9545\\ = 0.0455\end{aligned}\)

Thus, the probability that the number of cars is at least 25 is 0.0455.

Significantly high events are those which have probability 0.05 or less for the occurrence of events greater than or equal to the given event. Here, the probability that at least 25 cars are black is 0.0455, which is lesser than 0.05.

Therefore, 25 is significantly a high number of cars.